Question

我正在尝试通过以下代码将JSON数据发送到Web服务器。由于某种原因，请求似乎没有出去。我错过了什么？ NSURLConnection（retStr）的结果也总是空的？

NSDictionary *data = [NSDictionary dictionaryWithObject:@"test sending ios" forKey:@"value1"];
NSError *error;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:data options:kNilOptions error:&error];

    NSURL *url = [NSURL URLWithString:@"http://webserveraddress"];
NSMutableURLRequest *req = [NSMutableURLRequest requestWithURL:url cachePolicy:nil timeoutInterval:60];
[req setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[req setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[req setValue:[NSString stringWithFormat:@"%d", [jsonData length]] forHTTPHeaderField:@"Content-Length"];
[req setHTTPMethod:@"POST"];
[req setHTTPBody:jsonData];
NSString *retStr = [[NSString alloc] initWithData:[NSURLConnection sendSynchronousRequest:req returningResponse:nil error:nil] encoding:NSUTF8StringEncoding];

Answer 1

要将后期变量中的简单数据发送到运行php的网络服务器，您只需在

中执行此操作

示例

NSString * key = [NSString stringWithFormat:@"var1=%@&var2=%@&var3=%@",@"var1String" ,@"var2string" ,[NSnumber numberWithBool:YES]]; NSURL * url = [NSURL URLWithString:@"http://webserver.com/yourScriptPHP.php"]; NSMutableURLRequest * request = [NSMutableURLRequest requestWithURL:url]; [request setHTTPMethod:@"POST"]; [request setHTTPBody:[key dataUsingEncoding:NSUTF8StringEncoding]]; [[NSURLConnection alloc] initWithRequest:request delegate:self]; // this is for you to be able to get your server answer. // you will need to make your class a delegate of NSURLConnectionDelegate and NSURLConnectionDataDelegate myClassPointerData = [[NSMutableData data] retain];

<强>实施

-(void)connection:(NSURLConnection *)connection didReceiveData:(NSData *)data { [myClassPointerData appendData:data] } -(void)connection:(NSURLConnection *)connection DidFinishLoading { // do what you want with myClassPointerData the data that your server did send you back here // for info on your server php script you just need to do: echo json_encode(array('var1'=> $var1, 'var2'=>$var2...)); // to get your server sending an answer }

Answer 2

一次检查

ASIFormDataRequest * request = [ASIFormDataRequest requestWithURL：[NSURL URLWithString：@“ur url”]];

[request setPostValue:appdelegate.userid forKey:@"userid"];
[request setPostValue:self.nameLbl.text forKey:@"username"];
[request setPostValue:self.website.text forKey:@"website"];

NSLog(@"~~~~~~ request~~~~~ %@",request);

[request setTimeOutSeconds:5000];
[request startSynchronous];

Answer 3

不是像JSON一样发送，而是通过普通的方式发送它 - 通过POST方法设置请求的HTTPBody。

如果您需要拨打电话，则需要进行区分 'http://abc.example.com/user_profile.json?user [FIRST_NAME] = fdffdf＆安培;用户[姓氏] = dffdf'

是，你需要让你的字典键有一个前缀。也就是说，'first_name = fdffdf'需要更改为'user [first_name] = fdffdf'。

尝试使用这段代码将参数字典更改为JSON所需的格式..

for (id key in [self allKeys]) {
NSString *newKey = [NSString stringWithFormat:@"%@[%@]", parent, key];
[result setObject:[self objectForKey:key] forKey:newKey];
}

在Xcode中使用NSURLConnection发送JSON数据

3 个答案: