bash grep 命令静默响应
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grep command response silently
提问by Nimjox
I'm new to linux shell and am trying to do this, preferably in one line, with the following condition: It can't output anything to the terminal.
我是 linux shell 的新手，我正在尝试这样做，最好在一行中，条件如下：它无法向终端输出任何内容。
/var/folder/program.exe -L parameters | grep text_to_filter && echo SomeText >'/tmp/Log.txt'
The problem is the .exe spits out XML data to terminal. I can't figure out how to grep it and not have the shell output it. If I use /dev/null 2>&1, it pipes it quite but then I can't grep the data. Any idea's?
问题是 .exe 将 XML 数据输出到终端。我不知道如何 grep 并且不让 shell 输出它。如果我使用/dev/null 2>&1，它会很好地传输它，但是我无法 grep 数据。有任何想法吗？
回答by anubhava
Use grep -q(quiet)
使用grep -q（安静）
/var/folder/program.exe -L parameters | grep -q "text_to_filter" && echo 'SomeText' > '/tmp/Log.txt'
As per man grep:
根据man grep：
-q, --quiet, --silentQuiet; do not write anything to standard output. Exit immediately with zero status if any match is found, even if an error was detected. Also see the -s or --no-messages option.
-q, --quiet, --silent安静的; 不要向标准输出写入任何内容。如果找到任何匹配项，则立即以零状态退出，即使检测到错误也是如此。另请参阅 -s 或 --no-messages 选项。
回答by damienfrancois
Try using |& rather than just |. (needs bash 4)
尝试使用 |& 而不仅仅是 |。（需要 bash 4）
