php - 如何将表单数据插入mysql数据库

Question

我正在尝试将表单中的数据插入到我创建的数据库中。当我检查数据是否在数据库中时，我不断收到此错误:

Error Code: 1064. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'user_comments' at line 1

这是我的 php:

<?php
$db_connection = mysqli_connect('localhost','root','',"project_online_planner");
if (!$db_connection){
    die('Failed to connect to MySql:'.mysql_error());
}else{
    echo "it worked";
}

if(isset($_POST['insertComments'])){
    $name=$_POST['name'];
    $comment=$_POST['comment'];
    mysqli_query($db_connection,"INSERT INTO user_comments (name, comment) VALUES ($name, $comment)");
    Print "Your information has been successfully added to the database.";
}

?>

这是我的 html:

<div id="uploadComments">
    <form id="insertComments" name="insertComments" method="post">
        <label for="name">Name: </label><input type="text" id="name" name="name"><br/>
        <label for="comment">Comments: </label><textarea name="comment" id="comment"></textarea>
        <input type="submit" value="Submit">
    </form>
</div>

Answer 1

正如之前的人评论的那样，您应该单独构建“查询”，这样更容易处理和更改。此外，当将任何变量传递到查询中时，您需要将它们从字符串中取出，以便 PHP 可以正确处理它们。

以下是我建议您构建代码的方式:

...

$name=$_POST['name'];
$comment=$_POST['comment'];

$sql="INSERT INTO user_comments (name, comment)VALUES('{$name}','{$comment}')";

if (!mysqli_query($db_connection,$sql))
{
 die('Error: ' . mysqli_error($con));
 }
echo "1 record added";