Question

所以代码的作用是根据数据库的内容（项目表）向用户提供项目建议。

点击正确的项目建议后，下一个方法（ajax）现在将根据点击的建议检索另一列（项目价格），以在第二个文本框中显示该项目的价格。

当两个方法分开时它已经工作了，所以我猜错误在index.php脚本中。弹出也没有错误，所以我真的需要帮助

的index.php

<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Auto Complete Input box</title>

<link rel="stylesheet" type="text/css" href="jquery.autocomplete.css" />

<script type="text/javascript" src="jquery.js"></script>
<script type="text/javascript" src="jquery.autocomplete.js"></script>
<script>
$(document).ready(function(){
    $("#itemTravel").autocomplete("autocomplete.php", {
    selectFirst: true
    });

    function showRoom(){
    var itemNamed = $("#itemTravel").val();
    $.ajax({
        type:"POST",
        url:"process.php",
        data:{action:"showroom", itemName: itemNamed },
        success:function(data){
            $("#content").html(data);
        }
    });
}
showRoom();
});
</script>
</head>

<body>
    <label>Tag:</label>
    <input name="tag" type="text" id="itemTravel" size="20"/>
    <input name="number" type="text" id="itemPrice" size="20"/>
</body>
</html>

process.php

<?php
$link=mysqli_connect("localhost","root","","phpcollab");

if (mysqli_connect_errno())
    echo "Failed to connect to MySQL: " . mysqli_connect_error();

$action=$_POST["action"];
$itemName=$_POST["itemName"];
if($action=="showroom"){
    echo $itemName;
    $query="SELECT * FROM item WHERE itemName='".$itemName."'";
    $show=mysqli_query($link,$query) or die ("Error");
    $row=mysqli_fetch_array($show);
    echo $row['price'];
    echo "<script>
            document.getElementById('itemPrice').value = '".$row['price']."';
            </script>";

}
?>

autocomplete.php

    <?php
 $q=$_GET['q'];
 $my_data=mysql_real_escape_string($q);
 $mysqli=mysqli_connect('localhost','root','','phpcollab') or die("Database Error");
 $sql="SELECT name FROM tag WHERE name LIKE '%$my_data%' ORDER BY name";
 $result = mysqli_query($mysqli,$sql) or die(mysqli_error());

 if($result)
 {
  while($row=mysqli_fetch_array($result))
  {
   echo $row['name']."\n";
  }
 }
?>

自动填充方法

0 个答案: