Question

我在尝试获取项目的详细信息页面时遇到此错误。它对于学校而言我还不太了解PHP。 “您的SQL语法中有错误;请查看与您的MySQL服务器版本对应的手册，以便在第1行的''附近使用正确的语法”

以下是该页面的代码。

<?php
    require_once('connection.php');
    mysqli_select_db($conn, $dbname);
    $recordID = $_GET['recordID'];
    $query_Shoe_Details = "SELECT * FROM Products WHERE Shoe_Brand = $recordID";
    $Shoe_Details = mysqli_query($conn, $query_Shoe_Details) or die(mysqli_error(($conn)));
    $row_Shoe_Details= mysqli_fetch_assoc($Shoe_Details);
    $totalRows_Shoe_Details = mysqli_num_rows($Shoe_Details);
?>


<!DOCTYPE html>
    <html>
    <head>
        <title>details</title><?php include 'connection.php';?>
    </head>
    <body>
        <p>Product Name: <?php echo $row_Shoe_Details['Product_Name']; ?></p>
        <p><img src=
        "images/%3C?php%20echo%20$row_Shoe_Details['Image_Name'];%20?%3E"></p>
        <p>Description: <?php echo $row_Shoe_Details['Product_Description']; ?></p>
        <p>Price: $<?php echo $row_Shoe_Details['Product_Price']; ?></p><?php
            mysqli_free_result($Shoe_Details);
            ?>
    </body>
</html>

Answer 1

将您的查询更改为，使用单引号

Dynamic programming

同时删除此$query_Shoe_Details = "SELECT * FROM Products WHERE Shoe_Brand = '$recordID'";，无需再次添加

Answer 2

 input[type=radio]:not(old) + label:after{
 content: '';
 position: absolute;
 top: 0;
 left: 23px;
 width: 100%;
 height: 100%;
 z-index: -1;
 background   : url('radio.png') no-repeat 0 0;

PHP MSQL服务器语法错误

2 个答案: