Shell Programming and Scripting

I've got at least 30,000 XML files that I'm using the grep command to get their filename. Can I use the head command to grab just the beginning 8 lines and compare that instead of parsing the whole document? It would speed things up! or maybe grep -m?

If all of those files are in the current directory, then this might work:

Code:

for file in *.xml; do
  head -8 $file | grep <whatever you need to grep> >/dev/null && echo $file
done

If you're looking for a fixed string (rather than a match against a regular expression), use grep -F or fgrep (depending on which operating system you're using).

If you're trying to match the 1st few characters in a file and you have relatively large files and you have a lot of files that won't match, then using read or head to get the start of a file might make sense, but firing up grep and head for every input file is going to be slower than letting grep read the entire file unless the file you're processing is large.

If you show what you're doing right now, maybe we can speed it up.

I've got 30,000 files to process using

Code:

xmlFileNames=$(find . -name "*.xml" -exec grep -l "Status" {} \; 2>/dev/null)

The re is in the header of the xml files. Is there a way to grab just the header the pattern match instead of reading the whole files?

Try:

Code:

xmlFileNames=$(find . -name "*.xml" -exec grep -l "Status" {} '+' 2>/dev/null)

This should put more filenames into a single grep and make it more efficient.

It didn't work Corona688...