CMD="mysqldump -usomeuser -psomepass db_name \> /../SQL_$(date +'%m-%d-%Y').sql"
The above throws the following exception:
mysqldump: Couldn't find table: ">"
"&1>" also is seen as a table name.
I tried the option -all-database(s) and that doesn't work either.
Thanks!
You must not escape greater-than in the command:
CMD="mysqldump -usomeuser -psomepass db_name > /../SQL_$(date +'%m-%d-%Y').sql"
even then, if you try to execute the command just by invoking it, bash treats the greater-than as a positional parameter. You can execute it with a call to sh or bash:
bash -c "$CMD"
Related
This scipt gives proper result, but also gives mistake:$'\r': command not found in line with query.
#!/bin/bash
keyOrPass=$1
intercom=$2
flat=$3
number=$4
mysql -ulogin -ppass dbname -e "select cli.codeGuestEmail, cli.codePrivateEmail, cliKey.rf_id, cliKey.emailNotification from mbus_clients as cli join mbusClientKeys as cliKey on cliKey.id_client=cli.id WHERE cli.flat=${flat} and cli.domophone=${intercom};";
This is how I run the script:
sh sendEmailNotification.sh key 10001014 11 1
Create a version of your script without Windows line delimiter \r:
tr -d "\r" < sendEmailNotification.sh > sendEmailNotification_fixed.sh
Check and fix line-endings in your script, probably, it was saved with \r\n as line delimiters.
This link may be useful: How to convert DOS/Windows newline (CRLF) to Unix newline (\n) in a Bash script?
I am trying to execute an insert statement from linux shell where one of the columns has '$2a$10$zKjqmgld1gDYB/qkDuAS' in the value. When I see the inserted data the value is truncated and I get only 'aKjqmgld1gDYB/qkDuAS' as any digit followed by dollar is treated by linux as a parameter passed to the script.
This is how I am executing the script
mysql -u user --password=password -e "insert into users(id,name,password) values(1,'Some Name','\$2a\$10\$zKjqmgld1gDYB/qkDuAS')"
I have even tried escaping the $ like \$2a\$10\$zKjqmgld1gDYB/qkDuAS, but still it yields the same truncated data , however when I do echo '\$2a\$10\$zKjqmgld1gDYB/qkDuAS', I get the entire thing back.
Sameer
Single quotes don't nest in shell (do they anywhere else, anyway?). So, your string is effectively outside the quotes. Combine with double quotes and backslashes:
mysql -u user --password=password -e \
"insert into users(id,name,password) values(1,'Some Name','\$2a$10\$zKjqmgld1gDYB/qkDuAS')"
it was already in double quotes (sorry, for writing the wrong query). That didn't work either, however I found a workaround
echo "insert into users(id,name,password) values(1,'Some Name','\$2a\$10\$zKjqmgld1gDYB/qkDuAS')" >> temp.sql
mysql -u user --password=password < temp.sql
This finally worked.
I wrote a MySQL command in bash (Ubuntu) :
[XXXX:~]$ mysql -h localhost -u XXXX -pXXXX -e "DROP DATABASE IF EXISTS `f-XXXX`;"
I need backquote in this command, cause database name is variable.
That command doesn't work and it sends f-XXXX command not found
I think my problem is related to backquotes. How can I do?
You need not use backtick for variable substitution here.
[XXXX:~]$ mysql -h localhost -u XXXX -pXXXX -e "DROP DATABASE IF EXISTS ${DB};""
Bash takes the content of the backtick and runs another bash process with that as a command.
This is a backtick. Backtick is not a quotation sign, it has a very special meaning. Everything you type between backticks is evaluated (executed) by the shell before the main command (like chown in your examples), and the output of that execution is used by that command, just as if you'd type that output at that place in the command line.
Use $(commands) instead.
mysql -h localhost -u XXXX -pXXXX -e "DROP DATABASE IF EXISTS $('f-XXXX');"
I tried to execute the following mysql command in one of my scripts:
mysql -e 'show global status like 'open_files''
But it doesn't seem to work, because of the single quotes around the string 'open_files'.
How can I issue a command like this, that contains a single quote?
Use double quote outside.
mysql -h127.0.0.1 -uxxxxx -pxxxxx -A databasename -e "show global status like 'open_files'"
or the reverse way:
mysql -h127.0.0.1 -uxxxxx -pxxxxx -A databasename -e 'show global status like "open_files"'
or you could escape the single quote.
And you are using LIKE, didn't you miss the wild char %?
I'm trying to execute this:
$ mysql --user=XXX --password=XXX --batch --skip-column-names \
-e "SELECT userid, displayname FROM Users" stackoverflowdb | \
split -l 50 -a 5 - "result."
but bash complains about ) 'unexpected' character (i have this character and few other 'weird ones' in my mysql password). I tried to take my password in quotes --password="myweirdpasshere" but then mysql won't login me at all (probably password is incorrect?)
Some characters such as $, `, ", and sometimes \ retain their special meaning inside double quotes. If your password contains any of those characters, use single quotes instead. (Unless your password also contains single quotes, in which case you might just drop the quotes altogether and put a \ before each special character.)
You should be able to put the password in the $HOME/.my.cnf file, avoiding this issue as well as increasing the security.
Another option to using $HOME/.my.cnf is embedding the configuration file into the script. A comment on the documentation gives examples on this. For your case it should be:
$ mysql --user=XXX --defaults-file <(printf '[client]\npassword=XXX\n') \
--batch --skip-column-names \
-e "SELECT userid, displayname FROM Users" stackoverflowdb | \
split -l 50 -a 5 - "result."
If you don't give the password on the command line, it will bring up an interactive prompt. That might solve your error. Just use -p instead of --password=XXX.
Of course, if you require unattended access to the script, that's not a very useful answer. It would be more helpful if you could create a minimal example and tell us the exact error that Bash gives you.