Bash writing variable? : command not found

Question

 #!/bin/bash

    VAR019='priusr'
    VAR901='pripasswd'

    echo "CREATE DATABASE ftp;
    GRANT SELECT, INSERT, UPDATE, DELETE ON ftp.* TO '${VAR019}'@'localhost' IDENTIFIED BY '${VAR901}';

This will output

   : command not found
    : command not found
    : command not found
    : command not found

How can I fix this? What am I doing wrong?

There was a lowercase/uppercase problem with var019 as well — sehe
– sehe, Commented Sep 16, 2011 at 13:03

sehe · Accepted Answer · 2011-09-16 13:01:46Z

2

I suggest using a here document to reduce complexity of quoting.

#!/bin/bash

VAR019='priusr'
VAR901='pripasswd'

cat <<HERE 
CREATE DATABASE ftp;
GRANT SELECT, INSERT, UPDATE, DELETE ON ftp.* TO '${VAR019}'@'localhost' IDENTIFIED BY '${VAR901}';
HERE

To the skeptics, here is the output:

CREATE DATABASE ftp;
GRANT SELECT, INSERT, UPDATE, DELETE ON ftp.* TO 'priusr'@'localhost' IDENTIFIED BY 'pripasswd';

edited Sep 16, 2011 at 13:01

answered Sep 16, 2011 at 12:56

sehe

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8 Comments

Arnaud F. Over a year ago

Variable replacement will not be done using quotes, use double quotes instead

Blagovest Buyukliev Over a year ago

Or use a backslash to continue the current statement on the next line.

sehe Over a year ago

@Arnaud/@Blagovest: your comments don't have anything to do with my answer. Please move them to the question?

jmituzas Over a year ago

why is it I am still getting not found errors, and I know how to write variables... why are they acting like commands rather then variables? heres wht yours gave me as an output: @sehe this is what I got: : not found : : not found : cat: CREATE DATABASE ftp; : No such file or directory 'pripasswdN ftp.* TO 'priusr

sehe Over a year ago

How do you invoke it? My guess you are not actually using bash. Otherwise, copy the script Literally and retest

|

jakub.g · Accepted Answer · 2011-09-16 14:24:02Z

1

Make sure your file is using right EOL (newline separating).

I had exactly the same problem some time ago and it was caused by newlines being \r\n (Windows-style) instead of \n (Linux style).

In that case, Linux assumes that \r is a name of command and tries invoking it. But \r is "invisible" character, that's why the strange-looking error message.

See this link on how to convert CRLF to LF.

edited Sep 16, 2011 at 14:24

answered Sep 16, 2011 at 14:07

jakub.g

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Comments

David W. · Accepted Answer · 2011-09-16 14:11:15Z

I just cut and pasted what you put in your program. The only issue I had was the lack of a closing double-quote quotation mark.

I got the following output:

CREATE DATABASE ftp;
GRANT SELECT, INSERT, UPDATE, DELETE ON ftp.* TO 'priusr'@'localhost' IDENTIFIED BY 'pripasswd';

Which seems to be what you want.

The :command not found is strange. For example, here's what I get when I enter nonsense:

$ asdas
bash: asdas: command not found.

This looks like maybe BASH isn't executing your script, or you think it's executing your script, but it's really executing another script. Use the which command and make sure that your version of the script is being executed.

You can also use the bash command itself: bash myscript.sh. This way, you're verifying that your script is being executed with the BASH shell, and that it is your script that's being executed and not some other script in your path.

Another possibility is that /bin/bash is not where your bash executable is actually located. You might want to verify that too. I know I've seen this when running Perl scripts with #! /usr/bin/perl on top, and Perl was actually located in /bin/perl or /usr/local/bin/perl.

Also make sure your shebang is formatted correctly. For example, there is no whitespace before the #, and it's the first character on the line. That the ! is right next to the pound sign. Some processors or shells require one and only one space between the shebang and the name of the command interpretor.

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