Read in bash script and passing argument to script -

i have question. unfortunately, didn't find answer. how can pass arguments script result of command. example:

 ls | ./myscript.sh 

i want pass result of ls myscript. , if execute above command , script :

#!/bin/bash read arg in $@       grep $some $arg done 

then didn't wait read some variable , variable some gets value result passed ls command? how works ? want following: put script names of files (with command ls), after user enter string , want print every line contain entered word. didn't know why code above didn't work. in advance.

xargs utility use converting stdin input command-line arguments.

naively, following:

ls | xargs ./myscript.sh 

however, not work expected filenames have embedded spaces, split multiple arguments.
^{note ./myscript.sh $(ls) has same problem.}

if xargs implementation supports nonstandard -0 option parsing nul-separated input, can fix follows:

printf '%s\0' * | xargs -0 ./myscript.sh 

otherwise, use following, which, however, work if no filenames have embedded " characters (or embedded newlines):

printf '"%s" ' * | xargs ./myscript.sh 

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since stdin input comes filenames in case, can use find, has xargs functionality built in:

find . -type f -maxdepth 1 -exec ./myscript.sh {} + 

• note find . -type f -maxdepth 1 in essence same ls, there subtle differences, notably each matching filename prefixed ./ , filenames may not sorted.
• -exec ./myscript.sh {} + invokes script many filenames ({}) can fit on single command line (+) - xargs - typically all of them.

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note both xargs , find ... -exec ... + could result in multiple invocations of specified command, if not arguments fit on single command line.
however, given how long command lines allowed on modern platforms, happen - happens if have huge number of files in directory and/or names long.