使用 PHP 将 HTML 表单数据插入 MySQL

Question

我正在尝试制作一个简单的留言板 MySQL 数据库，您可以在其中撰写评论并通过 HTML 表单在一个页面上提交，并在提交评论后在单独的页面上查看所有评论。

我的问题是 HTML 表单中的两个字段没有插入到我的 MySQL 数据库中，这导致我查看所有评论页面缺少名称和标题。

Link to what the "Read all Reviews" page looks like.

当我只用 PHP 进行 MySQL 查询测试时，代码可以正常工作，但我需要我的 HTML 表单才能工作。

HTML 表单：

<form action ="process.php" method = "post">
            <fieldset>
                <legend>Review Field</legend>
                Reviewer Name: <br />
                <input type="text" name "name" id = "name"><br />
                Title of Review:<br />
                <input type="text" name "title" id = "title"><br />
                Enter your review below:
                 <!--Textbox start-->   
                 <textarea name="body" id = "body" rows="10" cols="100">
                 </textarea>
                  <!--Textbox end-->    
                <br />
                <input type="submit" name = "submit" id="submit">
                <br />
            </fieldset>
            </form>

process.php 的代码：

<?php // Create a database connection.
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "password";
$dbname = "ya_reviews";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);

//Test if connection occurred.
if (mysqli_connect_errno()) {
die("Database connection failed: " .
mysqli_connect_error() .
" (" . mysqli_connect_errno() . ")"
);
}

//Perform database query
$name = $_POST['name'];
$title = $_POST['title'];
$body = $_POST['body'];

//This function will clean the data and add slashes.
// Since I'm using the newer MySQL v. 5.7.14 I have to addslashes
$name = mysqli_real_escape_string($connection, $name);
$title = mysqli_real_escape_string($connection, $title);
$body = mysqli_real_escape_string($connection, $body);

//This should retrive HTML form data and insert into database
$query  = "INSERT INTO reviews (name, title, body) 
            VALUES ('".$_POST["name"]."','".$_POST["title"]."','".$_POST["body"]."')";

        $result = mysqli_query($connection, $query);
        //Test if there was a query error
        if ($result) {
            //SUCCESS
        header('Location: activity.php');
        } else {
            //FAILURE
            die("Database query failed. " . mysqli_error($connection)); 
            //last bit is for me, delete when done
        }

mysqli_close($connection);
?>

查看所有评论：

<?php

        //This will fetch the data from the database 
        $query = "SELECT * FROM reviews";
        $result = mysqli_query($connection, $query);
        //Test if there was a query error
        if (!$result) {
            die("Database query failed.");
        }

        // This will let me display the data.
        // The loop will be spilt so I can format with HTML
        while ($row = mysqli_fetch_assoc($result)) {
            //output data from each row
        ?>

        Name: <?php echo $row["name"] . "<br />"; ?>
        Title: <?php echo $row["title"] . "<br />"; ?>
        Review: <?php echo $row["body"] . "<br />";
            echo "<hr>"; ?>
        <?php
        } ?>

注意：我用上面代码之前在 process.php 中看到的相同代码连接到数据库，我排除了它以节省空间。

Answer 1

哟，你只是遗漏了一些语法，因此在从这些元素中收集数据时会产生错误，

<input type="text" name "title" id = "title">

您缺少名称参数中的“=”符号

Answer 2

问题是 HTML 中的 name 属性不正确。

<input type="text" name="name" id = "name"><br />

<input type="text" name="title" id = "title"><br />

Answer 3

我认为你搞砸了 HTML 的语法

<form action ="process.php" method = "post">
            <fieldset>
                <legend>Review Field</legend>
                Reviewer Name: <br />
                <input type="text" name="name" id = "name"><br />
                Title of Review:<br />
                <input type="text" name="title" id = "title"><br />
                Enter your review below:
                 <!--Textbox start-->   
                 <textarea name="body" id = "body" rows="10" cols="100">
                 </textarea>
                  <!--Textbox end-->    
                <br />
                <input type="submit" name = "submit" id="submit">
                <br />
            </fieldset>
            </form>

一定会成功的！

Answer 4

您的 HTML 属性语法不正确。它在attribute 和value 之间缺少= 符号。

将name "name" 更改为name="name" 和name "title" 更改为name="title"

<input type="text" name="name" id = "name"><br />
            Title of Review:<br />
<input type="text" name="title" id = "title"><br />

在insert 期间，您也没有使用转义值。

在插入查询中使用$name 而不是$_POST["name"]。标题和正文值也是如此。