I am trying to take the output from a MySQL query in bash and use it in a bash variable, but it keeps coming up blank when used in the script, but works perfectly from the terminal. What's wrong here?
I've tried changing the way the statement is written and changing the name of the variable just in case it was somehow reserved. I've also done a significant amount of searching but it turns out if you but 'bash', 'blank', and 'variable' in the search it usually comes up with some version of how to test for blank variables which I already know how to do.
tempo=$(mysql -u "$dbuser" -p"$dbpass" -D "$database" -t -s -r -N -B -e "select user from example where user='$temp' > 0;")
printf "the output should be: $tempo" # This is a test statement
The end result should be that the $tempo variable should either contain a user name from the database or be blank if there isn't one.
I think there is some error with your sql statement at user = '$temp' > 0.
But to get the result from MySql you have to redirect the standard error (stderr) to the standard output (stdout), you should use 2>&1.
Most probably you will run into MySql error but try running this on terminal.
tempo=$((mysql -u "$dbuser" -p"$dbpass" -D "$database" -t -s -r -N -B -e "select user from example where user='$temp' > 0;") 2>&1)
The solution was to echo the result of the sql query like this:
tempo=$(echo $(mysql -u "$dbuser" -p"$dbpass" -D "$database" -s -N -B -e "select user from example where user='$username' > 0;"))
Now I'm left with logic issues but I think I can handle that.
Related
I have the following script:
VAR=$(mysql -u root -e " use <database_name>; select column1,column2,column3 from <table_name>;")
dialog --title "something" --msgbox "$VAR" 50 50
and the output that I got is like this.
this
but i'm expecting the output to be like this instead inside the dialog command
this
I can get this output by not storing the sql commands in a variable.
where am I wrong here?
On your VAR=$(mysql ...), you can add --table to get the disired effect.
VAR=$(mysql -u root -t -e ...)
I don't have remote access to a MySQL server, so I am trying to do it via an SSH session.
It partly works, but not correctly.
sshpass -p $password ssh user#$IP /bin/bash << EOF
mysql -N -uroot -ppassword test -e "select id from client where user ='$user'"
EOF
This will show the result of the select statement, but I'd like to be able to use that result in another echo statement.
eg:
The user ID is: xxxxx
I tried to assign the output to a variable using:
sshpass -p $password ssh user#$IP /bin/bash << EOF
res=$(mysql -N -uroot -ppassword test -e "select id from client where user ='$user'")
echo $res
EOF
But that results in:
ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/run/mysql/mysql.sock' (2)
If I quote EOF like 'EOF' then I can get the result into res but I lose the ability to use $user
Is there anyway to do this so I can use my variable in the heredoc and get the result of the MySQL query to a new variable ?
If I quote EOF like 'EOF' then I can get the result in to res but I lose the ability to use $user
You can pass $user to bash as a positional parameter and still have the quoted EOF and its advantages. E.g:
sshpass -p "$password" ssh "user#$IP" /bin/bash -s "$user" << 'EOF'
res=$(mysql -N -uroot -ppassword test -e "select id from client where user ='$1'")
echo $res
EOF
Bash manual describes the -s option as follows.
If the -s option is present, or if no arguments remain after option processing, then commands are read from the standard input. This option allows the positional parameters to be set when invoking an interactive shell or when reading input through a pipe.
I want to connect to mysql databse and execute some queries and export its result to a varibale, and do all of these need to be done entirely by bash script
I have a snippet code but does not work.
#!/bin/bash
BASEDIR=$(dirname $0)
cd $BASEDIR
mysqlUser=n_userdb
mysqlPass=d2FVR0NA3
mysqlDb=n_datadb
result=$(mysql -u $mysqlUser -p$mysqlPass -D $mysqlDb -e "select * from confs limit 1")
echo "${result}" >> a.txt
whats the problem ?
The issue was resolved in the chat by using the correct password.
If you further want to get only the data, use mysql with -NB (or --skip-column-names and --batch).
Also, the script needs to quote the variable expansions, or there will be issues with usernames/passwords containing characters that are special to the shell. Additionally, uppercase variable names are usually reserved for system variables.
#!/bin/sh
basedir=$(dirname "$0")
mysqlUser='n_userdb'
mysqlPass='d2FVR0NA3'
mysqlDb='n_datadb'
cd "$basedir" &&
mysql -NB -u "$mysqlUser" -p"$mysqlPass" -D "$mysqlDb" \
-e 'select * from confs limit 1' >a.txt 2>a-err.txt
Ideally though, you'd use a my.cnf file to configure the username and password.
See e.g.
MySQL Utilities - ~/.my.cnf option file
mysql .my.cnf not reading credentials properly?
Do this:
result=$(mysql -u $mysqlUser -p$mysqlPass -D $mysqlDb -e "select * from confs limit 1" | grep '^\|' | tail -1)
The $() statement of Bash has trouble handling variables which contain multiple lines so the above hack greps only the interesting part: the data
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 6 years ago.
Lets say I have a query first that inserts certain values in a table and the next query is to display the maximum value of one of the columns and then store that in a variable. I then need to display that variable such that it shows the max value.
For example:
sudo mysql -u$mysql_user -p$mysql_pwd -h $mysql_host --database $db_name -e "INSERT INTO service_status_batch VALUES ();"
batch_id= sudo mysql -u$mysql_user -p$mysql_pwd -h $mysql_host --database $db_name -e "SELECT MAX(id) as maxid FROM service_status_batch;"
echo "Value of the id is:" $batch_id
This echo command should then show the value of the variable. However it ends up showing me the value of the select query in the form of a table and not the value of the variable.
Is there a particular way to assign the query value to a variable in shell script?
I have attached the select query value that it shows.
Use -s and -N options with mysql command like this.
sudo mysql -u$mysql_user -p$mysql_pwd -h $mysql_host --database $db_name -e "INSERT INTO service_status_batch VALUES ();"
batch_id=`sudo mysql -u$mysql_user -p$mysql_pwd -h $mysql_host --database $db_name -s -N -e "SELECT MAX(id) as maxid FROM service_status_batch;"`
echo "Value of the id is:" $batch_id
Refer the details for -s and -N :
--silent, -s
Silent mode. Produce less output. This option can be given multiple
times to produce less and less output.
This option results in nontabular output format and escaping of
special characters. Escaping may be disabled by using raw mode; see
the description for the --raw option.
--skip-column-names, -N
Do not write column names in results.
EDIT3: Bad explanation - I was trying to show how to get the value considering it could be used as necessary:
sudo echo $(echo "SELECT MAX(id) as maxid FROM service_status_batch" | mysql dbnamehere -uUser -pPassword)
EDIT1: variable version obviously
EDIT2: corrected variable assignment by using shellcheck.net as suggested. thanks.
EDIT3: one last edit to add sudo right before mysql command as it won't work without it for users other than root.
batch_id=$(echo "SELECT MAX(id) as maxid FROM service_status_batch" | sudo mysql dbnamehere -uUser -pPassword)
I'm trying to set variable in the cycle in Ubuntu bash, which is getting recordset from database, but this variable is setting to its previous value.
Here is a code:
#!/bin/bash
PREV_FILE_PATH="127"
while true
do
echo "$PREV_FILE_PATH"
mysql -h$DB_HOST -u $DB_USER -p$DB_PASS $DB_NAME --skip-column-names --default-character-set=UTF8 -e "here is a query" | while read "here is getting variables from recordset";
do
PREV_FILE_PATH="777"
done
done
And this code prints every time:
127
127
127
But whe I replaced this block-:
mysql -h$DB_HOST -u $DB_USER -p$DB_PASS $DB_NAME --skip-column-names --default-character-set=UTF8 -e "here is a query" | while read "here is getting variables from recordset";
with just while true and break at the end of cycle it works fine and prints:
127
777
777
777
Script creates some subshell and running that MySQL query in that subshell. So what should I do to make script change that variable?
As you noted the issue is due to the creation of a subshell which is being caused by piping the output of the mysql command to the while loop. A simple example:
PREV_FILE_PATH=127
echo test | while read _; do
PREV_FILE_PATH=777
done
echo $PREV_FILE_PATH
# output: 127
Since you're using BASH you can move the mysql command from being a pipe to a substituted process fed to the while loop via STDIN redirection. Using the previous simple example:
PREV_FILE_PATH=127
while read _; do
PREV_FILE_PATH=777
done < <(echo test)
echo $PREV_FILE_PATH
# output: 777
So to fix your code, you will want to move your mysql command in the same fashion that I moved the echo command above:
while read "here is getting variables from recordset"
do
PREV_FILE_PATH="777"
done < <(mysql -h$DB_HOST -u $DB_USER [..remaining options..])
Note that process substitution via <() is a BASH-ism and isn't POSIX compliant.