![]() ![]() The version I find most useful is this: find $(pwd) -name \*.mp3 > FullPlaylist. type f -name 'style' Output Now lets say we want to find files with a particular extension like. type d grep -E 'dir1dir2' xargs ls find. How to search files by name or extension Suppose we need to find files that contain 'style' in their name. If alphabetical order is needed, the find output may be piped through sort: find -maxdepth 1 -name \*.txt -printf "%f\n" | sort > Out_file.txt I know you specified that using find, but only to show other options that can be used, you can use xargs: find. This output will be in the same form as many of the other answers here, but won't necessarily be sorted in alphabetical order. option with find command to search filename using a pattern. name 'abc' will list the files that are exact match. If full recursion isn't wanted (or if "limited recursion" is needed), the maxdepth option is available: find -maxdepth 1 -name \*.txt -printf "%f\n" > Out_file.txt The above command will search the file that starts with abc under the current working directory. The output will be of the form filename.txt, one file per line. If only the bare filename is wanted in the output (no directory specs), find ( most versions) has a built-in printf option: find -name \*.txt -printf "%f\n" > Out_file.txt txt files at or below the user's current dir, and redirect the list to Out_file.txt: find -name \*.txt > Out_file.txt Recursion and the relative file specification are "free" with find, so this command will gather a list of all. ![]() Find strikes me as a good choice for the OP's question, and for many similar objectives: The basic syntax for the find command looks like this: find starting location criteria options action to take The starting location can be a directory name (e.g., /var/log), the. ![]()
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