Bash Database Backup to AWS S3











up vote
4
down vote

favorite












I've written a bash script that backs up databases and uploads the backups to an S3 bucket. I have the bucket configured to delete backups that are older than 90 days.



Is there anything that can be improved in this script?



#!/usr/bin/env bash

# Backup MySQL databases to S3

# Follow the installation instructions listed on the user guid doc:
# http://docs.aws.amazon.com/cli/latest/userguide/installing.html

# Note backups will be deleted after 90 days. This is setup in the management tab:
# https://s3.console.aws.amazon.com/s3/buckets/backups.databases/?tab=management

# DB details
DB_USER=""
DB_PASSWORD=""
DB_HOST=""
DB_IGNORED=(Database mysql information_schema performance_schema cond_instances)

# S3 details
S3_BUCKET=backups.databases
S3_FOLDER=backups/databases

# Temp storage location
TMP_DIR=/tmp

# Check if dependencies exist
dependencies=(aws mysql mysqldump tar rm echo)
for dependency in "${dependencies[@]}"
do
if [[ $(command -v "$dependency") = "" ]]
then
# Tell the user that the dependency is not installed, then exit the script
$(which echo) "Dependency $dependency isn't installed. Install this dependency to continue."
exit 0
fi
done

DB_DATABASES=($($(which echo) "show databases;" | $(which mysql) -h ${DB_HOST} -p${DB_PASSWORD} -u ${DB_USER} 2> /dev/null))

# Filename for the backups
time=$($(which date) +%b-%d-%y-%H%M)
filename="backup-$time.tar.gz"

for database in "${DB_DATABASES[@]}"
do
# Skip ignored databases
if [[ "${DB_IGNORED[@]}" =~ "$database" ]]
then
continue
fi

# Dump database SQL schema to an SQL file stored in the temp directory
$(which echo) "Backing up $database..."
$(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null
$(which tar) -cpzf "$TMP_DIR/$database-$filename" "$TMP_DIR/$database.sql" 2> /dev/null

# Upload the exported SQL file to S3
$(which echo) "Uploading to S3..."
$(which aws) s3 cp "$TMP_DIR/$database-$filename" "s3://$S3_BUCKET/$S3_FOLDER/$database-$filename"
$(which echo) "Database $database uploaded to S3"

# Remove the temporary files
$(which echo) "Cleaning up..."
$(which rm) -f "$TMP_DIR/$database-$filename"
$(which rm) -f "$TMP_DIR/$database.sql"
done


I had to add 2> /dev/null to the end of the mysql and mysqldump commands due to a warning message showing.




Warning: Using a password on the command line interface can be
insecure.




Is there a better way of handling this?



Any feedback would be greatly appreciated.










share|improve this question




















  • 2




    Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Ludisposed
    Nov 16 '17 at 13:03















up vote
4
down vote

favorite












I've written a bash script that backs up databases and uploads the backups to an S3 bucket. I have the bucket configured to delete backups that are older than 90 days.



Is there anything that can be improved in this script?



#!/usr/bin/env bash

# Backup MySQL databases to S3

# Follow the installation instructions listed on the user guid doc:
# http://docs.aws.amazon.com/cli/latest/userguide/installing.html

# Note backups will be deleted after 90 days. This is setup in the management tab:
# https://s3.console.aws.amazon.com/s3/buckets/backups.databases/?tab=management

# DB details
DB_USER=""
DB_PASSWORD=""
DB_HOST=""
DB_IGNORED=(Database mysql information_schema performance_schema cond_instances)

# S3 details
S3_BUCKET=backups.databases
S3_FOLDER=backups/databases

# Temp storage location
TMP_DIR=/tmp

# Check if dependencies exist
dependencies=(aws mysql mysqldump tar rm echo)
for dependency in "${dependencies[@]}"
do
if [[ $(command -v "$dependency") = "" ]]
then
# Tell the user that the dependency is not installed, then exit the script
$(which echo) "Dependency $dependency isn't installed. Install this dependency to continue."
exit 0
fi
done

DB_DATABASES=($($(which echo) "show databases;" | $(which mysql) -h ${DB_HOST} -p${DB_PASSWORD} -u ${DB_USER} 2> /dev/null))

# Filename for the backups
time=$($(which date) +%b-%d-%y-%H%M)
filename="backup-$time.tar.gz"

for database in "${DB_DATABASES[@]}"
do
# Skip ignored databases
if [[ "${DB_IGNORED[@]}" =~ "$database" ]]
then
continue
fi

# Dump database SQL schema to an SQL file stored in the temp directory
$(which echo) "Backing up $database..."
$(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null
$(which tar) -cpzf "$TMP_DIR/$database-$filename" "$TMP_DIR/$database.sql" 2> /dev/null

# Upload the exported SQL file to S3
$(which echo) "Uploading to S3..."
$(which aws) s3 cp "$TMP_DIR/$database-$filename" "s3://$S3_BUCKET/$S3_FOLDER/$database-$filename"
$(which echo) "Database $database uploaded to S3"

# Remove the temporary files
$(which echo) "Cleaning up..."
$(which rm) -f "$TMP_DIR/$database-$filename"
$(which rm) -f "$TMP_DIR/$database.sql"
done


I had to add 2> /dev/null to the end of the mysql and mysqldump commands due to a warning message showing.




Warning: Using a password on the command line interface can be
insecure.




Is there a better way of handling this?



Any feedback would be greatly appreciated.










share|improve this question




















  • 2




    Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Ludisposed
    Nov 16 '17 at 13:03













up vote
4
down vote

favorite









up vote
4
down vote

favorite











I've written a bash script that backs up databases and uploads the backups to an S3 bucket. I have the bucket configured to delete backups that are older than 90 days.



Is there anything that can be improved in this script?



#!/usr/bin/env bash

# Backup MySQL databases to S3

# Follow the installation instructions listed on the user guid doc:
# http://docs.aws.amazon.com/cli/latest/userguide/installing.html

# Note backups will be deleted after 90 days. This is setup in the management tab:
# https://s3.console.aws.amazon.com/s3/buckets/backups.databases/?tab=management

# DB details
DB_USER=""
DB_PASSWORD=""
DB_HOST=""
DB_IGNORED=(Database mysql information_schema performance_schema cond_instances)

# S3 details
S3_BUCKET=backups.databases
S3_FOLDER=backups/databases

# Temp storage location
TMP_DIR=/tmp

# Check if dependencies exist
dependencies=(aws mysql mysqldump tar rm echo)
for dependency in "${dependencies[@]}"
do
if [[ $(command -v "$dependency") = "" ]]
then
# Tell the user that the dependency is not installed, then exit the script
$(which echo) "Dependency $dependency isn't installed. Install this dependency to continue."
exit 0
fi
done

DB_DATABASES=($($(which echo) "show databases;" | $(which mysql) -h ${DB_HOST} -p${DB_PASSWORD} -u ${DB_USER} 2> /dev/null))

# Filename for the backups
time=$($(which date) +%b-%d-%y-%H%M)
filename="backup-$time.tar.gz"

for database in "${DB_DATABASES[@]}"
do
# Skip ignored databases
if [[ "${DB_IGNORED[@]}" =~ "$database" ]]
then
continue
fi

# Dump database SQL schema to an SQL file stored in the temp directory
$(which echo) "Backing up $database..."
$(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null
$(which tar) -cpzf "$TMP_DIR/$database-$filename" "$TMP_DIR/$database.sql" 2> /dev/null

# Upload the exported SQL file to S3
$(which echo) "Uploading to S3..."
$(which aws) s3 cp "$TMP_DIR/$database-$filename" "s3://$S3_BUCKET/$S3_FOLDER/$database-$filename"
$(which echo) "Database $database uploaded to S3"

# Remove the temporary files
$(which echo) "Cleaning up..."
$(which rm) -f "$TMP_DIR/$database-$filename"
$(which rm) -f "$TMP_DIR/$database.sql"
done


I had to add 2> /dev/null to the end of the mysql and mysqldump commands due to a warning message showing.




Warning: Using a password on the command line interface can be
insecure.




Is there a better way of handling this?



Any feedback would be greatly appreciated.










share|improve this question















I've written a bash script that backs up databases and uploads the backups to an S3 bucket. I have the bucket configured to delete backups that are older than 90 days.



Is there anything that can be improved in this script?



#!/usr/bin/env bash

# Backup MySQL databases to S3

# Follow the installation instructions listed on the user guid doc:
# http://docs.aws.amazon.com/cli/latest/userguide/installing.html

# Note backups will be deleted after 90 days. This is setup in the management tab:
# https://s3.console.aws.amazon.com/s3/buckets/backups.databases/?tab=management

# DB details
DB_USER=""
DB_PASSWORD=""
DB_HOST=""
DB_IGNORED=(Database mysql information_schema performance_schema cond_instances)

# S3 details
S3_BUCKET=backups.databases
S3_FOLDER=backups/databases

# Temp storage location
TMP_DIR=/tmp

# Check if dependencies exist
dependencies=(aws mysql mysqldump tar rm echo)
for dependency in "${dependencies[@]}"
do
if [[ $(command -v "$dependency") = "" ]]
then
# Tell the user that the dependency is not installed, then exit the script
$(which echo) "Dependency $dependency isn't installed. Install this dependency to continue."
exit 0
fi
done

DB_DATABASES=($($(which echo) "show databases;" | $(which mysql) -h ${DB_HOST} -p${DB_PASSWORD} -u ${DB_USER} 2> /dev/null))

# Filename for the backups
time=$($(which date) +%b-%d-%y-%H%M)
filename="backup-$time.tar.gz"

for database in "${DB_DATABASES[@]}"
do
# Skip ignored databases
if [[ "${DB_IGNORED[@]}" =~ "$database" ]]
then
continue
fi

# Dump database SQL schema to an SQL file stored in the temp directory
$(which echo) "Backing up $database..."
$(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null
$(which tar) -cpzf "$TMP_DIR/$database-$filename" "$TMP_DIR/$database.sql" 2> /dev/null

# Upload the exported SQL file to S3
$(which echo) "Uploading to S3..."
$(which aws) s3 cp "$TMP_DIR/$database-$filename" "s3://$S3_BUCKET/$S3_FOLDER/$database-$filename"
$(which echo) "Database $database uploaded to S3"

# Remove the temporary files
$(which echo) "Cleaning up..."
$(which rm) -f "$TMP_DIR/$database-$filename"
$(which rm) -f "$TMP_DIR/$database.sql"
done


I had to add 2> /dev/null to the end of the mysql and mysqldump commands due to a warning message showing.




Warning: Using a password on the command line interface can be
insecure.




Is there a better way of handling this?



Any feedback would be greatly appreciated.







bash amazon-web-services






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 16 '17 at 13:03









Ludisposed

6,79221959




6,79221959










asked Nov 16 '17 at 10:35









Enijar

957




957








  • 2




    Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Ludisposed
    Nov 16 '17 at 13:03














  • 2




    Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
    – Ludisposed
    Nov 16 '17 at 13:03








2




2




Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
– Ludisposed
Nov 16 '17 at 13:03




Welcome to Code Review! Please do not update the code in your question to incorporate feedback from answers, doing so goes against the Question + Answer style of Code Review. This is not a forum where you should keep the most updated version in your question. Please see what you may and may not do after receiving answers.
– Ludisposed
Nov 16 '17 at 13:03










2 Answers
2






active

oldest

votes

















up vote
3
down vote



accepted










$(which foo) ...



First of all, I completely agree with the other answer, this is pointless, inefficient and ugly.



Clearing variables



Instead of var="", you can write simply: var=



Exit code



When a dependency is missing, the script does exit 0.
Exit code 0 means success.
It would make more sense to exit with non-zero, to indicate failure,
for example exit 1.



Usability



Instead of exiting immediately when any dependency is missing,
it would be more user-friendly to collect all the missing dependencies,
and report them all at once.
A user might get irritated to run the script repeatedly when multiple dependencies are missing.



Here-strings



Instead of echo ... | mysql ... it's better to write mysql ... <<< ....



