PHP 无法发布到 MySQL 数据库

Question

我有一个 formText.php 文件，其中包含一个表单，表单代码如下:

<form action="insert.php" method="post">
    <p>
        <label for="theNames">Name:</label>
        <input type="text" name="theName" id="theName">
    </p>
    <p>
        <label for="theCitys">City:</label>
        <input type="text" name="theCity" id="theCity">
    </p>
    <p>
        <label for="theAges">Are you over eighteen?(Y/N)</label>
        <input type="text" name="theAge" id="theAge">
    </p>
    <p>
    <label for="theDates">Date:</label>
    <input type="text" name="theDate" id="theDate">
    </p>
    <input type="submit" value="Submit">
</form>

然后我有一个包含以下脚本的 insert.php 文件:

<?php
/* Attempt MySQL server connection. Assuming you are running MySQL
server with default setting (user 'root' with no password) */
$link = mysqli_connect("localhost", "root", "root","phpteste");

// Check connection
if($link === false){
    die("ERROR: Could not connect. " . mysqli_connect_error());
}

// Escape user inputs for security (EDITED)
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));


// attempt insert query execution
$sql = "INSERT INTO tabelateste (id, name, city, overeighteen, date) VALUES (NULL, '$theName', '$theCity', '$theAge', '$theDate')";
if(mysqli_query($link, $sql)){
    echo "Records added successfully.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// close connection
mysqli_close($link);
?>

我的数据库名为 phpteste，我的表名为 tabelateste。 我在这里做错了什么？ 每当我单击“提交”时，什么都不会出现，也不会向数据库中添加任何内容。

Answer 1

您的帖子数据名称字段有误。所以您需要更改以下行:

// Escape user inputs for security
$theName = mysqli_real_escape_string($link, $_POST['theName']);
$theCity = mysqli_real_escape_string($link, $_POST['theCity']);
$theAge = mysqli_real_escape_string($link, $_POST['theAge']);
$theDate = mysqli_real_escape_string($link, date("Y-m-d h:i:s",$_POST['theDate']));

您需要根据您的数据库表结构将 date 更改为 signup_date。

$sql = "INSERT INTO tabelateste (name, city, overeighteen, signup_date) VALUES ('$theName', '$theCity', '$theAge', '$theDate')";