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Failed to parse arguments: Argument to "--command/-e" is not a valid command: Text ended before matching quote was found for ". (The text was '"cpulimit')

this is what i get when i run the following script in terminal 

    #!/bin/bash
read -p "Which program u want to limit its processes?" ProgrameName
read -p "Which limitation percentage u want for it ?" limitationPercentage  

getAllPIDRunUnderThisProgram=$( ps -e | grep "$ProgrameName" | awk '{print $1;}')
for i in $getAllPIDRunUnderThisProgram
   do
    gnomeTab+="  --tab -e \"cpulimit -p $i -l $limitationPercentage \" "  
   done

gnome-terminal $gnomeTab

he cant parse the escape character "\" which it has to be used because of the double quote in line 8 gnomeTab+=" --tab -e \"cpulimit -p $i -l $limitationPercentage \" ",so is there a solution to use the double quote as they are mandatory to be used after --tab -e " some commands " and not to get the problem of parsing ?

Ahmed Zain El Dein
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1 Answers1

3

You can change the very first line to

#!/bin/bash -xv

to make the shell show you how it interprets arguments.

Rather than escaping (which leads to eval), you should use arrays to accumulate options:

for i in $getAllPIDRunUnderThisProgram ; do
    gnomeTab+=(--tab -e "cpulimit -p $i -l $limitationPercentage")  
done

echo "${gnomeTab[@]}"
choroba
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