Question

我是网络开发的新手抱歉，但我没有遇到问题

我刚刚创建了这个并且它不起作用。请帮助我不知道是什么问题。我没有得到任何结果。

但如果我去search.php它会显示所有名字。

HTML

<head>
    <script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
</head>

<input type="text" onKeyup="getName(this.value)"/><br>
<div id="result"></div>

<script type="text/javascript">

function getName(value){

$.post("search.php", {partialName:value}, function(data){

$("#result").html(data);

});

}

</script>

PHP（search.php中）

<?php
require 'includes/connection.php';

$partialName = $_POST['partialName'];

$query = "SELECT Name FROM Members WHERE Name LIKE '%$partialName%'";
$names = mysqli_query($dbc, $query);

while($namesArray = mysqli_fetch_array($names)){

echo "<div>".$namesArray['Name']."</div>";

}


?>

Answer 1

HTML：

<input type="text" id="search"/><br>

JS：

$("#search").on("keyup", function()
{
    var value = $(this).val();
    $.ajax({
        url: "search.php",
        type: "POST",
        data: "partialName="+value,
        success: function(data)
        {
            $("#result").html(data);
        }
    });
});

编辑：就像说明一样，可能值得逃避MySQL查询的输入：

$partialName = mysqli_real_escape_string( $_POST['partialName'] );

Answer 2

<input type="text" id="search" /><br>
<div id="result"></div>

<script src="//ajax.googleapis.com/ajax/libs/jquery/1.9.1/jquery.min.js"></script>
<script type="text/javascript">

$(document).ready(function(){
    $('#search').keyup(function(){
        var name = $('#search').val();
        var datastring = "partialName="+name;
        $.ajax({
            url: "search.php",
            type: "POST",
            data: datastring,
            success: function(data)
            {
                $("#result").html(data);
            }
        });
    });
});
</script>

<?php
require 'includes/connection.php';

$partialName = $_POST['partialName'];

$query = "SELECT Name FROM Members WHERE Name LIKE '%".$partialName."%' ";
$names = mysqli_query($dbc, $query);

while($namesArray = mysqli_fetch_array($names)){

echo "<div>".$namesArray['Name']."</div>";

}

?>

无法进行现场ajax搜索工作

2 个答案: