Question

所以我试图通过POST将参数发送到php服务器，该参数值必须是JSON值。参数的名称必须是“数据”

我的意思是，get方法类似于http：url.com/url/something.php？data = {“jsonkey1”：1，“jsonkey2”：[....或类似的东西

这是我现在的代码，但我无法让它工作。我不想使用外部库，而是使用本机方法。

// Writing JSON...
NSNumber *imagesCount = [[NSNumber alloc] initWithInt:0];
NSMutableArray *arrayList = [[NSMutableArray alloc] init]; // That's an empty array
//[arrayList addObject:@"image1.png"]; [arrayList addObject:@"image2.png"];
NSDictionary *dictionaryJSON = [NSDictionary dictionaryWithObjectsAndKeys:imagesCount, @"count",arrayList,@"list", nil];

NSString *param = @"data=";
NSMutableData *dataJSON = [[NSMutableData alloc]init];
[dataJSON appendData:[param dataUsingEncoding:NSUTF8StringEncoding]];

NSError *error = nil;
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:dictionaryJSON options:NSJSONWritingPrettyPrinted error:&error];;
[dataJSON appendData:jsonData];

NSString* jsonString = [[NSString alloc] initWithBytes:[jsonData bytes] length:[jsonData length] encoding:NSUTF8StringEncoding];
NSLog(@"jsonData as string:\n%@, %@", jsonString, error);

jsonString = [jsonString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];

// Request to server
NSString *url =[NSString stringWithFormat:@"http://url.com/ios/api/get_images_cache.php?"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:url] cachePolicy:NSURLRequestUseProtocolCachePolicy timeoutInterval:60.0];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
[request setValue:[NSString stringWithFormat:@"%i", [dataJSON length]] forHTTPHeaderField:@"Content-Length"];


NSString* jsonStringdat = [[NSString alloc] initWithBytes:[dataJSON bytes] length:[dataJSON length] encoding:NSUTF8StringEncoding];
jsonStringdat=[jsonStringdat stringByReplacingOccurrencesOfString:@" " withString:@""];
    jsonStringdat=[jsonStringdat stringByReplacingOccurrencesOfString:@"\n" withString:@""];
NSLog(@"Final jsonData as string:\n%@, %@", jsonStringdat, error);
//[request setValue:jsonStringdat forHTTPHeaderField:@"data"];
//[request setValue:jsonData forKey:@"data"];
[request setHTTPBody: [jsonStringdat dataUsingEncoding:NSUTF8StringEncoding]];

感谢您的帮助。

修改

我终于做到了，问题在于这两行：

[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setValue:@"application/json" forHTTPHeaderField:@"Accept"];

我删除了它们，最后我开始工作了，谢谢！

Answer 1

NSURLConnection* connection = [NSURLConnection alloc] initWithRequest:request delegate:self startImmediately:YES];

将它添加到最后，然后你需要使你的类符合它，然后它可以知道请求何时成功完成以及何时失败并出现错误等。另外，我在我的构建请求的方式应用程序是：

NSMutableURLRequest* request = [[NSMutableURLRequest alloc] init];
[request setTimeoutInterval:10];
[request setURL:[NSURL URLWithString:url]];
[request setHTTPMethod:@"POST"];
[request setValue:postLength forHTTPHeaderField:@"Content-Length"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];

[request setHTTPBody:postData];

我希望有所帮助！

编辑：此外，在发送之前将您拥有的JSON转换为字符串。我就这样做了：

NSData *dataFromJSON = [NSJSONSerialization dataWithJSONObject:jsonObjects options:NSJSONWritingPrettyPrinted error:nil];
NSString *jsonString = [[NSString alloc] initWithData:dataFromJSON encoding:NSUTF8StringEncoding];

Answer 2

这是您的预期答案

第一个编码部分仅用于将数据发送到服务器

//Here YOUR URL
   NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"http://url.com/ios/api/get_images_cache.php?"]];


//create the Method "GET" or "POST"
   [request setHTTPMethod:@"POST"];

//Pass The String to server(YOU SHOULD GIVE YOUR PARAMETERS INSTEAD OF MY PARAMETERS)
  NSString *userUpdate =[NSString strin gWithFormat:@"user_email=%@&user_login=%@&user_pass=%@&  last_upd_by=%@&user_registered=%@&",txtemail.text,txtuser1.text,txtpass1.text,txtuser1.text,datestr,nil];



//Check The Value what we passed
  NSLog(@"the data Details is =%@", userUpdate);

//Convert the String to Data
  NSData *data1 = [userUpdate dataUsingEncoding:NSUTF8StringEncoding];

//Apply the data to the body
  [request setHTTPBody:data1];

//Create the response and Error
  NSError *err;
  NSURLResponse *response;

  NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];

  NSString *resSrt = [[NSString alloc]initWithData:responseData encoding:NSASCIIStringEncoding];

//This is for Response
  NSLog(@"got response==%@", resSrt);
 if(resSrt)
 {
   NSLog(@"got response");
   /* ViewController *view =[[ViewController alloc]initWithNibName:@"ViewController" bundle:NULL];
   [self presentViewController:view animated:YES completion:nil];*/
 }
 else
 {
   NSLog(@"faield to connect");
 }

第二部分是通过JSON获得回应

    //just give your URL instead of my URL

        NSMutableURLRequest *request=[NSMutableURLRequest requestWithURL:[NSURL  URLWithString:@"http://api.worldweatheronline.com/free/v1/search.ashx?query=London&num_of_results=3&format=json&key=xkq544hkar4m69qujdgujn7w"]];

       [request setHTTPMethod:@"GET"];

       [request setValue:@"application/json;charset=UTF-8" forHTTPHeaderField:@"content-type"];

        NSError *err;

        NSURLResponse *response;

        NSData *responseData = [NSURLConnection sendSynchronousRequest:request   returningResponse:&response error:&err];

      //Converting data to String for seeing response
        NSString *resSrt = [[NSString alloc]initWithData:responseData encoding:NSASCIIStringEncoding];

      //This is for Response
        NSLog(@"got response==%@", resSrt);

      //You need to check response.Once you get the response copy that and paste in ONLINE JSON VIEWER.If you do this clearly you can get the correct results.    

      //After that it depends upon the json format whether it is DICTIONARY or ARRAY 

        NSDictionary *jsonArray = [NSJSONSerialization JSONObjectWithData:responseData options: NSJSONReadingMutableContainers error: &err];

        NSArray *array=[[jsonArray objectForKey:@"search_api"]objectForKey:@"result"];

Answer 3

您可以在POST请求中使用application/x-www-form-urlencoded，您可以在其中发送许多参数（键/值对）。

请注意，必须正确编码参数键和值。特别是，JSON通常包含许多不能用作参数值的字符。

因此，给定一个代表您的JSON的对象 jsonRep ，它是NSArray或NSDictionary对象，您将得到JSON：

NSError* error;
NSData* json = [NSJSONSerialization dataWithJSONObject:jsonRep 
                                               options:0 
                                                 error:&error];

现在，您需要编码 json。所需的编码算法在此处定义： The application/x-www-form-urlencoded encoding algorithm

这是NSDictionary实现方法dataFormURLEncoded的类别，它返回一个NSData对象，表示编码参数，以NSDictionary形式给出（在另一个SO答案中显示））：How to send multiple parameterts to PHP server in HTTP post，例如：

NSData* paramsData = [@{@"key": @"value"} dataFormURLEncoded];

＆＃34;参数词典＆＃34;需要NSString作为键和值。因此，我们需要将JSON（NSData）转换为NSString（我们知道它的编码是UTF-8）：

NSString* jsonString = [[NSString alloc] initWithData:json 
                                             encoding:NSUTF8StringEncoding];

现在我们可以创建＆＃34;参数字典：＆＃34;

NSDictionary* paramsDict = @{@"data": jsonString};

鉴于，链接的SO答案中显示了NSDictionary类别，我们将参数作为NSData对象获取，正确编码：

NSData* paramsData = [paramsDict dataFormURLEncoded];

剩下的部分就是创建一个POST请求，这很简单：

NSMutableURLRequest* request = [[NSMutableURLRequest alloc] init];
[request setURL: url];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setValue: [NSString stringWithFormat:@"%i", [paramsData length]] forHTTPHeaderField:@"Content-Length"];
[request setHTTPBody: paramsData];

Answer 4

POST和GET的代码

    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"http://url.com/ios/api/get_images_cache.php?"]];

    [request setHTTPMethod:@"POST"];

    //Here GIVE YOUR PARAMETERS instead of my PARAMETER
    NSString *userUpdate =[NSString stringWithFormat:@"email_id=%@&pass=%@",txtEmail.text,txtPass.text,nil];

    NSData *data1 = [userUpdate dataUsingEncoding:NSUTF8StringEncoding];

    [request setHTTPBody:data1];

    NSError *err;
    NSURLResponse *response;

    NSData *responseData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&err];

    NSString *resultStr=[[NSString alloc]initWithData:responseData encoding:NSASCIIStringEncoding];
    NSLog(@"resultstr==%@",resultStr);

    //You need to check response.Once you get the response copy that and paste in ONLINE JSON VIEWER.If you do this clearly you can get the correct results.    

    //After that it depends upon the json format whether it is DICTIONARY or ARRAY 

    NSDictionary *dict = [NSJSONSerialization JSONObjectWithData:responseData options:NSJSONReadingMutableContainers error:nil];
    NSString *equalilstr=[dict objectForKey:@"message"];

将HTTP POST JSON参数发送到PHP页面 - iOS

4 个答案: