Question

我正在尝试学习如何为php文档创建一个http请求，然后将其插入到mysql表中。问题是它没有返回任何东西。没有错误，甚至没有mysql中的空行。我做错了什么？

目标c

              NSString *post = [NSString stringWithFormat:@"name=%@&id=%@&userid=%@",@"lol", @"2", @"1"];

                NSData *postData = [post dataUsingEncoding:NSASCIIStringEncoding allowLossyConversion:YES];

                NSString *postLength = [NSString stringWithFormat:@"%d",[postData length]];

                NSMutableURLRequest *request = [[NSMutableURLRequest alloc] init];
                [request setURL:[NSURL URLWithString:[NSString stringWithFormat:@"http://www.ratemyplays.com/api/post_user.php"]]];

                [request setHTTPMethod:@"POST"];
                [request setValue:postLength forHTTPHeaderField:@"Content-Length"];
                [request setValue:@"application/x-www-form-urlencoded" forHTTPHeaderField:@"Current-Type"];
                [request setHTTPBody:postData];

php代码：

<?php

$link = mysqli_connect("mysql12.gigahost.dk","dirts","password_example","dirts_api");


$name = (string)$_POST['name'];
$id = (int)$_POST['id'];
$userid = (int)$_POST['userid'];



$query = "INSERT INTO User_friends (name, profileID, userID) VALUES ('$name', '$id', '$userid')";
$result = mysqli_query($link,$query);
?>

Answer 1

如何添加提交然后用php检查

NSString *post = [NSString stringWithFormat:@"name=%@&id=%@&userid=%@submit=",@"lol", @"2", @"1"];


<?php

    if (isset($_POST['submit'])
    {

        $link = mysqli_connect("mysql12.gigahost.dk","dirts","password_example","dirts_api");


        $name = (string)$_POST['name'];
        $id = (int)$_POST['id'];
        $userid = (int)$_POST['userid'];



        $query = "INSERT INTO User_friends (name, profileID, userID) VALUES ('$name', '$id',          '$userid')";
        $result = mysqli_query($link,$query);

        if (!$result)
        {
            die(mysql_error());
       }
    ?>

Http请求什么都不返回

1 个答案: