Question

关注视频，应该正常工作，而不是给我一个错误。我从！mysqli_stmt_prepare得到了WHOOPS错误。也许有一些我错过的东西？视频让我们删除了代替“？”的值，但这对我来说从来没有用过，所以我只是将$ var放回去.BTW：个人陈述中只有特殊字符（撇号和引号）当他们形成提交。我以为声明会逃脱那些没问题。感谢。

<?php

    header('Refresh:3; url=/Collaborate/');

$link = mysqli_connect("*****","*****","*****", "*****");

if (!$link) {
    echo "Error: Unable to connect to MySQL." . PHP_EOL;
    echo "Debugging errno: " . mysqli_connect_errno() . PHP_EOL;
    echo "Debugging error: " . mysqli_connect_error() . PHP_EOL;
    exit;
 }

    $Fname = $_POST['Fname'];

    $Lname = $_POST['Lname'];

    $Website = $_POST['Website'];

    $Phone = $_POST['Phone'];

    $Email = $_POST['Email'];

    $Interests1 = $_POST['Int1'];

    $Interests2 = $_POST['Int2'];

    $Interests3 = $_POST['Int3'];

    $PersonalStatement = $_POST['PersonalStatement'];

    // BAD ESCAPE. BAD!
        // array_walk_recursive($link, $_POST, 'mysqli_real_escape_string' );

    $sql = "INSERT INTO User(Fname, Lname, Website, Phone, Email, Interest1, Interest2, Interest3, PersonalStatement) VALUES(?, ?, ?, ?, ?, ?, ?, ?, ?);";

    $stmt = mysqli_stmt_init($link);
    if (!mysqli_stmt_prepare($stmt, "sssssssss", $sql)) {
        echo "WHOOPS!";
    } else {
        mysqli_stmt_bind_param($stmt, $Fname, $Lname, $Website, $Phone, $Email, $Interests1, $Interests2, $Interests3, $PersonalStatement);
        mysqli_stmt_execute($stmt);
        echo "<h2>We got you $Fname!</h2>";

    }
    ?>

Answer 1

将您的sql字符串值替换为?。在mysqli_stmt_bind_param()中 - 您需要第二个参数为 - "sssssssss"

准备好的声明会出错

1 个答案: