将 HTTP POST JSON 参数发送到 PHP 页面 - iOS答案

【问题标题】：Sending HTTP POST JSON parameter to PHP page - iOS将 HTTP POST JSON 参数发送到 PHP 页面 - iOS
【发布时间】：2014-03-27 14:59:00
【问题描述】：

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

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

这是我现在拥有的代码，但我无法让它工作。我不喜欢使用外部库，而是使用本地方法。

// 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。

编辑：

我终于做到了，问题出在这两行：

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

我删除了它们，终于让它工作了，谢谢！

【问题讨论】：

标签： php ios objective-c json

【解决方案1】：

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

将其添加到最后，然后您需要使您的类符合应用是：

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];

希望对你有帮助！

编辑：此外，在发送之前将您拥有的 JSON 转换为字符串。我就是这样做的：

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

【讨论】：

我已经建立了连接，但是服务器响应不是我所期望的，我只是没有把它说出来，因为那部分代码运行良好。我还尝试在发送之前将 JSON 转换为 NSSTRING，但没有结果。无论如何感谢您的快速回复！
您是否尝试过向服务器发送虚拟字符串？只是想看看你是否得到回应？

【解决方案2】：

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，参数键和值必须正确编码。特别是，JSON 通常包含许多不能按原样用作参数值的字符。

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

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

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

这里是NSDictionary 实现方法dataFormURLEncoded 的类别，它返回一个NSData 对象，表示作为NSDictionary 给出的编码参数（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];

“参数字典”需要有NSStrings 作为键和值。因此我们需要将 JSON（如 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“参数字典：”

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，正确编码：

NSData* paramsData = [paramsDict dataFormURLEncoded];

剩下的部分只是创建一个 POST 请求，这是直截了当的：

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];

【讨论】：

【解决方案3】：

下面的POST和GET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"];

【讨论】：

【解决方案4】：

这里是你的预期答案

第一个编码部分仅用于将数据发布到服务器

//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 获得响应

    //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"];

【讨论】：