blocks|key|444117|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|444118|，这是因为你的JSONObject格式是错误的。参见How+to+convert+String+to+JSONObject+in+Java。只是"results"不是JSON。试着这样做：|unordered-list-item|offset|length|style|CODE|444119|444120|{"result":"blahblah"}|444121|Or错误地，您在编写代码时在result中包含了双引号|444122|444123|JSONObject+responseObject+=+new+JSONObject("results");|444124|由于result已经是一个字符串，请尝试将该行替换为：|444125|JSONObject+responseObject+=+new+JSONObject(results);|444126|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/5245840/how-to-convert-string-to-jsonobject-in-java^0|0|7|A|20|9|2B|4|Q|17|0|0|0|0|L|0|E|6|0|0|0|1I|0|2|6|0|0|1G|0^^$0|@$1|2|3|-4|4|5|6|11|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|12|7|@$D|13|E|14|F|G]|$D|15|E|16|F|G]|$D|17|E|18|F|G]]|8|@$D|19|E|1A|1|1B]]|9|$]]|$1|H|3|-4|4|5|6|1C|7|@]|8|@]|9|$]]|$1|I|3|J|4|5|6|1D|7|@$D|1E|E|1F|F|G]]|8|@]|9|$]]|$1|K|3|L|4|C|6|1G|7|@$D|1H|E|1I|F|G]]|8|@]|9|$]]|$1|M|3|-4|4|5|6|1J|7|@]|8|@]|9|$]]|$1|N|3|O|4|5|6|1K|7|@$D|1L|E|1M|F|G]]|8|@]|9|$]]|$1|P|3|Q|4|5|6|1N|7|@$D|1O|E|1P|F|G]]|8|@]|9|$]]|$1|R|3|S|4|5|6|1Q|7|@$D|1R|E|1S|F|G]]|8|@]|9|$]]|$1|T|3|-4|4|5|6|1T|7|@]|8|@]|9|$]]]|U|$V|$4|W|X|Y|9|$Z|10]]]]

<ul>
<li>That's because your format for <code>JSONObject</code> is wrong. See
<a href="https://stackoverflow.com/questions/5245840/how-to-convert-string-to-jsonobject-in-java">How to convert String to JSONObject in Java</a>
. Just <code>"results"</code> is not a <code>JSON</code>. Try and have something like:

<code>{"result":"blahblah"}</code></li>
<li>Or by mistake you have included double quotes in the <code>result</code> while
writing 

<code>JSONObject responseObject = new JSONObject("results");</code>

As <code>result</code> is already a string, try replacing the line with:

<code>JSONObject responseObject = new JSONObject(results);</code></li>
</ul>

blocks|key|5591104|text|++++String+obj=JSONObject.quote(YourData);
++++JSONArray+lArray=new+JSONArray(obj);

++++//+or+simply++Delete+the+prefix+'results'++from+your+php+Code
++++//+$res2=array("results"=>$response);
++++//+and+you+will+retrive+directelly+your+JsonArray+like+

++++JSONArray+lArray=new+JSONArray(YouData);|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5591105|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> String obj=JSONObject.quote(YourData);
 JSONArray lArray=new JSONArray(obj);

 // or simply Delete the prefix 'results' from your php Code
 // $res2=array("results"=&gt;$response);
 // and you will retrive directelly your JsonArray like 

 JSONArray lArray=new JSONArray(YouData);
</code></pre>

blocks|key|444132|text|看起来你把这两行弄混了，这样试试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|444133|//turn+the+result+into+a+JSON+object
JSONObject+responseObject+=+new+JSONObject(result);

//+get+the+JSON+array+named+"results"
JSONArray+resultsArray+=+responseObject.getJSONArray("results");|code-block|syntax|javascript|444134|这是假设你从某个地方得到的JSON响应包含一个结果，而这个结果又包含了一个‘JSONObject’。|444135|从您的代码示例中的注释判断，情况就是这样，这只是一个由于相似命名而导致的简单混淆。|444136|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Looks like you just mixed up these two lines, try it this way:

<pre><code>//turn the result into a JSON object
JSONObject responseObject = new JSONObject(result);

// get the JSON array named "results"
JSONArray resultsArray = responseObject.getJSONArray("results");
</code></pre>

This is assuming the JSON response you get from somewhere contains a JSONObject, which in turn contains a JSONArray in 'results'.

Judging by the comments in your code sample, this is the case, it just is a case of a simple mixup due to similar naming.

blocks|key|5591189|text|检查JSON字符串本身是否以[例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5591190|[{"hello":"goodbye","name":"bob","age":26}]|code-block|syntax|javascript|5591191|如果是，这意味着它不是一个JSONObject，而是一个JSONArray。试着改变|5591192|JSONObject+responseObject+=+new+JSONObject("results");

++++++++//+get+the+JSON+array+named+"results"
JSONArray+resultsArray+=+responseObject.getJSONArray(result);|5591193|类似这样的东西|5591194|JSONArray+responseArray+=+new+JSONArray("results");
JSONObject+resonseObject+=+responseArray.toJSONObject(responseArray);|5591195|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Check if the JSON string itself begins with [
e.g.

<pre><code>[{"hello":"goodbye","name":"bob","age":26}]
</code></pre>

if it does this means it is not a JSONObject but a JSONArray.
Try changing

<pre><code>JSONObject responseObject = new JSONObject("results");

 // get the JSON array named "results"
JSONArray resultsArray = responseObject.getJSONArray(result);
</code></pre>

to something like

<pre><code>JSONArray responseArray = new JSONArray("results");
JSONObject resonseObject = responseArray.toJSONObject(responseArray);
</code></pre>

I'm having problem with string cannot convert to JSONObject.Anyone could help in solving this problem? Thanks and very appreciate for helping.

<pre><code>protected void onPostExecute(String result) { 
if (result==null || result.length()==0){
 // no result:
 return;
}

//clear the list
moviesList.clear();

try {
 //turn the result into a JSON object
 JSONObject responseObject = new JSONObject("results");

 // get the JSON array named "results"
 JSONArray resultsArray = responseObject.getJSONArray(result);

 // Iterate over the JSON array: 
 for (int i = 0; i &lt; resultsArray.length(); i++) {
 // the JSON object in position i 
 JSONObject messageObject = resultsArray.getJSONObject(i);

 // get the primitive values in the object
 String title = messageObject.getString("title");
 String details = messageObject.getString("synopsis");

 //put into the list:
 Movie movie = new Movie(title, details, null,null);
 moviesList.add(movie);
 }
 } catch (JSONException e) {
 e.printStackTrace();
 }

 //refresh listView:
 adapter.notifyDataSetChanged();

 }
} 
</code></pre>

result has value

the error is in the following line : 

<pre><code>JSONObject responseObject = new JSONObject("results");
</code></pre>

Value results of type java.lang.String cannot be converted to JSONObject

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。

页面排版不美观，阅读体验差。

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我遇到了字符串无法转换为JSONObject.Anyone的问题，可以帮助解决这个问题吗？谢谢，非常感谢你的帮助。protected void onPostExecute(String result) {           if (result==null || result.length()==0){       ...

问java.lang.String类型的值结果不能转换为JSONObject
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