Question

<?php

session_start();

// initializing variables
$title = "";
$categories = "";
$profile = "";
$description = "";
$errors = array();

// connect to the database
$conn = mysqli_connect("localhost", "root", "", "blogs");

// CREATE BLOGS
if (isset($_POST['blog_submit'])) {
    // receive all input values from the form
    $title = mysqli_real_escape_string($conn, $_POST['title']);
    $categories = mysqli_real_escape_string($conn, $_POST['categories']);
    $profile = mysqli_real_escape_string($conn, $_FILES['profile']['name']);
    $description = mysqli_real_escape_string($conn, $_POST['description']);

    // by adding (array_push()) corresponding error unto $errors array
    if (empty($title)) {
        array_push($errors, "Title is required");
    }
    if (empty($categories)) {
        array_push($errors, "Category is required");
    }
    if (empty($profile)) {
        array_push($errors, "Profile is required");
    }
    if (empty($description)) {
        array_push($errors, "Description is required");
    }
    if (count($errors) == 0) {
        $q2 = "INSERT INTO category (title, categories, profile, description) 
                  VALUES('$title', '$categories','$profile', '$description')";
        mysqli_query($conn, $q2);
        // $q2="INSERT INTO categories (title, categories, profile, description) 
        // VALUES('$title', '$categories','$profile', '$description')";
        // $run=mysqli_query($conn, $q2);
        // while($row=mysqli_fetch_array($run))
    }
}
?>

它给出了错误，通知未定义：第19行上的类别和未定义的通知：广告内容描述行，如何获得定义类别和描述？帮助我如何初始化说明和类别。

Answer 1

您可以这样做：

if( !empty($_POST['description']) ){
$description = mysqli_real_escape_string($conn, $_POST['description']);
}
else{
$description = '';
}

对未定义的其他索引进行相同的设置。这是因为您的表单未发送此POST数据。

注意：未定义的索引：类别和注意：未定义的索引：描述

1 个答案: