我在 objective-c 中有两个数组，它们的数据结构如下:

{"index":"1","lastName":"Brown","firstName":"Kathy","company":"ABC inc."},
{"index":"2","lastName":"Smith","firstName":"Mike","company":"XYZ inc."}

和

{"index":"1","make":"Toyota","model":"RAV4","year":"2009"},
{"index":"2","make":"Honda","model":"Pilot","year":"2012"}

我的任务是将这两个数组放入一个 json NSData 对象中。同时，在 json 数据中，我应该将名称“People”分配给第一种类型的数组实体，将“Car”分配给第二种类型的数组实体，所以稍后在 php 端，我应该能够分辨出哪个分别放在不同的表中。你能帮我找到生成所需 json 对象的正确方法吗？

最佳答案

您可以获取两个数组，然后使用两个键构建一个新字典，people 和 cars，使用您的两个数组作为与这两个键关联的值。例如:

NSArray *people = @[@{@"index":@"1",@"lastName":@"Brown",@"firstName":@"Kathy",@"company":@"ABC inc."},
                    @{@"index":@"2",@"lastName":@"Smith",@"firstName":@"Mike",@"company":@"XYZ inc."}];

NSArray *cars = @[@{@"index":@"1",@"make":@"Toyota",@"model":@"RAV4",@"year":@"2009"},
                  @{@"index":@"2",@"make":@"Honda",@"model":@"Pilot",@"year":@"2012"}];

NSDictionary *fullDictionary = @{@"people": people, @"cars": cars};

NSError *error;
NSData *data = [NSJSONSerialization dataWithJSONObject:fullDictionary options:0 error:&error];

生成的 JSON(经过格式化以便您可以更轻松地阅读)如下所示:

{
  "cars" : [
    {
      "model" : "RAV4",
      "year" : "2009",
      "make" : "Toyota",
      "index" : "1"
    },
    {
      "model" : "Pilot",
      "year" : "2012",
      "make" : "Honda",
      "index" : "2"
    }
  ],
  "people" : [
    {
      "lastName" : "Brown",
      "firstName" : "Kathy",
      "company" : "ABC inc.",
      "index" : "1"
    },
    {
      "lastName" : "Smith",
      "firstName" : "Mike",
      "company" : "XYZ inc.",
      "index" : "2"
    }
  ]
}

要发送该请求，您可以执行以下操作:

NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
request.HTTPBody = data;
request.HTTPMethod = @"POST";

NSURLSessionTask *task = [[NSURLSession sharedSession] dataTaskWithRequest:request completionHandler:^(NSData *data, NSURLResponse *response, NSError *error) {
    // check for fundamental networking error (e.g. not connected to Internet)

    if (error) {
        NSLog(@"error = %@", error);
        return;
    }

    // also check to see if the server reported some HTTP error

    if ([response isKindOfClass:[NSHTTPURLResponse class]]) {
        NSInteger statusCode = [(NSHTTPURLResponse *)response statusCode];
        if (statusCode != 200) {
            NSLog(@"statusCode = %ld", (long)statusCode);
            return;
        }
    }

    // OK, if we've gotten here, let's parse the response; I'm assuming you'll send JSON response; so parse it

    NSError *parseError;
    NSDictionary *responseObject = [NSJSONSerialization JSONObjectWithData:data options:0 error:&parseError];
    if (parseError) {
        NSString *responseString = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
        NSLog(@"responseString = %@", responseString);
    } else {
        NSLog(@"responseObject = %@", responseObject);
    }
}];
[task resume];

在 PHP 中，要解析此响应，您可以执行以下操作:

<?php

header("Content-Type: application/json");

$handle = fopen("php://input", "rb");
$raw_post_data = '';
while (!feof($handle)) {
    $raw_post_data .= fread($handle, 8192);
}
fclose($handle);

$json_data = json_decode($raw_post_data, true);

$people = $json_data["people"];
$cars = $json_data["cars"];

foreach ($people as $person) {
    $index      = $person["index"];
    $last_name  = $person["lastName"];
    $first_name = $person["firstName"];
    $company    = $person["company"];

    // use these four variables to insert row of data here; note, make sure you either use 
    // `mysqli::real_escape_string` or that you manually bind these values to `?` placeholders
}

foreach ($cars as $car) {
    $index = $car["index"];
    $make  = $car["make"];
    $model = $car["model"];
    $year  = $car["year"];

    // again, use these four variables to insert row of data here; note, make sure you either use 
    // `mysqli::real_escape_string` or that you manually bind these values to `?` placeholders
}

// obviously, if we had errors above, we'd send something like Array("success" => false, ...) with error messages and error codes
$response = Array("success" => true);
echo json_encode($response);

?>

关于php - 如何将两个具有不同类型实体的数组组合成一个 json 对象？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/34469205/