How to use 'cat' command with -exec in bash?

Question

I search for some files and I want to replace old content with new content from other files, so I have

#!/bin/bash
find /home/old -path "*$1" -exec cat /home/new/$1 > {} \;

When I execut:

> sh find.sh 'blam.php'

Problem is the script is creating a file called {} in the same dir with find.sh

Why run it under sh when your `#!` if bash? Try `./find.sh 'blam.php'` — Nic3500, Apr 24 '18 at 10:57
Possible duplicate of [find and copy file using Bash](https://stackoverflow.com/q/1562102/608639), [How to move or copy files listed by 'find' command in unix?](https://stackoverflow.com/q/17368872/608639), [Find and copy files](https://stackoverflow.com/q/5241625/608639), etc. — jww, Apr 25 '18 at 00:12

Nahuel Fouilleul · Accepted Answer · 2018-04-24T11:54:24.833

2

The problem you get is that > is part of shell syntax, and should be literal for find command and syntactic in exec command.

what you are trying to do may be done in two steps

generating script
executing the script

.

find /home/old -path "*$1" -printf 'cat /home/new/'"$1"' > %p\n' > script.sh
sh script.sh

this way allows to check the commands, or using pipe

find /home/old -path "*$1" -printf 'cat /home/new/'"$1"' > %p\n' | sh

another way less efficient because spwans a shell for each file

find /home/old -path "*$1" -exec sh -c 'cat "/home/new/$1" > {}' _ "$1" \;

edited Apr 24 '18 at 11:54

answered Apr 24 '18 at 11:30

Nahuel Fouilleul

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Good solutions, but can I pass argumnet $1 in -sh c like `-exec sh -c 'echo $1'`, is possible? – Dobi Caps Apr 24 '18 at 11:43
It's safer to pass it as an argument to the `sh` command, rather than embed it in the command: `find ... -exec sh -c 'echo $1' _ "$1"`. There are two different `$1`s here; the one in single quotes is the value of the original shell's (in double quotes). – chepner Apr 24 '18 at 11:44
2

So `-exec sh -c 'cat /home/new/$1 > {}' _ "$1" \;`. – chepner Apr 24 '18 at 11:45
updated from comment – Nahuel Fouilleul Apr 24 '18 at 11:55
Hey @chepner , is possible to pass a variable in -exec sh -c? Like `-exec sh -c 'echo $VAR' _ "$VAR"` – Dobi Caps Apr 25 '18 at 08:30
1

@DobiCaps, the variables are inherited from caller process, `env` can be used to set variable for a process, command passed by arguments : `-exec env VAR=123 sh -c 'echo "$VAR"' \;`, the syntax `_ "$1"` is used to set positional parameters from `sh -c` or `bash -c` command line arguments the `_` is for `$0` – Nahuel Fouilleul Apr 25 '18 at 11:55
from man bash `-c string If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.` – Nahuel Fouilleul Apr 25 '18 at 11:59

score 1 · Answer 2 · edited Apr 24 '18 at 11:45

The problem here is that > {} is not part of the find command. The script is first interpreted by the shell, and the meaning of > is handled by the shell before passing the stuff around it as arguments to execute:

$ set -- blam.php
$ set -x
$ find /home/old -path "*$1" -exec cat "/home/new/$1" > {} \;
+ find /home/old -path '*blam.php' -exec cat /home/new/blam.php ';'

If you want a copy of your file somewhere else, use cp.

How to use 'cat' command with -exec in bash?

2 Answers2