3

在过去的几天里，我一直在为此苦苦挣扎；我正在尝试将数组发布到 PHP。我可以成功发送它，但它没有使用后变量（我正在尝试使用变量键“json”...使用此代码，我在 php 中收到数组：

Objective-C

NSError *error;
NSDictionary *jsonDictionary = [NSDictionary dictionaryWithObjects:[NSArray arrayWithObjects: @"one", @"two", @"three", nil] forKeys: [NSArray arrayWithObjects: @"a", @"b", @"c", nil]];
NSArray *jsonArray = [NSArray arrayWithObject:jsonDictionary];
NSData *jsonData = [NSJSONSerialization dataWithJSONObject:jsonArray options:NSUTF8StringEncoding error:&error];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:[NSURL URLWithString:@"somewebservicelocation/arrayTest.php?json="]];
[request setValue:@"application/json" forHTTPHeaderField:@"Content-Type"];
[request setHTTPMethod:@"POST"];
[request setHTTPBody:jsonData];
NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:nil error:nil];
NSString *response = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];
NSLog(@"response: %@",response);

PHP

$handle = fopen('php://input','r');
    $array = fgets($handle);
    echo $array;
    if(isset($array))
    {
        echo "success";
    }
    else
    {
        echo "failure";
    }

如果我使用这个 PHP，使用 _POST，我不会得到爱：

$rawJsonData = $_POST['json'];
$array = json_decode(stripslashes($rawJsonData),true);
echo $array;
if(isset($array))
{
    echo "success";
}
else
{
    echo "failure";
}

...我已经用了好几天了——在 Stack Overflow 上，我知道我需要在请求的正文中包含变量和数据，但我就是无法让它工作。我究竟做错了什么？你如何以不同的方式解决这个问题？让我摆脱这种头痛...

4

2 回答 2

6

在 PHP 方面，我使用了类似于您的第一个示例的内容：

<?php

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

// do what you want with it
//
// For diagnostic purposes, I'm just going to decode, make sure I got an array, 
// and respond with JSON that includes status, code, and the original request

$post_data = json_decode($http_raw_post_data,true);

if (is_array($post_data))
    $response = array("status" => "ok", "code" => 0, "original request" => $post_data);
else
    $response = array("status" => "error", "code" => -1, "original_request" => $post_data);

$processed = json_encode($response);
echo $processed;

?>

然后在 iOS 端，我使用：

// create the dictionary (or array)

NSDictionary *dictionary = @{@"a": @"One", @"b": @"Two", @"c": @"Three"};
NSError *error = nil;
NSData *data = [NSJSONSerialization dataWithJSONObject:dictionary options:0 error:&error];
if (error)
    NSLog(@"%s: JSON encode error: %@", __FUNCTION__, error);

// create the request

NSURL *url = [NSURL URLWithString:@"your.url.here"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];
[request setHTTPMethod:@"POST"];
[request setValue:@"application/json; charset=utf-8" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:data];

// issue the request

NSURLResponse *response = nil;
NSData *returnData = [NSURLConnection sendSynchronousRequest:request returningResponse:&response error:&error];
if (error)
    NSLog(@"%s: NSURLConnection error: %@", __FUNCTION__, error);

// examine the response

NSString *responseString = [[NSString alloc] initWithData:returnData encoding:NSUTF8StringEncoding];
NSLog(@"responseString: %@",responseString);

我刚刚测试了这次往返，它工作正常。

如果您决心使用该_POST技术，对我有用的是将数据设置为json=%@，例如：

NSDictionary *dictionary = ...
NSData *data = [NSJSONSerialization dataWithJSONObject:dictionary options:0 error:&error];
NSString *string = [[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
if (error)
    NSLog(@"%s: JSON encode error: %@", __FUNCTION__, error);

NSURL *url = [NSURL URLWithString:@"your.url.here"];
NSMutableURLRequest *request = [NSMutableURLRequest requestWithURL:url];

[request setHTTPMethod:@"POST"];
NSString *params = [NSString stringWithFormat:@"json=%@",
                    [string stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
NSData *paramsData = [params dataUsingEncoding:NSUTF8StringEncoding];
[request addValue:@"8bit" forHTTPHeaderField:@"Content-Transfer-Encoding"];
[request addValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Content-Type"];
[request setHTTPBody:paramsData];

// now send the request, like before

解析它的 PHP 很像你：

$http_raw_post_data = $_POST['json'];

$post_data = json_decode(stripslashes($http_raw_post_data),true);

if (is_array($post_data))
    $response = array("status" => "ok", "code" => 0, "original request" => $post_data);
else
    $response = array("status" => "error", "code" => -1, "original_request" => $post_data);

$processed = json_encode($response);
echo $processed;
于 2013-06-04T07:37:41.507 回答
0

最好在下面使用 PHP 函数响应将在stdobject 

parse_str(file_get_contents("php://input"), $data);

$d = json_decode(json_encode($data));
于 2016-02-24T06:28:22.130 回答