blocks|key|35781|text|响应上的string方法将读取输入(网络)流并将其转换为字符串。因此，它动态构建字符串并将其返回给您。第二次调用时，网络流已经被消耗，不再可用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|35782|您应该将string的结果保存到一个字符串变量中，然后根据需要多次访问它。|35783|entityMap^0|4|6|0|4|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|P|8|@]|D|@]|E|$]]]|I|$]]

The <code>string</code> method on the response will read the input (network) stream and convert it into a string. So it dynamically builds the string and returns it to you. The second time you call it, the network stream has already been consumed and is no longer available. 

You should save the result of <code>string</code> into a String variable, and then access it as many times as needed.

blocks|key|35814|text|顺便说一句，除了Greg+Ennis的回答之外，我还可以说出有一次当我在监视窗口中忘记了response.body().string()时发生了这种情况。因此，在调试器下，主体将读取到手表中，然后关闭网络流。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|35815|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|$]]

BTW in addition to Greg Ennis' answer I can tell what onetime this happened to me when I forgot response.body().string() in the watch window. So under the debugger the body was reading into the watch and the network stream was closed after that.

blocks|key|35831|text|对user2011622和Greg+Ennis的回答进行了一点扩展，我创建了一种方法，可以帮助您创建身体的完整克隆，允许您单独使用身体的每个副本。我自己在Retrofit2中使用了这种方法|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|35832|/**
+*+Clones+a+raw+buffer+so+as+not+to+consume+the+original
+*+@param+rawResponse+the+original+{@link+okhttp3.Response}+as+returned
+*++++++++++++++++++++by+{@link+Response#raw()}
+*+@return+a+cloned+{@link+ResponseBody}
+*/
private+ResponseBody+cloneResponseBody(okhttp3.Response+rawResponse)+{
++++final+ResponseBody+responseBody+=+rawResponse.body();
++++final+Buffer+bufferClone+=+responseBody.source().buffer().clone();
++++return+ResponseBody.create(responseBody.contentType(),+responseBody.contentLength(),+bufferClone);
}|code-block|syntax|javascript|35833|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/33862068/984830|1|https://stackoverflow.com/a/27922818/984830^0|1|B|0|D|A|1|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|10|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]|Q|$5|L|M|N|C|$O|R]]]]

Expanding a little on the answers of <a href="https://stackoverflow.com/a/33862068/984830">user2011622</a> and <a href="https://stackoverflow.com/a/27922818/984830">Greg Ennis</a>, I created a method which helps you to create a complete clone of the body, allowing you to consume each copy of the body separately. I use this method myself with Retrofit2

<pre><code>/**
 * Clones a raw buffer so as not to consume the original
 * @param rawResponse the original {@link okhttp3.Response} as returned
 * by {@link Response#raw()}
 * @return a cloned {@link ResponseBody}
 */
private ResponseBody cloneResponseBody(okhttp3.Response rawResponse) {
 final ResponseBody responseBody = rawResponse.body();
 final Buffer bufferClone = responseBody.source().buffer().clone();
 return ResponseBody.create(responseBody.contentType(), responseBody.contentLength(), bufferClone);
}
</code></pre>

blocks|key|35841|text|您可以调用response.peekBody(Long.MAX_VALUE);在内存中缓冲整个响应，并获得它的轻量级副本。这样就不会浪费太多了。See+this+issue+on+GitHub。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|35842|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/square/okhttp/issues/2869^0|5|Y|20|O|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

You can call <code>response.peekBody(Long.MAX_VALUE);</code> to buffer the entire response in memory and get a lightweight copy of it. It'll be much less wasteful. <a href="https://github.com/square/okhttp/issues/2869" rel="nofollow noreferrer">See this issue on GitHub</a>.

blocks|key|4543126|text|ResponseBody+body+=+response.peekBody(Long.MAX_VALUE);
String+content+=+body.string();
//do+something|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4543127|这段代码获取响应体，不会消耗缓冲区。这是this+issue中新增的接口|unstyled|offset|length|4543128|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/square/okhttp/issues/1740^0|0|K|A|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>ResponseBody body = response.peekBody(Long.MAX_VALUE);
String content = body.string();
//do something
</code></pre>

this code get the response body and won't consume the buffer. it is a new api added in <a href="https://github.com/square/okhttp/issues/1740" rel="noreferrer">this issue</a>

I implement my http calls via the OkHttp library. Everything works fine, but I noticed that, when I access the body as a string of the response twice an <code>IllegalStateException</code> will be thrown. 
That is, I do (for example): <code>Log.d("TAG", response.body().string())</code> and after that I actually want to use that string like <code>processResponse(response.body().string())</code>. But that second call throws the exception with the message <code>closed</code>. 

How can it be possible that accessing a string twice results in a failure? I want to process that response without the need to add a wrapper/dummy object just for saving some values (like header, body, statuscode).

Accessing body string of an OkHttp Response twice results in IllegalStateException: closed

调试器

我通过OkHttp库实现http调用。一切正常，但我注意到，当我两次以响应字符串的形式访问正文时，将抛出一个IllegalStateException。也就是说，我做了(例如)：Log.d("TAG", response.body().string())，然后我实际上想要使用像processResponse(respo...

问访问OkHttp响应的正文字符串两次会产生IllegalStateException: closed
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