按请求okhttp newbuilder
尝试以正确的方式实现OKHTTP。我了解必须共享OkhttpClient（Singleton），但是我不清楚.newBuilder（）;
示例代码：
// Instantiated once
private static OkHttpClient client = new OkHttpClient.Builder()
.readTimeout(readTime, TimeUnit.MILLISECONDS)
.connectionPool(new ConnectionPool(200, connectTimeout, TimeUnit.MILLISECONDS));
.build();
public static String makeRestCall(String url, String data, Interceptor customInterceptor) {
// Questions on the line below
OkHttpClient newClient = client.newBuilder()
.addInterceptor(customInterceptor)
.build();
....
try (Response response = newClient.newCall(httpRequest).execute()) {
final ResponseBody body = response.body();
return body.string();
}
return "NO_DATA";
}
有一些问题
- 我在
.newbuilder（）时 > newclient ，原始
客户端是否也通过参考更新？ 类调用
MakerestCall确定他们需要的自定义感应器。可以打电话.newbuilder（）为每个请求吗？
我一直在搜索文档并播放实施，但对上述情况并没有清楚。
任何援助/指针都将不胜感激。
Trying to implement okhttp the correct way. I understand the OkHttpClient must be shared (Singleton), however I am not clearly understanding .newBuilder();
Sample Code:
// Instantiated once
private static OkHttpClient client = new OkHttpClient.Builder()
.readTimeout(readTime, TimeUnit.MILLISECONDS)
.connectionPool(new ConnectionPool(200, connectTimeout, TimeUnit.MILLISECONDS));
.build();
public static String makeRestCall(String url, String data, Interceptor customInterceptor) {
// Questions on the line below
OkHttpClient newClient = client.newBuilder()
.addInterceptor(customInterceptor)
.build();
....
try (Response response = newClient.newCall(httpRequest).execute()) {
final ResponseBody body = response.body();
return body.string();
}
return "NO_DATA";
}
I have a few questions around .newBuilder()
When we add a new interceptor to
newClient, does the originalclientalso get updated by reference?Classes calling
makeRestCalldecide on what customInteceptor they need. Is it ok to call.newBuilder()for every request?
I have been searching the documentation and playing with the implementation but haven't had clarity on the above.
Any assistance/pointers are appreciated.
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不，原件没有变化。它的配置是不可变的。
绝对地。该操作很便宜，因为它仅复制配置。没有复制资源密集型的内容，例如连接池和高速缓存。
No, the original is unchanged. Its configuration is immutable.
Absolutely. That operation is cheap because it only duplicates the configuration. Resource-intensive stuff like the connection pool and cache are not duplicated.