Php ios将数据发布到web服务

我一直在关注这个。除了把名字插入数据库之外，我的一切都正常了。乐谱插入得很好。下面是我在xCode中的代码

-(void)postData:(NSString *)userName withScore:(NSNumber *)gamesWon
{
// build up the request that is to be sent to the server
NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:@"http://indie-world.com/iOS/index.php"]];
[request setHTTPMethod:@"POST"];
[request addValue:@"postValues" forHTTPHeaderField:@"METHOD"];
NSLog(@"%@", @"Request set up");

//create data that will be sent in the psot
NSMutableDictionary *dictionary = [[NSMutableDictionary alloc] init];
[dictionary setValue:userName forKey:@"name"];
[dictionary setValue:gamesWon forKey:@"score"];
NSLog(@"%@", @"values set up");

// serialize the dictionary data as json
NSData *data = [[dictionary copy] JSONValue];

NSLog(@"%@", dictionary);

[request setHTTPBody:data];
[request addValue:[NSString stringWithFormat:@"%d", data.length] forHTTPHeaderField:@"Content-Length"];

NSURLConnection *connection = [[NSURLConnection alloc] initWithRequest:request delegate:self];
if(!connection)
{
    NSLog(@"Connection Failed");
}
NSLog(@"%@", @"connection granted");
}

这是我在php页面上的代码

<?php
/*Simple Service
This is just a simple php script that will return values ,the 
method is selected using the value of HTTP_METHOD
*/
$con=mysql_connect("localhost","iOSDevelopment","IosDevelopment");
// Check connection
if (!$con)
{
echo("Could not connect to web service.");
}

$db_selected = mysql_select_db("iosDevelopment");

if ($_SERVER['HTTP_METHOD'] === 'postValues'){ 
$body;
/*Sometimes the body data is attached in raw form and is not attached 
to $_POST, this needs to be handled*/

if($_POST == null){
  $handle  = fopen('php://input', 'r');
  $rawData = fgets($handle);
  $body = json_decode($rawData);
}
else
{
    if (isset($_POST["name"]) && isset($_POST["score"]))
    {
        $name = $_POST["name"];
        $score = $_POST["score"];
                    // Check to see if the info is all here. return $body to iOS app.
        $body = $_POST;
    }
    else 
    {
        $data['error'] = 'No Post values';
        echo json_encode($data);
    }
}
 // add to mysql
 $sql = "INSERT INTO `iosdevelopment`.`rw_high_score` (`score`, `name`) VALUES ('$score',     '$name')";

if (!mysql_query($sql, $con))
{
    $data['error'] = 'Score was not added';
    echo json_encode($data);
}
$data['scoreAdded'] = $body;
echo json_encode($data);

}
else 
{
$data['error'] = 'The Service you asked for was not recognized';
echo json_encode($data);
}
?>

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我错过了什么？echo返回名称和分数。我不确定我遗漏了什么。

我认为错误在于您的PHP脚本检查这部分

$data['scoreAdded'] = $body;
echo json_encode($data);

我想应该是这样

$data['scoreAdded'] = $body;
echo json_encode($data['scoreAdded']);

---

我不知道你为什么要这样做，希望这会有所帮助：

这不是回声部分。是SQL没有插入名称。比分很好。该名称不会输入MySQL。或者echo正在执行插入MySQL表？数据库表中名称的数据类型是什么？