iPhone POST请求总是被PHP中的$ _SERVER ['REQUEST_METHOD']视为GET

Question

我正在iPhone上做项目。在这里，我必须发送一个POST请求到一个Web服务。我使用NSMutableURLRequest和NSURLConnection发送数据，而服务器使用PHP。我发现检查if ($_SERVER['REQUEST_METHOD'] != "POST")总是失败，因为在返回'GET'时，即使我在本地主机上尝试它。

我POST请求如下所示（通过Wireshark的嗅探）：

POST tester.php HTTP/1.1 
Host: localhost 
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10.7; rv:10.0.2) Gecko/20100101 Firefox/10.0.2 
Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8 
Accept-Language: en-us,en;q=0.5 
Accept-Encoding: gzip, deflate 
Connection: keep-alive 
Referer: http://localhost/tester.php 
Content-Type: application/x-www-form-urlencoded 
Content-Length: 422 

protocol=4&msgtype=authorize&merchant=86189919&language=da&ordernumber=1274168351&amount=100&currency=DKK&continueurl=http%3A%2F%2Fquickpay.net%2Ffeatures%2Fpayment-window%2Fok.php&cancelurl=http%3A%2F%2Fquickpay.net%2Ffeatures%2Fpayment-window%2Ferror.php&callbackurl=http%3A%2F%2Fquickpay.net%2Ffeatures%2Fpayment-window%2Fcallback.php&autocapture=0&cardtypelock=&splitpayment=1&md5check=d2259d9db077e0f5a41b4bf271c3c549 

当然存在有大量的虚拟文本时，通常无法从iPhone在那里，但我试图让通过HTML表单调用相同的调用，并且工作。因此，我已经复制了所有的信息，看看是否会有所作为，但无济于事。

我比较了两个请求（一个来自iPhone，另外从我的PHP页面），他们看起来一模一样。

我大吃一惊 - 有没有人有什么好的建议？

编辑：

您的测试没有为邮寄或交VS GET测试，是测试POST具体来说，你可以有一个情况下，问题出现......您是否尝试过检查的要求本身在PHP？ - Grady Player

是的，这可能是真的，但行总是返回GET，所以案例问题被覆盖，如果这就是你的意思。

另外，请不要的print_r的结果（$ _ SERVER）从当iPhone发送请求的服务器 - 阿莱克斯ģ

阵列（HTTP_HOST => 10.11.105.254 HTTP_USER_AGENT =>的Mozilla/5.0 （iPhone模拟器; CPU iPhone OS 5_1，例如Mac OS X）为AppleWebKit/534.46（KHTML，例如Gecko）移动/ 9B176 HTTP_ACCEPT => text/html的，应用/ XHTML + xml的，应用/ XML; q = 0.9，/; q = 0.8 HTTP_ACCEPT_LANGUAGE => zh_CN HTTP_ACCEPT_ENCODING => gzip，deflate HTTP_CONNECTION => keep-alive PATH =>/usr/bin：/ bin：/ usr/sbin：/ sbin SERVER_SIGNATURE => Apache/2.2.21（Unix） DAV/2 PHP/5.3.8与Suhosin-Patch Server a吨10.11.105.254端口80 SERVER_SOFTWARE =>的Apache/2.2.21（UNIX）DAV/2 PHP/5.3.8用了Suhosin贴片SERVER_NAME => 10.11.105.254 SERVER_ADDR => 10.11.105.254 SERVER_PORT => 80 REMOTE_ADDR => 10.11 .105.254 DOCUMENT_ROOT => /库/ Web服务器/文件SERVER_ADMIN => you@example.com SCRIPT_FILENAME => /Users/tester.php REMOTE_PORT => 51492 GATEWAY_INTERFACE => CGI/1.1 SERVER_PROTOCOL => HTTP/1.1 REQUEST_METHOD => GET QUERY_STRING = > REQUEST_URI => /tester.php SCRIPT_NAME => /tester.php PHP_SELF => /tester.php REQUEST_TIME => 1334211212 ARGV =>阵列的argc => 0）

你尝试检查，看看是否有已知的变量，例如$ _POST ['known_to_be_set_by_client']是否设置？如果请求是POST或GET，它不是PHP，它是Web服务器。 - Tim Post

当我尝试从iPhone请求中回显已知变量时，它当然是空的。这是有道理的。当我从形式上回应同样的东西时，它就在那里。

发表您的使用，使实际要求iOS的代码。 - 克里夫Ribaudo

下面是创建该请求，并将其发送代码：

//variables 
NSString *protocol = @"4"; 
NSString *msgtype = @"authorize"; 
NSString *merchant = @"86189919"; 
NSString *language = @"da"; 
NSString *orderNo = @"1274168351"; 
NSString *amount = @"100"; 
NSString *currency = @"DKK"; 
NSString *continueURL = @"http://quickpay.net/features/payment-window/ok.php"; 
NSString *cancelURL = @"http://quickpay.net/features/payment-window/error.php"; 
NSString *callbackURL = @"http://quickpay.net/features/payment-window/callback.php"; 
NSString *autoCapture = @"0"; 
NSString *cardtypeLock =@""; 
NSString *splitText = @"1"; 
NSString *md5check = @"d2259d9db077e0f5a41b4bf271c3c549"; 

//setting up the request 
NSMutableString *post = [[NSMutableString alloc] initWithString:@"protocol="]; 
[post appendString:protocol]; 

[post appendString:@"&msgtype="]; 
[post appendString:msgtype]; 

[post appendString:@"&merchant="]; 
[post appendString:merchant]; 

[post appendString:@"&language="]; 
[post appendString:language]; 

[post appendString:@"&ordernumber="]; 
[post appendString:orderNo]; 

[post appendString:@"&amount="]; 
[post appendString:amount]; 

[post appendString:@"&currency="]; 
[post appendString:currency]; 

[post appendString:@"&continueurl="]; 
[post appendString:continueURL]; 

[post appendString:@"&cancelurl="]; 
[post appendString:cancelURL]; 

[post appendString:@"&callbackurl="]; 
[post appendString:callbackURL]; 

[post appendString:@"&autocapture="]; 
[post appendString:autoCapture]; 

[post appendString:@"&cardtypelock="]; 
[post appendString:cardtypeLock]; 

[post appendString:@"&splitpayment="]; 
[post appendString:splitText]; 

[post appendString:@"&md5check="]; 
[post appendString:md5check]; 


//escape characters 
NSString *tmp = (__bridge NSString *)CFURLCreateStringByAddingPercentEscapes(
             NULL, 
             (__bridge CFStringRef)post, 
             NULL, 
             (CFStringRef)@";/:@+$,~", 
             kCFStringEncodingUTF8 
             ); 


NSData *body = [tmp dataUsingEncoding:NSUTF8StringEncoding]; 
//setup url 
NSString *urlString =@"http://10.10.225.255/tester.php"; 
NSURL *quickUrl = [NSURL URLWithString:urlString]; 

//setup URL request 
NSMutableURLRequest *request = [[NSMutableURLRequest alloc]init]; 
//method 
[request setHTTPMethod:@"POST"]; 
//url 
[request setURL:url]; 

//body data to string 
NSString *strData = [[NSString alloc]initWithData:body encoding:NSUTF8StringEncoding]; 

//content length 
NSString *postLenght = [NSString stringWithFormat:@"%d", [body length]]; 

//set body 
[request setHTTPBody:body]; 

//Dummy data to make the request look like the same from the HTML form. 
[request setValue:@"localhost" forHTTPHeaderField:@"Host"]; 
[request setValue:@"Mozilla/5.0 (Macintosh; Intel Mac OS X 10.7; rv:10.0.2) Gecko/20100101 Firefox/10.0.2" forHTTPHeaderField:@"User-Agent"]; 
[request setValue:@"text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8" forHTTPHeaderField:@"Accept"]; 
[request setValue:@"en-us,en;q=0.5" forHTTPHeaderField:@"Accept-Language"]; 
[request setValue:@"gzip, deflate" forHTTPHeaderField:@"Accept-Encoding"]; 
[request setValue:@"http://localhost/tester.php" forHTTPHeaderField:@"Referer"]; 
[request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"content-type"]; 
[request setValue:postLenght forHTTPHeaderField:@"content-length"]; 

//send the request to server 
if([[NSURLConnection alloc] initWithRequest:request delegate:self]){ 
    NSLog(@"sending data to quickpay"); 
} 

Answer 1

我想通了。如预期的那样，这是我自己的疏忽！ 当我将url加载到UIWebView中时，在委托中，我使用了loadRequest方法，这当然使用GET请求来返回内容。相反，我应该使用loadHTMLString：baseURL：方法来加载返回的数据。 这个错误意味着应用程序在我的初始POST后发出了另一个GET请求，这导致了错误。很明显，我没有注意到这个额外的要求一个looong时间。

Answer 2

（这是一个多回答评论）如果

if ($_SERVER['REQUEST_METHOD'] != "POST") 

总是失败（如你写在你的问题中），这个必须意思是即：

$_SERVER['REQUEST_METHOD'] == "POST" 

所以要么你问错了问题，要么你没有正确验证输入值。请将var_dump($_SERVER)的输出添加到您的问题中。