Nsryjdtyk

I am trying to assign bash scripts to variables for use in a wrapper script, yet every method I have tried has resulted to a "No such file or directory" error:

#!/bin/bash 

textgen=$(< textcreator.sh)

tablegen=$(< tablecreator.sh)

htmlgen=$(< webcreator.sh)





 exec $textproces

 if [[ $textproces -eq 0 ]]

    then

        echo "Text Processing has ran successfully"

        exec $listgen

        if [[ $listgen -eq 0 ]]

          then

              echo "List Generation has ran successfully"

              exec $pagegen

              if [[ $pagegen -eq 0 ]]

                then

                    echo "HTML Page Generation has ran successfully"

                    echo "The command wrapper has finished successfully"

                    exit 0

                else

                  echo "Error: HTML Page Generation was unsuccessful"

                  exit 1

              fi

          else

             echo "Error: Text Processing was unsuccessful"

             exit 1

        fi

    else

        echo "Error: Text Processing was unsuccessful"

        exit 1

 fi

The methods I have tried: cat filename, export, source and standard assigning. The idea is that it is to work without having to give a path name, all of these shells are in the same directory anyway. All help is vastly appreciated!

NOTE: I know its horrifying, but it was the way I thought of doing it with the IF statements

It actually depends upon which directory you were in when you executed the root script. So them being in the same directory does not help. You can cd into the directory containing the sub scripts before executing them from the root script.

There are several serious problems with the code in the question. Some are identified by Shellcheck. Always run Shellcheck on new and modified code.

Problems include:

1. The variables assigned at the start are not the variables that are used later.


2. The variables are set to the contents of the programs, not the names of the files containing the programs. (The only way to use the contents would be to pass them to eval. That would not be a good idea.)


3. 
   exec is used to try to run the programs. exec replaces the current program with a new program, determined by its arguments. It's roughly equivalent to running another program and exiting immediately afterwards. Once exec has been used, no code after it in the program is executed.


4. The paths to the programs to be run are not specified. Since the programs are in the current directory (.), and . is usually not in the PATH (for very good reasons), the shell will issue 'command not found` errors for the attempts to run them.


5. The code tries to get the exit status of programs by checking the value of the variable containing the program name. The exit status of a program can be obtained via the $? built-in variable. However, the simplest way to check if a command failed is to use it directly in an if.

The Shellcheck-clean code below attempts to fix the problems. It also eliminates the nested if statements in an attempt to make the code clearer.

#!/bin/bash



textgen=textcreator.sh

listgen=tablecreator.sh

pagegen=webcreator.sh



if ! "./$textgen" ; then

    echo "Error: Text Processing was unsuccessful"

    exit 1

fi



echo "Text Processing has ran successfully"



if ! "./$listgen" ; then

    echo "Error: List Generation was unsuccessful"

    exit 1

fi



echo "List Generation has ran successfully"



if ! "./$pagegen" ; then

    echo "Error: HTML Page Generation was unsuccessful"

    exit 1

fi



echo "HTML Page Generation has ran successfully"

echo "The command wrapper has finished successfully"



exit 0

The code could be used in various other ways (e.g. send error output to standard eror (... >&2), reduce duplication by using a function to output the error message and exit the program).