multiple files as argument in bash script

Question

I have a script that exports log files, usually I get between 1-3 separate files depending on the size of the log for a particular day. After the export I run a report generator to spit out an html document of those logs. My question is how to call the command that generates the reports depending on the number of files, I know I can use if statements and do:

./generateReport -i file1 -o output.html

./generateReport -i file1 file2 -o output.html

./generateReport -i file1 file2 file3 -o output.html

is there a way to loop over the files and include them as an input??

It would be more conventional to use -i file1 -i file2 -o output.html, or to remove -i altogether (file1 file2 -o output.html) — Charles Duffy
– Charles Duffy, Commented Sep 4, 2013 at 15:21
Use positional parameters rather than flags for multiple input files. — Todd A. Jacobs
– Todd A. Jacobs, Commented Sep 4, 2013 at 15:58
Was the question how to implement generateReport (to parse names off the command line when called through the given convention), or how to call generateReport? I read it as the former. — Charles Duffy
– Charles Duffy, Commented Mar 25, 2017 at 17:34

Emery Lapinski · Accepted Answer · 2013-09-04 15:45:40Z

3

Is it necessary to use flags?

./generateReport file1 file2 file3 > output.html

answered Sep 4, 2013 at 15:45

Emery Lapinski

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1 Comment

laitha0 Over a year ago

Yes it is, since I am using a tool to generate logs and it uses -i to start reading input files to it

Charles Duffy · Accepted Answer · 2013-09-04 15:23:53Z

2

The following will collect an array of inputFiles, and a single variable with the output file name:

inputFiles=( )
outputFile=
while (( $# )); do
  if [[ $1 = -o ]]; then
    outputFile=$2; shift
  elif [[ $1 = -i ]]; then
    inputFiles+=( "$2" ); shift
  else
    inputFiles+=( "$1" )
  fi
  shift
done

...then, you could do something like this:

# redirect stdout to the output file, if one was given
[[ $outputFile ]] && exec >"$outputFile"

# loop over the input files and process each one
for inputFile in "${inputFiles[@]}"; do
  process "$inputFile"
done

answered Sep 4, 2013 at 15:23

Charles Duffy

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Comments

svante · Accepted Answer · 2017-03-29 09:34:50Z

1

Try filename expansion :

./generateReport -i file? -o output.html

or using find for all log files created in the last 24 hours :

./generateReport -i `find . -name "file*" -mtime -1` -o output.html

edited Mar 29, 2017 at 9:34

answered Sep 4, 2013 at 16:24

svante

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1 Comment

Charles Duffy Over a year ago

The second one is buggy, if any of your filenames contain spaces in them. See also BashPitfalls #1, which applies.

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