bash:将整个命令(带参数)传递给函数
发布时间:2020-12-15 17:01:22 所属栏目:安全 来源:网络整理
导读:我本质上是试图实现一个函数,它断言另一个命令的失败(非零退出代码),并在失败时打印一条消息. 这是我的功能: function assert_fail () { COMMAND=$@ if [ `$COMMAND; echo $?` -ne 0 ]; then echo "$COMMAND failed as expected." else echo "$COMMAND didn
我本质上是试图实现一个函数,它断言另一个命令的失败(非零退出代码),并在失败时打印一条消息.
这是我的功能: function assert_fail () { COMMAND=$@ if [ `$COMMAND; echo $?` -ne 0 ]; then echo "$COMMAND failed as expected." else echo "$COMMAND didn't fail" fi } # This works as expected assert_fail rm nonexistent # This works too assert_fail rm nonexistent nonexistent2 # This one doesn't work assert_fail rm -f nonexixtent 只要我向命令添加选项,它就不起作用.以下是上述输出: rm: cannot remove `nonexistent': No such file or directory rm nonexistent failed as expected. rm: cannot remove `nonexistent': No such file or directory rm: cannot remove `nonexistent2': No such file or directory rm nonexistent nonexistent2 failed as expected. rm -f nonexistent didn't fail 我试过在命令周围加双引号,但没有用.我希望上面的第三次调用产生与其他两种相似的输出. 我感谢任何/所有的帮助!
rm -f永远不会在不存在的文件上失败.它与你的包装器无关.见男人:
OPTIONS -f,--force ignore nonexistent files,never prompt (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |