我正在查找文件列表。
我如何将它输送到另一个实用程序,如cat,以便cat显示所有这些文件的内容?
然后,我将使用grep来搜索这些文件中的一些文本。
我正在查找文件列表。
我如何将它输送到另一个实用程序,如cat,以便cat显示所有这些文件的内容?
然后,我将使用grep来搜索这些文件中的一些文本。
当前回答
对我来说,听起来像是shell脚本的工作:
for file in 'find -name *.xml'
do
grep 'hello' file
done
或者类似的东西
其他回答
对我来说,听起来像是shell脚本的工作:
for file in 'find -name *.xml'
do
grep 'hello' file
done
或者类似的东西
我使用的方法是这样的:
find . -name <filename> -print0 | xargs -0 cat | grep <word2search4>
"find"的"-print0"参数和"xargs"的"-0"参数需要正确处理文件路径/名称中的空白。
Piping to another process (although this won't accomplish what you said you are trying to do): command1 | command2 This will send the output of command1 as the input of command2. -exec on a find (this will do what you want to do, but it's specific to find): find . -name '*.foo' -exec cat {} \; Everything between find and -exec are the find predicates you were already using. {} will substitute the particular file you found into the command (cat {} in this case); the \; is to end the -exec command. Send output of one process as command line arguments to another process: command2 `command1` For example: cat `find . -name '*.foo' -print` Note these are backquotes not regular quotes (they are under the tilde ~ on my keyboard). This will send the output of command1 into command2 as command line arguments. It's called command substitution. Note that file names containing spaces (newlines, etc) will be broken into separate arguments, though.
这对我很有用
find _CACHE_* | while read line; do
cat "$line" | grep "something"
done
下面是我找到包含一些我感兴趣的内容的文件名的方法,只是一个简单的bash行,很好地处理文件名中的空格:
find . -name \*.xml | while read i; do grep '<?xml' "$i" >/dev/null; [ $? == 0 ] && echo $i; done