blocks|key|1738581|text|不是，因为新的throw不是直接在try块中。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1738582|entityMap^0|7|5|H|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

No, since the new <code>throw</code> is not in the <code>try</code> block directly.

blocks|key|1523586|text|不--正如Chris+Jester-Young所说，它将被抛出到层次结构中的下一个try-catch。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1523587|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No -- As Chris Jester-Young said, it will be thrown up to the next try-catch in the hierarchy.

blocks|key|1523666|text|如上所述。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1523667|我要补充的是，如果你很难看到发生了什么，如果你不能在调试器中重现这个问题，你可以在重新抛出新的异常之前添加一个跟踪(使用好的旧System.out.println更糟糕，使用像log4j这样的好的日志系统)。|1523668|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

As said above... 
I would add that if you have trouble seeing what is going on, if you can't reproduce the issue in the debugger, you can add a trace before re-throwing the new exception (with the good old System.out.println at worse, with a good log system like log4j otherwise).

blocks|key|1523673|text|它不会被第二个catch块捕获。只有在try块中才能捕获每个异常。你可以嵌套尝试(通常这不是一个好主意)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1523674|try+{
++++doSomething();
}+catch+(IOException)+{
+++try+{
+++++++doSomething();
+++}+catch+(IOException+e)+{
+++++++throw+new+ApplicationException("Failed+twice+at+doSomething"+%2B
+++++++e.toString());
+++}++++++++++
}+catch+(Exception+e)+{
}|code-block|syntax|javascript|1523675|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

It won't be caught by the second catch block. Each Exception is caught only when inside a try block. You can nest tries though (not that it's a good idea generally):

<pre><code>try {
 doSomething();
} catch (IOException) {
 try {
 doSomething();
 } catch (IOException e) {
 throw new ApplicationException("Failed twice at doSomething" +
 e.toString());
 } 
} catch (Exception e) {
}
</code></pre>

blocks|key|1738746|text|不会，因为捕获都引用相同的try块，所以从catch块中抛出的异常将被封闭的try块捕获(可能在调用此块的方法中)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738747|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No, since the catches all refer to the same try block, so throwing from within a catch block would be caught by an enclosing try block (probably in the method that called this one)

blocks|key|1738767|text|如果你想从catch块抛出一个异常，你必须通知你的方法/类等，它需要抛出这个异常。如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738768|public+void+doStuff()+throws+MyException+{
++++try+{
++++++++//Stuff
++++}+catch(StuffException+e)+{
++++++++throw+new+MyException();
++++}
}|code-block|syntax|javascript|1738769|现在你的编译器不会对你大喊大叫了:)|1738770|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you want to throw an exception from the catch block you must inform your method/class/etc. that it needs to throw said exception. Like so:

<pre><code>public void doStuff() throws MyException {
 try {
 //Stuff
 } catch(StuffException e) {
 throw new MyException();
 }
}
</code></pre>

And now your compiler will not yell at you :)

blocks|key|1523779|text|旧post，但"e“变量必须是唯一的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1523780|try+{
++//+Do+something
}+catch(IOException+ioE)+{
++throw+new+ApplicationException("Problem+connecting+to+server");
}+catch(Exception+e)+{
++//+Will+the+ApplicationException+be+caught+here?
}|code-block|syntax|javascript|1523781|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Old post but "e" variable must be unique:

<pre><code>try {
 // Do something
} catch(IOException ioE) {
 throw new ApplicationException("Problem connecting to server");
} catch(Exception e) {
 // Will the ApplicationException be caught here?
}
</code></pre>

This may seem like a programming 101 question and I had thought I knew the answer but now find myself needing to double check. In this piece of code below, will the exception thrown in the first catch block then be caught by the general Exception catch block below?

<pre><code>try {
 // Do something
} catch(IOException e) {
 throw new ApplicationException("Problem connecting to server");
} catch(Exception e) {
 // Will the ApplicationException be caught here?
}
</code></pre>

I always thought the answer would be no, but now I have some odd behaviour that could be caused by this. The answer is probably the same for most languages but I'm working in Java.

Exception thrown inside catch block - will it be caught again?

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Java

这可能看起来像是一个编程101的问题，我曾以为我知道答案，但现在发现自己需要再次检查。在下面的这段代码中，第一个catch块中抛出的异常会被下面的通用异常catch块捕获吗？try {  // Do something} catch(IOException e) {  throw new ApplicationExce...

问在catch块中抛出异常-它会被再次捕获吗？
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