bash 使用命令的输出作为下一个命令的输入
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Use the output of a command as input of the next command
提问by r2b2
So I call this PHP script from the command line:
所以我从命令行调用这个 PHP 脚本：
/usr/bin/php /var/www/bims/index.php "projects/output"
and its output is:
它的输出是：
file1 file2 file3
What I would like to do is get this output and feed to the "rm" command but I think im not doing it right:
我想做的是获取此输出并提供给“rm”命令，但我认为我做得不对：
/usr/bin/php /var/www/bims/index.php "projects/output" | rm
My goal is to delete whatever file names the PHP script outputs. What should be the proper way to do this?
我的目标是删除 PHP 脚本输出的任何文件名。这样做的正确方法应该是什么？
Thanks!
谢谢！
回答by Victor Sorokin
/usr/bin/php /var/www/bims/index.php "projects/output" | xargs rm
回答by Felix
Simplest solution:
最简单的解决方案：
rm `/usr/bin/php /var/www/bims/index.php "projects/output"`
What is between the backticks (`` ) is run and the output is passed as argument torm`.
反引号之间是什么 (`` ) is run and the output is passed as argument torm`.
回答by ghostdog74
you can try xargs
你可以试试 xargs
/usr/bin/php /var/www/bims/index.php "projects/output" | xargs rm
or just simply use a loop
或者只是简单地使用一个循环
/usr/bin/php /var/www/bims/index.php "projects/output" | while read -r out
do
rm $out
done
回答by TestBud
I guess this could help>>
我想这可能会有所帮助>>
grep -n "searchstring" filename | awk 'BEGIN { FS = " " };{print $1}'
grep -n "searchstring" 文件名 | awk 'BEGIN { FS = " " };{print $1}'
