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Top Forums UNIX for Dummies Questions & Answers Getting file without mentioning the path Post 302971481 by Don Cragun on Wednesday 20th of April 2016 05:05:21 PM
Old 04-20-2016
Quote:
Originally Posted by ashish_samarth
Try Below

Code:
find ./* -type f -iname "source_file.txt" -exec dirname {} \; 2> /dev/null | cd `awk '{print $0}'`

Moderator's Comments:
Mod Comment Please use CODE tags (not ICODE tags) for full-line and multi-line sample input, output, and code segments.
This might work on some systems using some shells, but there are a few problems:
  1. Using find ./* ... can give you argument list too long errors when invoked in directories containing lots of files and/or several files with long filenames. It would be better to use find . ....
  2. If there is a file named source_file.txt in more than one directory in the file hierarchy rooted in the current working directory, you will be invoking cd with two or more operands instead of the single operand you're assuming will be specified.
  3. The standards say that commands in a pipeline can be executed in a subshell environment. If cd is executed in a subshell environment, the current working directory of the shell execution environment you will be in when the pipeline completes will not have been changed.
If the intent is to move into the directory containing a file named example.txt and do something in that directory for each directory that contains a file by that name, something more like:
Code:
#!/bin/ksh
spot=$PWD
find . -name 'example.txt' | while read -r path
do	file=${path##*/}
	dir=${path%/*}
	if ! cd "$dir"
	then	continue
	fi
	printf 'Process file "%s" in directory "%s"\n' "$file" "$dir"
	# do whatever you want to do in this directory...
	cd "$spot"	# return to the directory in which we started
done

This was written and tested using a Korn shell, but this should work with any shell that performs the basic parameter expansions specified in the POSIX standards.
 

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