测试来自给定模式的文件是否存在于 bash 中
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/15907914/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
test if file(s) from a given pattern exists in bash
提问by RSFalcon7
How do I test if a file (or many) named from given pattern exists and do stuff based on that test?
如何测试从给定模式命名的文件（或多个）是否存在并根据该测试执行操作？
This is my failed attempt:
这是我失败的尝试：
[[ { ls ../outputListWorkerPid_* | wc -l } -ge "1" ]] && echo "ARRRRRR" || && echo "FAIL"
But I got the following error:
但我收到以下错误：
bash: conditional binary operator expected
bash: syntax error near `ls'
Also it would be nice to avoid the No such file or directoryfrom lsin case the doesn't exists.
如果No such file or directoryfromls不存在，也可以避免使用from 。
回答by Matt W
Use an if-statement:
使用 if 语句：
if [ "$(ls *.txt | wc -l)" -ge "1" ];
then
echo "Yes";
else
echo "No";
fi;
Does that answer your question?
这是否回答你的问题？
