php - 如何在单列、不同行的 mysql 数据库中存储 3 个单选按钮列表？

Question

这是我的ratings.php(html代码)

<input type="radio" name="selectThree" value="1">
<input type="radio" name="selectThree" value="2">
<input type="radio" name="selectThree" value="3">
<input type="radio" name="selectThree" value="4">
<input type="radio" name="selectThree" value="5">

<input type="radio" name="selectTwo" value="1">
<input type="radio" name="selectTwo" value="2">
<input type="radio" name="selectTwo" value="3">
<input type="radio" name="selectTwo" value="4">
<input type="radio" name="selectTwo" value="5">

<input type="radio" name="selectOne" value="1">
<input type="radio" name="selectOne" value="2">
<input type="radio" name="selectOne" value="3">
<input type="radio" name="selectOne" value="4">
<input type="radio" name="selectOne" value="5">

所以当用户选择值时，它会在这里生成下面的代码插入到数据库中:

<?php
include_once "mysqli.connect.php";
include_once "config.php";

if(isset($_POST['Click']))      
{

$rating = explode($_POST['selectOne'], $_POST['selectTwo'], $_POST['selectThree']);
$_SESSION['commentInput'] = array();
$_SESSION['commentInput'][] = $_POST['comment'][0];
$_SESSION['commentInput'][] = $_POST['comment'][1];
$_SESSION['commentInput'][] = $_POST['comment'][2];

if(isset($_REQUEST["comment"]))
{

$merge = array_combine ($_SESSION['product'],$_SESSION['commentInput']);
foreach($merge as $key => $value)
{

$sqlComment = "INSERT into comment (comment, product) VALUES ('".$value."', '".$key."')";
$result = $mysqli->query($sqlComment);
}

echo"<script type='text/javascript'>alert('Thank you for your comment!' )</script>";
}

else
{

echo "<script type='text/javascript'>alert('Please comment!')</script>";    
}   

}

我想这样存储在mysql数据库中->

product|rating
--------------
shirt  | 2
pants  | 3
dress  | 5

但现在它是这样存储的:

product|rating
--------------
shirt  | Array
pants  | Array
dress  | Array

在我使用这个之后->

$rating = explode($_POST['selectOne'], $_POST['selectTwo'], $_POST['selectThree']);

//mysql
$sqlRating = "INSERT into ratings (product, rating) VALUES ('".$key."', '".$rating."')";
$result = $mysqli->query($sqlRating);

如何将值存储到 mysql 中？请帮忙谢谢!

Answer 1

你的$rating是一个数组

您应该存储像 $rating[0] 或 $rating[1] 或 $rating[2] 这样的值，以便像这样存储它们，你可以在 php 中管理它们选择或单击按钮，然后将它们存储在你的表中