I am currently working on a shell script.
I want to modify the contains of the column "balance" in "Student" table as "balance=balance-$deduct" in shell script.
I am executing following line in shell
mysql -h "$host" -u "$user" --password="$pass" --database="$db" -e "UPDATE student SET balance=balance-$deduct WHERE pid=$pid";
Here i specified the host, username and password properly and i have one numeric value stored in $deduct
but when i execute the script i am getting following error:
ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'WHERE pid=132054' at line 1
where I am wrong here?
thanks
You don't need --database=
mysql -h "$host" --user="$user" --password="$pass" "$db" -e "UPDATE student SET balance=balance-$deduct WHERE pid=$pid";
You can also do Use $db in you query
Just remove --database=
It should work.
Related
How do I pass variables that I would have created in linux shell to the mysql command.
For example:
# carname=tesla-new
# echo $carname
I want to use the $car-name variable when passing arguments to MySQL commands.
This variable will come in as input from loop while iterating through a list of values.
mysql -u wp-supportuser -h xx.xxx.xxx.xxx -p<mypassword>! -e 'DROP DATABASE ${carname};'
I need the hyphon(-) in the variable's value, else my application will accept. We get the below error when the above command is run:
mysql: > [Warning] Using a password on the command line interface can be insecure. ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-wp' at line 1
How will I be able to achieve this?
It is a MySQL error, because MySQL does not support the usage of hyphen by itself. Instead, you need to enclose it into
`
characters. You can use it if you wrap your command inside apostrophes, like:
mysql -u wp-supportuser -h xx.xxx.xxx.xxx -p<mypassword> -e 'drop database `$carname`;
I need a one-liner to provision users in my databases. I'm running MySQL 5.6. Here is what I'm setting the password as a variable so that I can pass it dynamically (obviously it won't always be 'password').
mysql_password="password"
mysql -u ted -e "SET PASSWORD FOR 'ted'#'localhost' = PASSWORD($mysql_password);"
I'm getting the following error when I run this:
ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'password)' at line 1
What is wrong with the statement? Can it be corrected or is another solution available so I accomplish this via Bash?
You need to surround the variable by single quotes:
mysql -u root -e "SET PASSWORD FOR 'ted'#'localhost' = PASSWORD('$mysql_password');"
I'm trying to run an update query on a db server via bash command.
I have to update an IP field (which is sorted as a string) but i'm getting a syntax error...
ssh admin#192.168.3.240 "/usr/local/mysql/bin/mysql -D SMARTPARK -u parkuser -ppass -e 'update client SET online=0 where client_ip='192.168.42.11''"
I'm getting as error
ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.42.11' at line 1
which is the error?
Try to escape " character with \" to make sure you don't escape from the string you're sending to your DB.
In other words, try to put the following in the bash file you're executing:
ssh admin#192.168.3.240 "/usr/local/mysql/bin/mysql -D SMARTPARK -u parkuser -ppass -e \"update client SET online=0 where client_ip='192.168.42.11'\""
I'm tryng to create a code in raspbian system to update a value in my database mySQL stored into my hosting server using a bash ".sh" scrypt.
this line won't to work but I don't know why...
mysql -h eliuslab.com -u user -pPassWord -D mydatabase -e "INSERT INTO 'home-IP' ('IP') VALUES ('192.168.1.2')"
And the output:
'ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ''home-IP' ('IP') VALUES ('192.168.1.2')' at line 1 '
this query works well in a GUI mySQL client, but what I want to do is to launch this action with my ".sh" bash scrypt.
Thank you so much :)
Okay I got the solution. I ran your script in my mysql database and got the same error. It's not the fault of script but it's a parsing error.
The correct command is
mysql -h eliuslab.com -u user -pPassWord -D mydatabase -e "INSERT INTO tablename (column) VALUES ('BLABLASTRING')"
Notice there is no need of using quotation marks around the table name and the column name. The quotation marks treat them as string constant rather than a container.
I have a trouble with my bash script here, the mysql select to be more specific. It runs fine without a variable, but when I am declaring variable the script fails, please advise:
#!/bin/bash
table="user_vpn_account"
1="2" # when I introduce this variable script fails , if I will use value "2" directly script works
pwd="password"
D="database"
mysql -uroot -p$pwd -D$D -s -N -e "SELECT port FROM user_vpn_account where user_vpn_id=$1;"
/usr/bin/mysql -uroot -p$pwd -D$D<< eof
eof
ERROR 1064 (42000) at line 1: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1
Thanks in advance.
You can put $1 parameter into a variable, like this:
variable=$1
and then use it like $variable.
Bash Guide - Variables