Quoting command line arguments



This is not safe:




$(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null



Variables and the result of $(...) should be double-quoted to protect from parameter expansion and globbing:



"$(which mysqldump)" -h "${DB_HOST}" -u "${DB_USER}" -p"${DB_PASSWORD}" "${database}" > "$TMP_DIR/$database.sql" 2> /dev/null





share|improve this answer




























    up vote
    4
    down vote













    is there a reason for $(which foo) ?



    If foo is in your path, it will be found, if not which will give you an error.



    Expressions like command will have a return code, so do not write



    if [[ $(command -v "$dependency") = "" ]]


    just use



    if command -v "$dependency" > /dev/null
    then ...





    share|improve this answer























    • Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
      – Enijar
      Nov 16 '17 at 12:57








    • 2




      @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
      – Gao
      Nov 16 '17 at 23:52











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    $(which foo) ...



    First of all, I completely agree with the other answer, this is pointless, inefficient and ugly.



    Clearing variables



    Instead of var="", you can write simply: var=



    Exit code



    When a dependency is missing, the script does exit 0.
    Exit code 0 means success.
    It would make more sense to exit with non-zero, to indicate failure,
    for example exit 1.



    Usability



    Instead of exiting immediately when any dependency is missing,
    it would be more user-friendly to collect all the missing dependencies,
    and report them all at once.
    A user might get irritated to run the script repeatedly when multiple dependencies are missing.



    Here-strings



    Instead of echo ... | mysql ... it's better to write mysql ... <<< ....



    Quoting command line arguments



    This is not safe:




    $(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null



    Variables and the result of $(...) should be double-quoted to protect from parameter expansion and globbing:



    "$(which mysqldump)" -h "${DB_HOST}" -u "${DB_USER}" -p"${DB_PASSWORD}" "${database}" > "$TMP_DIR/$database.sql" 2> /dev/null





    share|improve this answer

























      up vote
      3
      down vote



      accepted










      $(which foo) ...



      First of all, I completely agree with the other answer, this is pointless, inefficient and ugly.



      Clearing variables



      Instead of var="", you can write simply: var=



      Exit code



      When a dependency is missing, the script does exit 0.
      Exit code 0 means success.
      It would make more sense to exit with non-zero, to indicate failure,
      for example exit 1.



      Usability



      Instead of exiting immediately when any dependency is missing,
      it would be more user-friendly to collect all the missing dependencies,
      and report them all at once.
      A user might get irritated to run the script repeatedly when multiple dependencies are missing.



      Here-strings



      Instead of echo ... | mysql ... it's better to write mysql ... <<< ....



      Quoting command line arguments



      This is not safe:




      $(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null



      Variables and the result of $(...) should be double-quoted to protect from parameter expansion and globbing:



      "$(which mysqldump)" -h "${DB_HOST}" -u "${DB_USER}" -p"${DB_PASSWORD}" "${database}" > "$TMP_DIR/$database.sql" 2> /dev/null





      share|improve this answer























        up vote
        3
        down vote



        accepted







        up vote
        3
        down vote



        accepted






        $(which foo) ...



        First of all, I completely agree with the other answer, this is pointless, inefficient and ugly.



        Clearing variables



        Instead of var="", you can write simply: var=



        Exit code



        When a dependency is missing, the script does exit 0.
        Exit code 0 means success.
        It would make more sense to exit with non-zero, to indicate failure,
        for example exit 1.



        Usability



        Instead of exiting immediately when any dependency is missing,
        it would be more user-friendly to collect all the missing dependencies,
        and report them all at once.
        A user might get irritated to run the script repeatedly when multiple dependencies are missing.



        Here-strings



        Instead of echo ... | mysql ... it's better to write mysql ... <<< ....



        Quoting command line arguments



        This is not safe:




        $(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null



        Variables and the result of $(...) should be double-quoted to protect from parameter expansion and globbing:



        "$(which mysqldump)" -h "${DB_HOST}" -u "${DB_USER}" -p"${DB_PASSWORD}" "${database}" > "$TMP_DIR/$database.sql" 2> /dev/null





        share|improve this answer












        $(which foo) ...



        First of all, I completely agree with the other answer, this is pointless, inefficient and ugly.



        Clearing variables



        Instead of var="", you can write simply: var=



        Exit code



        When a dependency is missing, the script does exit 0.
        Exit code 0 means success.
        It would make more sense to exit with non-zero, to indicate failure,
        for example exit 1.



        Usability



        Instead of exiting immediately when any dependency is missing,
        it would be more user-friendly to collect all the missing dependencies,
        and report them all at once.
        A user might get irritated to run the script repeatedly when multiple dependencies are missing.



        Here-strings



        Instead of echo ... | mysql ... it's better to write mysql ... <<< ....



        Quoting command line arguments



        This is not safe:




        $(which mysqldump) -h ${DB_HOST} -u ${DB_USER} -p${DB_PASSWORD} ${database} > "$TMP_DIR/$database.sql" 2> /dev/null



        Variables and the result of $(...) should be double-quoted to protect from parameter expansion and globbing:



        "$(which mysqldump)" -h "${DB_HOST}" -u "${DB_USER}" -p"${DB_PASSWORD}" "${database}" > "$TMP_DIR/$database.sql" 2> /dev/null






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 17 '17 at 20:54









        janos

        96.6k12124350




        96.6k12124350
























            up vote
            4
            down vote













            is there a reason for $(which foo) ?



            If foo is in your path, it will be found, if not which will give you an error.



            Expressions like command will have a return code, so do not write



            if [[ $(command -v "$dependency") = "" ]]


            just use



            if command -v "$dependency" > /dev/null
            then ...





            share|improve this answer























            • Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
              – Enijar
              Nov 16 '17 at 12:57








            • 2




              @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
              – Gao
              Nov 16 '17 at 23:52















            up vote
            4
            down vote













            is there a reason for $(which foo) ?



            If foo is in your path, it will be found, if not which will give you an error.



            Expressions like command will have a return code, so do not write



            if [[ $(command -v "$dependency") = "" ]]


            just use



            if command -v "$dependency" > /dev/null
            then ...





            share|improve this answer























            • Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
              – Enijar
              Nov 16 '17 at 12:57








            • 2




              @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
              – Gao
              Nov 16 '17 at 23:52













            up vote
            4
            down vote










            up vote
            4
            down vote









            is there a reason for $(which foo) ?



            If foo is in your path, it will be found, if not which will give you an error.



            Expressions like command will have a return code, so do not write



            if [[ $(command -v "$dependency") = "" ]]


            just use



            if command -v "$dependency" > /dev/null
            then ...





            share|improve this answer














            is there a reason for $(which foo) ?



            If foo is in your path, it will be found, if not which will give you an error.



            Expressions like command will have a return code, so do not write



            if [[ $(command -v "$dependency") = "" ]]


            just use



            if command -v "$dependency" > /dev/null
            then ...






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday









            Sᴀᴍ Onᴇᴌᴀ

            7,98561750




            7,98561750










            answered Nov 16 '17 at 12:19









            Archemar

            1413




            1413












            • Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
              – Enijar
              Nov 16 '17 at 12:57








            • 2




              @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
              – Gao
              Nov 16 '17 at 23:52


















            • Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
              – Enijar
              Nov 16 '17 at 12:57








            • 2




              @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
              – Gao
              Nov 16 '17 at 23:52
















            Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
            – Enijar
            Nov 16 '17 at 12:57






            Thank for the feedback. The reason I use $(which command) is because I ran into an issue where the script said rm: command not found when running it from a cronjob. I've updated the code to your suggestion.
            – Enijar
            Nov 16 '17 at 12:57






            2




            2




            @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
            – Gao
            Nov 16 '17 at 23:52




            @Enijar Then that probably indicates something is seriously wrong with your setup. Try adding . ~/.bash_profile or . ~/.bashrc at the top of your script and see if it works. Also, you're not supposed to edit your question to reflect changes suggested by the answers. It'll make your post super confusing.
            – Gao
            Nov 16 '17 at 23:52


















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