Question

我正在尝试执行准备好的SELECT查询，并且在准备好的部分失败并出现此错误...

Admin Help Email SELECT prepare() failed: Operand should contain 1 column(s)

我基本上理解，但我不知道如何改变它来做我想要完成的事情。我正在尝试SELECT来自users表的id，其中id是用户的会话ID。然后SELECT我尝试选择的所有其他字段而不是全部。我被告知，无论是个人多次，这是更好的练习，所以我试图以正确的方式做到这一点。

我怎样才能改变这一点，使其适用于我想要完成的任务？

$userid = ( isset( $_SESSION['user'] ) ? $_SESSION['user'] : "" );
$help_message = $_POST['help_message'];

 $con = mysqli_connect("localhost", "", "", "");
 $stmt = $con->prepare("SELECT(`id`, `firstname`, `lastname`, `email`, `phone_number`, `username`) FROM users WHERE id=?");
    if ( !$stmt || $con->error ) {
        // Check Errors for prepare
         die('Admin Help Email SELECT prepare() failed: ' . htmlspecialchars($con->error));
    }
    if(!$stmt->bind_param('i', $userid)) {
        // Check errors for binding parameters
        die('Admin Help Email SELECT bind_param() failed: ' . htmlspecialchars($stmt->error));
    }
    if(!$stmt->execute()) {
        die('Admin Help Email SELECT execute() failed: ' . htmlspecialchars($stmt->error));
    }

    while($row = mysqli_fetch_assoc($q)) { 
       $firstname = $row['firstname'];
       $lastname = $row['lastname'];
       $email = $row['email'];
       $phone_number = $row['phone_number'];
       $username = $row['username'];
    }

新代码

 $con = mysqli_connect("localhost", "", "", "");
 $stmt = $con->prepare("SELECT id, firstname, lastname, email, phone_number, username FROM users WHERE id=?");
    if ( !$stmt || $con->error ) {
        // Check Errors for prepare
         die('Admin Help Email SELECT prepare() failed: ' . htmlspecialchars($con->error));
    }
    if(!$stmt->bind_param('i', $userid)) {
        // Check errors for binding parameters
        die('Admin Help Email SELECT bind_param() failed: ' . htmlspecialchars($stmt->error));
    }
    if(!$stmt->execute()) {
        die('Admin Help Email SELECT execute() failed: ' . htmlspecialchars($stmt->error));
    }

    while($row = mysqli_fetch_assoc($stmt)) { 
       $firstname = $row['firstname'];
       $lastname = $row['lastname'];
       $email = $row['email'];
       $phone_number = $row['phone_number'];
       $username = $row['username'];
    }

SELECT查询的困难

0 个答案: