Question

将我的代码更改为mysqli程序中的预准备语句，其中在提交到我的php文件后显示此错误。

致命错误：无法通过引用传递参数8

这是我的PHP代码。

我刚刚将我的插入数据中的代码复制到没有数组的db，但是这个php代码会获得数组。

<?php  
include 'admin/db/database_configuration.php';
if(isset($_POST['submit'])){
    if (empty($_POST['title'])){$job_title = 'NULL'; } else{ $job_title ="'". mysqli_real_escape_string($conn, $_POST['title']) . "'";}
    if (empty($_POST['desc'])){$job_desc = 'NULL'; } else{ $job_desc ="'". mysqli_real_escape_string($conn, $_POST['desc']) . "'";}

    $qualifications ="";
        if(isset($_POST["quali"]) && is_array($_POST["quali"])){
        $qualifications = implode("\n", $_POST["quali"]); 
        }
    if (empty($_POST['name_cont'])){$name_contact = 'NULL'; } else{ $name_contact ="'". mysqli_real_escape_string($conn, $_POST['name_cont']) . "'";} 
    if (empty($_POST['contact'])){$contact_num = 'NULL'; } else{ $contact_num ="'". mysqli_real_escape_string($conn, $_POST['contact']) . "'";}
    if (empty($_POST['email_add'])){$email_cont = 'NULL'; } else{ $email_cont ="'". mysqli_real_escape_string($conn, $_POST['email_add']) . "'";}

    $stmt = $conn->prepare("INSERT INTO `tbljoba` (job_title, job_desc, job_qualifications, cont_name, contact_info, employer_email, job_status) VALUES(?,?,?,?,?,?,?)") or die(mysqli_error($conn));
    $stmt->bind_param("sssssss", $job_title, $job_desc, $qualifications, $name_contact, $contact_num, $email_cont, 'pending'); //bind to param

    if($stmt->execute()){
        $stmt->close();
        $conn->close();
        echo '<script>alert("Successfully Sent")</script>';
        echo '<script>window.location = "employer_contact_us.php"</script>';
    }else{
        echo '<script>alert("Error")</script>';
    }
}
$conn->close();
?>

在这行中我遇到了错误

$stmt->bind_param("sssssss", $job_title, $job_desc, $qualifications, $name_contact, $contact_num, $email_cont, 'pending'); //bind to param

Answer 1

错误在'pending'来电中bind_param。

bind_param的所有参数必须通过引用传递。字符串是原始值，不能通过引用传递。

您可以通过创建变量并将其作为参数传递来解决此问题：

$status = 'pending';
$stmt->bind_param("sssssss", $job_title, $job_desc, $qualifications, $name_contact, $contact_num, $email_cont, $status); //bind to param

或者，如果状态始终挂起，则可以将其硬编码到查询中。

// add 'pending' into the VALUES part of the query
$stmt = $conn->prepare("INSERT INTO `tbljoba` (job_title, job_desc, job_qualifications, cont_name, contact_info, employer_email, status) VALUES(?, ?, ?, ?, ?, 'pending')") or die(mysqli_error($conn));
// no need to bind 'pending'
$stmt->bind_param("ssssss", $job_title, $job_desc, $qualifications, $name_contact, $contact_num, $email_cont); //bind to param

致命错误：无法通过PHP中的引用传递参数8

1 个答案: