Revisions to How to read the whole shell script before executing it?

small style change

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edit approved Dec 3, 2024 at 18:31

1

Use:

{
  ... your code ...

  exit
}

Bash will read the whole block {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

For scripts that are "sourced" rather than executed, use:

{
  ... your code ...

  return 2>/dev/null || exit
}

Use:

{
  ... your code ...

  exit
}

Bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

For scripts that are "sourced" rather than executed, use:

{
  ... your code ...

  return 2>/dev/null || exit
}

Use:

{
  ... your code ...

  exit
}

Bash will read the whole block {} before executing it, and the exit directive will make sure nothing will be read outside of the code block.

For scripts that are "sourced" rather than executed, use:

{
  ... your code ...

  return 2>/dev/null || exit
}

simplify my answer

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edited Oct 10, 2023 at 5:40

VasyaNovikov

1.6k
2
17
26

Simplest solutionUse:

{
  ... your code ...

  exit
}

This way, bashBash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

If you do not want to "execute" the script, butFor scripts that are "sourced" rather to "source" itthan executed, you need a different solution. This should work thenuse:

{
  ... your code ...

  return 2>/dev/null || exit
}

Or if you want direct control over exit code:

{
  ... your code ...

  ret="$?";return "$ret" 2>/dev/null || exit "$ret"
}

Voilà! This script is safe to edit, source and execute. You still have to be sure you don't modify it in those milliseconds when it is initially being read.

Simplest solution:

{
  ... your code ...

  exit
}

This way, bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

If you do not want to "execute" the script, but rather to "source" it, you need a different solution. This should work then:

{
  ... your code ...

  return 2>/dev/null || exit
}

Or if you want direct control over exit code:

{
  ... your code ...

  ret="$?";return "$ret" 2>/dev/null || exit "$ret"
}

Voilà! This script is safe to edit, source and execute. You still have to be sure you don't modify it in those milliseconds when it is initially being read.

Use:

{
  ... your code ...

  exit
}

Bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

For scripts that are "sourced" rather than executed, use:

{
  ... your code ...

  return 2>/dev/null || exit
}

deleted 44 characters in body

Source Link

edited Jun 14, 2018 at 14:09

VasyaNovikov

1.6k
2
17
26

Simplest solution:

{
  sleep 10s
... your echocode "test"...

  exit
}

The code block {} and the exit directive is what matters. This way, bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

If you do not want to "execute" the script, but rather to "source" it, you need a different solution. This should work then:

{
  sleep 10s
... your echocode "test"...

  ret="$?";return "$ret"return 2>/dev/null || exit "$ret"
}

Or thisif you want direct control over exit code:

# presuming bash is run with -e (errexit)
{
  sleep 10s
... your echocode "test"...

  returnret="$?";return "$ret" 2>/dev/null || exit 0"$ret"
}

Voilà! This script is safe to edit, source and execute. You still have to be sure you don't modify it in those milliseconds when it is initially being read.

Simplest solution:

{
  sleep 10s
  echo "test"

  exit
}

The code block {} and the exit directive is what matters. This way, bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

If you do not want to "execute" the script, but rather to "source" it, you need a different solution. This should work then:

{
  sleep 10s
  echo "test"

  ret="$?";return "$ret" 2>/dev/null || exit "$ret"
}

Or this:

# presuming bash is run with -e (errexit)
{
  sleep 10s
  echo "test"

  return 2>/dev/null || exit 0
}

Voilà! This script is safe to edit, source and execute. You still have to be sure you don't modify it in those milliseconds when it is initially being read.

Simplest solution:

{
  ... your code ...

  exit
}

This way, bash will read the whole {} block before executing it, and the exit directive will make sure nothing will be read outside of the code block.

If you do not want to "execute" the script, but rather to "source" it, you need a different solution. This should work then:

{
  ... your code ...

  return 2>/dev/null || exit
}

Or if you want direct control over exit code:

{
  ... your code ...

  ret="$?";return "$ret" 2>/dev/null || exit "$ret"
}

Voilà! This script is safe to edit, source and execute. You still have to be sure you don't modify it in those milliseconds when it is initially being read.