bash - Running multiple commands with xargs

Question

cat a.txt | xargs -I % echo %

In the example above, xargs takes echo % as the command argument. But in some cases, I need multiple commands to process the argument instead of one. For example:

cat a.txt | xargs -I % {command1; command2; ... }

But xargs doesn't accept this form. One solution I know is that I can define a function to wrap the commands, but I want to avoid that because it is complex. Is there a better solution?

Answer 1

cat a.txt | xargs -d $'
' sh -c 'for arg do command1 "$arg"; command2 "$arg"; ...; done' _

...or, without a Useless Use Of cat:

<a.txt xargs -d $'
' sh -c 'for arg do command1 "$arg"; command2 "$arg"; ...; done' _

---

To explain some of the finer points:

• The use of "$arg" instead of % (and the absence of -I in the xargs command line) is for security reasons: Passing data on sh's command-line argument list instead of substituting it into code prevents content that data might contain (such as $(rm -rf ~), to take a particularly malicious example) from being executed as code.

• Similarly, the use of -d $' ' is a GNU extension which causes xargs to treat each line of the input file as a separate data item. Either this or -0 (which expects NULs instead of newlines) is necessary to prevent xargs from trying to apply shell-like (but not quite shell-compatible) parsing to the stream it reads. (If you don't have GNU xargs, you can use tr ' ' '' <a.txt | xargs -0 ... to get line-oriented reading without -d).

• The _ is a placeholder for $0, such that other data values added by xargs become $1 and onward, which happens to be the default set of values a for loop iterates over.