blocks|key|3325825|text|如果discountCode.length+>+2不为true，则不会返回值，因此将返回undefined。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3325826|3325827|我还要补充的是，如果你把它放在另一个函数alert(checkDiscountCode())中，并且你关闭了firebug，或者在另一个没有打开控制台的浏览器中打开它，如果你仍然在代码中保留console.log(...)行，你的浏览器将会遇到javascript错误。|3325828|必须打开控制台窗口才能登录到该窗口。|3325829|entityMap^0|2|N|0|0|K|Q|2N|G|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|P|8|@]|D|@]|E|$]]|$1|G|3|H|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|I|3|J|5|6|7|V|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|L|$]]

If <code>discountCode.length &gt; 2</code> is not true, you aren't returning a value, thus it will return undefined.

<hr>

I'll also add that if you did put it in another function <code>alert(checkDiscountCode())</code> and you closed firebug up, or opened in another browser that doesn't have console open, your browser will encounter a javascript error, if you left the <code>console.log(...)</code> line still in your code.

A console window has to be open in order to log to it.

blocks|key|3325848|text|您的checkDiscountCode()函数没有return语句，因此它默认返回undefined。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3325849|您所包含的return语句位于作为$.getJSON()回调传递的匿名函数中，因此它从该函数返回给调用它的jQuery代码，而不是返回给您的代码(即使这样，如果message不是'OK!'，它也会返回undefined+)。(此外，除非使用discountCode.length+>+2，否则您的$.getJSON()不会运行，尽管您的checkDiscountCode()函数不会尝试以这两种方式返回值。)|3325850|entityMap^0|2|J|P|6|15|9|0|5|6|H|B|28|7|2H|5|3C|N|44|B|4O|J|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|I|$]]

Your <code>checkDiscountCode()</code> function doesn't have a <code>return</code> statement, so it returns <code>undefined</code> by default.

The <code>return</code> statement that you've included is inside the anonymous function passed as a callback for <code>$.getJSON()</code>, so it returns from that function back to the jQuery code that called it, not back to your code (and even then it returns undefined if <code>message</code> is not <code>'OK!'</code>). (Also your <code>$.getJSON()</code> doesn't run unless <code>discountCode.length &gt; 2</code>, though your <code>checkDiscountCode()</code> function doesn't try to return a value either way.)

blocks|key|2282467|text|您正在从传递给异步工作的函数的回调中返回。您必须调整代码以接受回调，并更改调用方以处理异步回调这一事实。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2282468|这意味着，而不是这样：|2282469|//+won't+work
function+checkSomething()+{
++++someAsynchronousFunction(function(result)+{
++++++++return+result+==+"success";
++++});
}
if(checkSomething())+{
++++alert("It+worked.");
}else{
++++alert("It+didn't+work.");
}|code-block|syntax|javascript|2282470|您必须将您的代码更改为如下所示：|2282471|function+checkSomething(callback)+{
++++someAsynchronousFunction(function(result)+{
++++++++callback(result+==+"success");
++++});
}
checkSomething(function(result)+{
++++if(result)+{
++++++++alert("It+worked.");
++++}else{
++++++++alert("It+didn't+work.");
++++}
});|2282472|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You're returning from a callback passed to a function that does work asynchronously. You'll have to adapt the code to take a callback and change the caller to deal with the fact that it's asynchronous.

That means that rather than this:

<pre><code>// won't work
function checkSomething() {
 someAsynchronousFunction(function(result) {
 return result == "success";
 });
}
if(checkSomething()) {
 alert("It worked.");
}else{
 alert("It didn't work.");
}
</code></pre>

You'll have to change your code to look like this:

<pre><code>function checkSomething(callback) {
 someAsynchronousFunction(function(result) {
 callback(result == "success");
 });
}
checkSomething(function(result) {
 if(result) {
 alert("It worked.");
 }else{
 alert("It didn't work.");
 }
});
</code></pre>

blocks|key|3325907|text|JQuery的getJSON()方法异步运行。您应该向checkDiscountCode()添加一个回调函数；该回调函数将运行应用程序逻辑的其余部分。例如：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3325908|function+sampleCallback(outputData)+{
++console.log('outputData+=+'+%2B+outputData);
}

function+checkDiscountCode(callback)+{
++$.getJSON(+'calc.php?code='+%2B+discountCode,+function(+json+)+{
++++//+get+output+data,+and+then+call+callback.
++++callback(outputData);
++});
}

checkDiscountCode(sampleCallback);|code-block|syntax|javascript|3325909|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|$]]

JQuery's getJSON() method runs asynchronously. You should add a callback function to checkDiscountCode(); the callback function will run the rest of your app's logic. For example:

<pre><code>function sampleCallback(outputData) {
 console.log('outputData = ' + outputData);
}

function checkDiscountCode(callback) {
 $.getJSON( 'calc.php?code=' + discountCode, function( json ) {
 // get output data, and then call callback.
 callback(outputData);
 });
}

checkDiscountCode(sampleCallback);
</code></pre>

blocks|key|370792|text|如果为discountCode.length+>+2，您将调用$.getJSON函数，但这将接受一个回调，并且不会立即返回任何内容。因此，在这两种情况下，您都会得到undefined。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|370793|entityMap^0|3|N|V|9|2A|9|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|O|8|@]|D|@]|E|$]]]|G|$]]

And if <code>discountCode.length &gt; 2</code> you are invoking the <code>$.getJSON</code> function, but that takes a callback, and returns nothing immediately. Hence, in either case, you get <code>undefined</code>.

I got a wierd issue with a javascript script.

This is my code:

<pre><code>function checkDiscountCode()
{

 var discountCode = $('#korcode').val();
 var errorMessage = $('#errorMessage').text();

 if(discountCode.length &gt; 2)
 {

 $.getJSON( 'calc.php?code=' + discountCode, function( json ) 
 { 
 var color = json.color; 
 var soort = json.soort;
 var waarde = json.waarde; 
 var message = json.message;

 if(errorMessage != message)
 {
 $('#error').html('&lt;font color="' + color + '" id="errorMessage"&gt;' + message + '&lt;/font&gt;');
 }

 if(message == 'OK!')
 {
 var outputData = soort + '-' + waarde;
 console.log('outputData = ' + outputData);
 return outputData;
 }

 }); 
 } 
}
</code></pre>

The console log give the right data, for example 0-12.00 but when I request this function in another function, for example just a alert(checkDiscountCode()); it says undefined..

Can anyone please help me? I don't see the issue..

Thanks in advance!

Javascript return doesn't work? returns undefined

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

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我在javascript脚本上遇到了一个奇怪的问题。这是我的代码：function checkDiscountCode(){  var discountCode = $('#korcode').val();  var errorMessage = $('#errorMessage').text();  if(disco...

问Javascript返回不起作用吗？返回未定义的
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