Why Does My BASH Script Fail When Adding Additional Arguments?

Question

I have a BASH script to go through my Git projects recursively and pull them. We use two branches, master and development. When I change the script to include && git checkout development, my script fails to run and I get this error:

find: missing argument to `-exec'

When I remove the && git checkout development part again, the error remains the same even though the script is reverted to its initial state.

Here is the script:

#!/bin/bash
find . -type d -name .git -exec sh -c "cd \"{}\"/../ && pwd && git pull && git status" \;

And here it is with the Git checkout call (and removing git status):

#!/bin/bash
find . -type d -name .git -exec sh -c "cd \"{}\"/../ && pwd && git checkout development && git pull" \;

What is going on here? Is it possibly a red herring that the issue only occurs after changing the script?

Try this: find . -type d -name .git -exec sh -c 'cd "$1"/../ && pwd && git checkout develop && git pull' _ '{}' \; — anubhava
– anubhava, Commented Aug 17, 2017 at 8:34
That's worked. Thanks! Could you explain why the change makes it work in an answer please? Is the "$1" part of it pulling in a command line argument, or doing something different? — Donglecow
– Donglecow, Commented Aug 17, 2017 at 8:39
No need to use cd: find . -type d -name .git -exec sh -c 'readlink -f "$1"/../ && git -C "$1"/../ checkout develop && git -C "$1"/../ pull' _ '{}' \; — user2394284
– user2394284, Commented Aug 25, 2017 at 11:07

Donglecow · Accepted Answer · 2017-08-25 10:56:54Z

2

Don't use placeholder {} inside command string after sh -c. You can use this find:

find . -type d -name .git -exec sh -c \
'cd "$1"/../ && pwd && git checkout develop && git pull' _ '{}' \;

Here we are passing dummy _ to populate $0 in command line and {} is being passed to populate $1 positional argument.

Edit based on comment: After tinkering to pass in arguments to the script, I've found it's easiest to export them for the subshell first:

#!/bin/bash
BRANCH=$2
export BRANCH
echo "Checking out branch $BRANCH"
find . -type d -name .git -exec sh -c 'cd "$1"/../ && pwd && git checkout "$BRANCH" && git pull; echo "Branch being checked out: $BRANCH"' _ '{}' "$BRANCH" \;

edited Aug 25, 2017 at 10:56

Donglecow

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answered Aug 17, 2017 at 8:41

anubhava

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2 Comments

Donglecow Over a year ago

Many thanks. So if I want to pass in a command line argument to tell it which branch to check out, will the "$1" interfere, or have to be changed in any way?

anubhava Over a year ago

You can pass additional parameters e.g.: -exec sh -c 'cd "$1"/../ && pwd && git checkout develop && git pull; echo "second value: $2"' _ '{}' "value" \;

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