Ios JSON数据不是来自本地主机

我正在尝试从本地主机获取json数据。我已经做过很多次了。但这一次，它不是在获取数据

Json数据

{ 
    "swimming_pool":"0", 
    "security":"0", 
    "lift":"0",
    "gym":"0",
    "reserved_parking":"0",
    "visitor_parking":"0",
    "power_backup":"0",
    "servant_room":"0",
    "tennis_court":"0",
    "rainwater_harvesting":"0",
    "waste_management":"0",
    "club_house":"0",
    "desc":"Dkkd",
    "city":"City",
    "pincode":"Pin Co",
    "locality":"locality",
    "no_of_beds":"1",
    "no_of_baths":"4"
}

客户端代码

 {
        NSString *selectQuery=[NSString stringWithFormat:@"http://localhost/FilterQuery.php?swimming_pool=%li&&security=%li&&lift=%li&&gym=%li&&visitor_parking=%li&&power_backup=%li&&servant_room=%li&&rainwater_harvesting=%li&&waste_management=%li&&clubhouse=%li&&Posdesc=%@&&no_of_baths=%li&&no_of_beds=%li&&pincode=%li&&locality=%@&&protypedesc=%@",(long)swimpoolb,(long)securityb,(long)liftb,(long)gymb,(long)visparkingb,(long)pbu,(long)servantroom,(long)rainwaterh,(long)wastemanagement,(long)clubHouse,possesion,(long)bathrooms,(long)bedrooms,(long)zipcode,locality,propertyType];

        NSString *newInsrStr = [selectQuery stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding];
        NSData *dataaa=[NSData dataWithContentsOfURL:[NSURL URLWithString:newInsrStr]];
        NSString *rr=[[NSString alloc]initWithData:dataaa encoding:NSUTF8StringEncoding];
        NSLog(@"%@",rr);
        jsondataa=[NSJSONSerialization JSONObjectWithData:dataaa options:0 error:nil];
        //jsondataa is dictionary
        swimmingPool=@"";
        swimmingPool=[jsondataa objectForKey:@"swimming_pool"];
        security=@"";
        security=[jsondataa objectForKey:@"security"];
        lift=@"";
        lift=[jsondataa objectForKey:@"lift"];
        gym=@"";
        gym=[jsondataa objectForKey:@"gym"];
        reserved_parking=@"";
        reserved_parking=[jsondataa objectForKey:@"reserved_parking"];
        visitor_parking=@"";
        visitor_parking=[jsondataa objectForKey:@"visitor_parking"];
        power_backUp=@"";
        power_backUp=[jsondataa objectForKey:@"power_backup"];
        NSLog(@"%@,%@,%@,%@,%@,%@,%@",swimmingPool,security,lift,gym,reserved_parking,visitor_parking,power_backUp);
  }

输出：

2015-06-29 15:20:51.874 NexGV1[1684:60b]
通知： 未定义变量：中的网球场 /Applications/XAMPP/xamppfiles/htdocs/FilterQuery.php在线 21
{“游泳池”：“0”，“安全”：“0”，“电梯”：“0”，“健身房”：“0”，“预留停车场”：“0”，“访客停车场”：“0”，“电源备用”：“0”，“仆人室”：“0”，“网球场”：“0”，“雨水收集”：“0”，“废物管理”：“0”，“俱乐部之家”：“0”，“描述”：“Dkkd”，“城市”：“城市”，“pincode”：“Pin” Co，“地点”：“地点”，“无床”：“1”，“无浴池”：“4”} 2015-06-29 15:20:51.875 NexGV1[1684:60b] （零），（零），（零），（零），（零），（零），（零），（零），（零）

---

它显示空值。为什么？

这应该是一条评论，但评论太长了。

因此，您的方法是通过url进行json请求，它不适用于类似的情况，因为它令人困惑且难以阅读

我懒得检查那个很长的url，所以我将向您介绍这种方法

NSString *selectQuery=[NSString stringWithFormat:@"http://localhost/FilterQuery.php?swimming_pool=%li&&security=%li&&lift=%li&&gym=%li&&visitor_parking=%li&&power_backup=%li&&servant_room=%li&&rainwater_harvesting=%li&&waste_management=%li&&clubhouse=%li&&Posdesc=%@&&no_of_baths=%li&&no_of_beds=%li&&pincode=%li&&locality=%@&&protypedesc=%@",(long)swimpoolb,(long)securityb,(long)liftb,(long)gymb,(long)visparkingb,(long)pbu,(long)servantroom,(long)rainwaterh,(long)wastemanagement,(long)clubHouse,possesion,(long)bathrooms,(long)bedrooms,(long)zipcode,locality,propertyType];

NSURLRequest *request = [[NSURLRequest alloc] initWithURL:[NSURL URLWithString:selectQuery]];

您可以将该url转换为请求，然后使用下面的

NSURLConnection

---

此外，如果您喜欢使用上面的

NSURLConnection

发出请求的简单方式，也可以

- (NSURLRequest *)convertToRequest:(NSString *)stringURL withDictionary:(NSDictionary *)dictionary
{

    NSError *error = nil;

    NSData *JSONData = [NSJSONSerialization dataWithJSONObject:dictionary options:0 error:&error];

    NSURL *url = [NSURL URLWithString:stringURL];

    NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];

    [request setHTTPMethod:@"POST"];
    [request setHTTPBody: JSONData];
    [request setValue:@"application/json" forHTTPHeaderField:@"Accept"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
    [request setValue:@"application/json" forHTTPHeaderField:@"Accept-Encoding"];

    [request setValue:[NSString stringWithFormat:@"%lu", (unsigned long)[JSONData length]] forHTTPHeaderField:@"Content-Length"];


    return request;
}

并且像这样使用它：

NSDictionary *jsonDictionary = @{
                                  @"swimming_pool": [NSNumber numberWithLong:(long)swimpoolb],
                                  @"security" : [NSNumber numberWithLong:(long)securityb],
                                  .. and so on
                                }; 
NSURLRequest *request = [ImplementationClass convertToRequest:YourServerURL withDictionary: jsonDictionary];

在服务器中，它应该是：

$handle = fopen('php://input','r');
$jsonInput = fgets($handle);
$json_decoded = json_decode($jsonInput,true);

$json_decoded['swimming_pool']; 
$json_decoded['security'];

---

希望这是信息和帮助

你可以使用AFNetworking，你可以很容易地从这个库中获取JSON数据。因为它给出了

NSLog（@“%@”，rr）的输出但jsondataa给出的是空值。@SukruK`使用
+（instancetype）dataWithContentsOfURL:（NSURL*）url选项：（NSDataReadingOptions）ReadOptions掩码错误：（NSError**）errorPtr
并将错误对象传递给该方法，以便查看是否得到响应。同时检查您的本地服务器是否正在运行。@BhupeshKumar如果您转到
http://localhost/FilterQuery.php
您应该能够看到它是否正在运行。我检查了它，它正在运行。@sbarow
$handle = fopen('php://input','r');
$jsonInput = fgets($handle);
$json_decoded = json_decode($jsonInput,true);

$json_decoded['swimming_pool']; 
$json_decoded['security'];