LinuxQuestions.org - [SOLVED] Executing jobs as stored in database

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Executing jobs as stored in database

Hi everyone,

I am coping with a problem for a few weeks already. After extensive searching I still can't find the solution.

Here is my problem:

I have a MySQL database with a table that has job commands, like: $SCRIPTDIR/path/to/script.sh "param1" "param2"

I have a script that executes a query and retrieves the value from that column. I echo'd the value (stored in $JOBCOMMAND) and it is correct. So that part works.

Now I try to execute the job as:

`$JOBCOMMAND` --> fails
$($JOBCOMMAND) --> fails
`echo $JOBCOMMAND` --> fails
`$($JOBCOMMAND)` --> fails

The error: No such file or directory.

The actual problem (as I see it) is that I have a $ in the job command, which does not get parsed. I am trying hard to avoid eval, but I am completely lost.

Does anyone have any suggestions?

Thank you for your time and effort.

I am unsure where you are saying the $ is. Is it part of the command itself stored in $JOBCOMMAND, or the variable name is $JOBCOMMAND instead of JOBCOMMAND?

It is part of the command itself stored in $JOBCOMMAND

When doing: echo $JOBCOMMAND, the result is:

$SCRIPTDIR/path/to/script.sh "param1" "param2"

Like this?

Code:

lm ~ # SCRIPTDIR=/tmp

+ SCRIPTDIR=/tmp

lm ~ # JOBCOMMAND='ls -al $SCRIPTDIR'

+ JOBCOMMAND='ls -al $SCRIPTDIR'

lm ~ # $(echo $JOBCOMMAND | sed  -e 's/$SCRIPTDIR/\'$SCRIPTDIR'/')

++ echo ls -al '$SCRIPTDIR'

++ sed -e 's/$SCRIPTDIR/\/tmp/'

+ ls -al /tmp

total 4292

drwxrwxrwt 15 root    root      20480 Jan 16 15:50 .

drwxr-xr-x 23 root    root        4096 Dec 17 15:47 ..

-rw-r--r--  1 root    root          0 Jan 16 00:00 access.log

drwx------  2 root    root        4096 Jan 14 11:19 cvcd

...

That would work, except ... ;-)

At this moment I have 1 variable, $SCRIPTDIR. There will be more in the future and I don't want to update the script for each new variable.

Can you post any of your script then?

I solved it for now:

Code:

#!/bin/bash



PART="home/my_account"

SUBFOLDER="test"



SUB_COMMAND(){

  PART_IN=$1

# Add backslashes to all forward slashes

  PART_OUT=$(echo $PART_IN | sed -e 's/\//\\\//g')

# If there are 2 backslashes in front of a forward slash, change to 1 backslash

  PART_OUT=$(echo $PART_OUT | sed -e 's/\\\\\//\\\//g')

# Return end result

  echo $PART_OUT

}



CREATE_COMMAND(){

  SQL='ls /{@PART}/apps/{@SUBFOLDER}'

# Per variable substitute value

# Variable: PART

  PART_SUB=$(SUB_COMMAND $PART)

  COMMAND_OUT=$(echo $SQL | sed -e "s/{@PART}/$PART_SUB/g")

# Variable: SUBFOLDER

  COMMAND_OUT=$(echo $COMMAND_OUT | sed -e "s/{@SUBFOLDER}/$SUBFOLDER/g")

# Return end command

  echo $COMMAND_OUT

}



echo "Part: $PART"

COMMAND=$(CREATE_COMMAND)



echo "Command: $COMMAND"



exit 0

Three notes on this:
1. I will have to create a substitute value for each parameter.
2. I will have to update the CREATE_COMMAND function for each parameter.
3. I changed the variable names from $PART to {@PART} in the value, since I could not find a way to parse $PART.

There was nothing I could think of that was clean and elegant.

Thank you for your time and effort!