php - 将 HTTP POST JSON 参数发送到 PHP 页面 - iOS

Question

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。参数的名称必须是“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]];

谢谢你的帮忙。

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，最后我让它工作了，谢谢!

Answer 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 ，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。此外，我在我的应用程序中构建请求的方式是:

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];

我希望这会有所帮助!

编辑:
此外，在发送之前将您拥有的 JSON 转换为字符串。我是这样做的:

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