首页 \ 问答 \ 在测试退出状态的上下文中调用时，set -e在函数中被忽略了吗？(set -e ignored in function when invoked in context where exit status is tested?)

在测试退出状态的上下文中调用时，set -e在函数中被忽略了吗？(set -e ignored in function when invoked in context where exit status is tested?)

 我试图退出shell函数，只要其中运行的任何命令都使用set -e失败  
foo () (
    set -e
    echo $SHELLOPTS
    false
    echo set -e failed
)

echo test1:
foo
test $? -ne 0 && echo died

echo
echo test2:
foo && echo died
 
 这是我使用bash 4.2.37获得的输出：  
test1:
braceexpand:errexit:hashall:interactive-comments
died

test2:
braceexpand:errexit:hashall:interactive-comments
set -e failed
died
 
 如果将foo称为简单命令（ test1 ），则foo在echo命令之前退出，正如我所期望的那样。 但是，如果在复合命令（ test2 ）中调用，则似乎忽略set -e并继续执行失败命令。  
 我知道除了管道的最后一个元素之外，除了-e故障之外的所有部分都会被忽略，但我不希望如果子管道作为管道的一部分运行，它的exiterr状态将被覆盖，特别是如果明确设置的话。 请注意， SHELLOPTS表示即使在第二次测试中也始终设置errexit 。  
 我错过了什么吗？  
 谢谢，  
 迪亚布 

I'm attempting to exit out of a shell function whenever any of the commands run in it fail using set -e 
foo () (
    set -e
    echo $SHELLOPTS
    false
    echo set -e failed
)

echo test1:
foo
test $? -ne 0 && echo died

echo
echo test2:
foo && echo died
 
Here's the output I get using bash 4.2.37: 
test1:
braceexpand:errexit:hashall:interactive-comments
died

test2:
braceexpand:errexit:hashall:interactive-comments
set -e failed
died
 
If foo is called as a simple command (test1), foo exits before the echo command, as I expect. However, if called in a compound command (test2), the set -e seems to be ignored and execution continues past the failing command. 
I know that with -e failures are ignored for all but the last element of a pipeline, but I don't expect that if a subshell is run as part of a pipeline that its exiterr status would be overridden especially if explicitly set. Note that SHELLOPTS indicates that errexit is always set, even in the second test. 
Am I missing something? 
Thanks, 
Diab

原文：https://stackoverflow.com/questions/35706525

更新时间：2021-11-26 22:11

最满意答案

 你可以使用.siblings()  
 
 
  
  $("li").on("click", function(){ 
  $(this).siblings().removeClass('active');
  $(this).addClass("active");
}); 
  li:hover{
  background-color: #eee;
  cursor: pointer;
}

.active{
  color: red;
} 
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul>
  <li>One</li>
  <li class="active">Two</li>
  <li>Three</li>
  <li>Four</li>
  <li>Five</li>
</ul> 
  
 

You can use .siblings() 
 
 
  
  $("li").on("click", function(){ 
  $(this).siblings().removeClass('active');
  $(this).addClass("active");
}); 
  li:hover{
  background-color: #eee;
  cursor: pointer;
}

.active{
  color: red;
} 
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<ul>
  <li>One</li>
  <li class="active">Two</li>
  <li>Three</li>
  <li>Four</li>
  <li>Five</li>
</ul>

在测试退出状态的上下文中调用时，set -e在函数中被忽略了吗？(set -e ignored in function when invoked in context where exit status is tested?)

最满意答案

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