php - php参数传递

Question

我尝试将一些文本从我的应用程序发送到 php，但不知何故我从未收到文本。这里有什么问题？

NSString *name = [[NSString alloc]initWithString: @"Hello World"];
NSString *email = [[NSString alloc]initWithString: @"Ohai2u"];
NSString *urlString = [NSString stringWithFormat:@"http://www.myhosturlxyz.com/tracktxt.php?name=%@ email=%@",
                       [name stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding],
                       [email stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];

NSURL *url = [[NSURL alloc] initWithString:urlString];

NSURLRequest *urlRequest = [NSURLRequest requestWithURL:url];

NSData *urlData;
NSURLResponse *response;
urlData = [NSURLConnection sendSynchronousRequest:urlRequest returningResponse:&response error:nil];
[url release];

我做的php是

<?php
    $name = $_REQUEST['name'];
    $email = $_REQUEST['email'];
    $myFile = "tracktxtf.txt";
    $fh = fopen($myFile, 'a') or die("can't open file");
    fwrite($fh, "my posted var is " .$name $email "\n");
    fclose($fh);   
 ?>

Answer 1

只是一个疯狂的猜测，但你可能需要&在name=%@and之间email=%a，制作它name=%@&email=%@吗？

或者，您可以尝试使用$_GET['name']and$_GET['email']代替 REQUEST 吗？也许尝试写入print_r($_REQUEST, true); print_r($_GET, true);您的调试文件并查看它包含的内容。