如何从文件集合中仅选择shell脚本

Question

I want to find all my bash scripts (i have accumulated many of them now) and run them all through bash -n automatically.

What's a quick way to do this? I'd like to have grep match only the files whose first non blank lines start with #!/bin/sh or #!/usr/bin/env sh or #!/usr/bin/env bash or #!/bin/bash ...

A serviceable answer is of course something along the lines of

for file in *; do 
    if head -n 5 $file | grep "#!.*sh$" > /dev/null; then
        bash -n $file
    fi
done

But for the sake of "correctness" how might I reasonably do my grep on only the first non-whitespace (or alternatively non-blank) line?

Answer 1

使用find：

     find . -type f -exec grep -e "^#\!\/bin\/.*sh$" {} +

Answer 2

With GNU awk

awk '
FNR==1 && $0~/#!\/bin\/sh|#!\/usr\/bin\/env sh|#!\/usr\/bin\/env bash|#!\/bin\/bash/ { 
    print FILENAME " is shell script";
}
FNR>1 {
    nextfile
}' *

• You can play around with the above regex and reduce it to just #! .
• FNR==1 along with the regex would ensure that the first line is checked for she-bang line.
• nextfile would ensure that no file is looked beyond first line.
• print if just for logging.
• FILENAME will print the name of the file under inspection.
• * will glob to all files under working directory.