Question

我正在尝试通过ajax将表单中的数据传递到后端php脚本以与数据库进行交互。运行verifyUN函数有效，它会检查是否存在没有类似用户并继续执行下一个函数create。不幸的是，create函数没有从它调用的php脚本中传回任何内容。运行两个不同的$ .post函数不起作用吗？我有什么简单的东西吗？

    function verifyUN(username,password) {
        $.post('verify.php', { login: username}, function(result) {
                console.log("passed from ajax to verify, returned: "+ result);
                if (result==0) {
                    create(username,password);
                }
                else if(result == 1) {
                    alert("Username already exists!"); //until a better error
                }
                else {
                    alert("Error:" +result);
                }
        });

    }

此功能创建，不打印任何返回的数据。即使我粘贴了vertifyUN的内部，使它与verifyUN完全相同。然而，它仍然成功地触及数据库，因为它传递的数据，甚至将数据插入数据库。没有任何东西可以返回显示它是成功的......

    function create(username,password){
        console.log("This is inside create function");
        $.post('create.php', { login: username, password: password}, function(returned) {
                console.log("passed from create.php to verify, returned: "+ returned);

        });
    }

verify.php代码，通过verifyUN成功运行。

//needs cleaned
require '../database/info.php';

$dbname = "test";

$conn = new mysqli($servername, $username, $password, $dbname);

if($conn->connect_error) {

    die("connection not created: ". $conn->connect_error);

}

$username = htmlspecialchars($_POST['login']);

if($query = $conn->prepare("SELECT username FROM User WHERE username = ?") ) { 

    $query->bind_param("s", $username);
    $query->execute();
    $query->bind_result($un);

    $query->fetch();
    $query->close();
    if ( $username != $un ) { //double checking

        $result = 0; //doesn't exist, sends to ajax

    }
    else {
        $result =1; //username exists, sends to ajax
    }

    echo json_encode($result);

}
else {
    echo "query failed\n";
}

$conn->close();

?>

如果需要，

create.php

require '../database/info.php';

$dbname = "test";

$conn = new mysqli($servername, $username, $password, $dbname);

if($conn->connect_error) {

    die("connection not created: ". $conn->connect_error);

}

// if (!isset($_POST['login'])) {
//  $_POST['login'] = "testing";
// }
// if (!isset($_POST['password'])) {
//  $_POST['password'] = "password";
// }

$username = htmlspecialchars($_POST['login']);
$password = htmlspecialchars($_POST['password']);
$result = 0;
//echo "first result: $result\n";
$q1 = "INSERT INTO User (username) VALUES (?)";
$q2 = "INSERT INTO Security (username, shpassword) VALUES (?, ?)";

if( $stmt = $conn->prepare($q1) ) { 

    $stmt->bind_param("s", $username);
    $stmt->execute();
    $stmt->close();

    $result++;

}
else {
    $result = 0;

    echo json_encode("Username failed\n");
}

if ( $stmt = $conn->prepare($q2) ) {
    $stmt->bind_param("ss", $username, $password);
    $stmt->execute();
    $stmt->close();

    $result++;

}
else {
    echo json_encode("Failed to add user and password to database");
}

echo json_encode($result);

$conn->close();

?>

提前致谢。

Answer 1

我认为问题在于php脚本create.php，而且很可能在于未声明变量$un和$hash

两个sql语句是相关的，所以使用prepare方法的返回值来继续处理或完全放弃 - 真正的代码应该/将比下面显示的更干净地执行它但你应该得到想法

<?php

    $result = false;
    $username = !empty( $_POST['login'] ) ? $_POST['login'] : false;
    $password = !empty( $_POST['password'] ) ? $_POST['password'] : false;

    if( !$username or !$password )exit('bad foo');

    require '../database/info.php';

    $dbname = "test";

    $conn = new mysqli($servername, $username, $password, $dbname);
    if($conn->connect_error) die("connection not created: ". $conn->connect_error);




    $q1 = "insert into `user` ( `username` ) values (?)";
    $q2 = "insert into `security` ( `username`, `shpassword` ) values (?, ?)";

    $stmt=$conn->prepare( $q1 );
    if( $stmt ) {

        $stmt->bind_param("s", $username);
        $stmt->execute();
        $stmt->close();

        $stmt = $conn->prepare( $q2 );
        if( $stmt ){

            /* assumed algorithm here is default */
            $hash=password_hash( $password, PASSWORD_DEFAULT );

            $stmt->bind_param("ss", $username, $hash );
            $stmt->execute();
            $stmt->close();
        } else {
            exit('all is lost-abandon ship');
        }
        $conn->close();
    } else {
        exit('prepare failed');
    }
?>

调用成功的ajax post函数不会返回任何数据

1 个答案: