Question

我试图从表中获取值，我面临的问题是无论输入是什么，它都会给我两行的结果。我想将输入ID与表中的ID匹配。这是我试图使用的代码Image of the table is attached文件名与javascript文件链接，其中我有一个click函数，这个javascript（global.js）文件链接到另一个有文本框的文件。我的主要文件代码在

下面

  <!DOCTYPE html>
<html>
    <head>
        <title>products</title>
    </head>
    <body>
            Item: 
            <input type="text" required id="item">
            <input type="submit" id="item-submit" value="Grab"> 
            <div id="item-data"></div>

            <script src="http://code.jquery.com/jquery-2.2.0.min.js"></script> 
            <script src="js/global.js"></script> 

    </body>
</html>

global.js文件代码

 $(document).ready(function(){
    $('input#item-submit').click(function(){     

        var item = $('input#item').val();    
        if ($.trim(item) != ''){             
            $.post('ajax/name.php',{item:name}, function(data){ 
                $('div#item-data').text(data);  

            });

        }

    })
});

name.php文件代码在

下面

 require '../db/connect.php';

    $sql = "SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID = ($_POST['name']) ";

    $result = mysqli_query($con,$sql);
        while($row = mysqli_fetch_array($result)) {
            echo " " . $row["BarcodeID"]. " shirt color:  " . $row["shirts"]. " price: " . $row["price"];
        }
    mysqli_close($con);

Answer 1

尝试此查询：

"SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID = '{$_POST['name']}'"

为什么要将BarcodeID与名称进行比较？

旁注：您的查询不安全。读这个 How can I prevent SQL injection in PHP?

Answer 2

    $id=$_POST['item']; // your post variable
    $sql = "SELECT BarcodeID, shirts, price FROM clothes WHERE BarcodeID=".$id;

     $result = mysqli_query($con,$sql);
if(mysqli_num_rows($result)>0)
{
        while($row = mysqli_fetch_array($result)) {
            echo " " . $row["BarcodeID"]. " shirt color:  " . $row["shirts"]. " price: " . $row["price"];
        }
}
else
{
echo "No results found";
}
    mysqli_close($con);

查询表

2 个答案: