AJAX :为什么自动建议(给定）不起作用

AJAX : Why Autosuggest(given) is not working?

本文关键字：给定 不起作用 为什么 AJAX      更新时间：2023-09-26

我从网上得到了插件并进行了一些必要的更改。但我认为PHP没有回应。

exp1.html代码 

<html>
<head>
<meta charset="UTF-8">
<title>AutoSuggestion using Ajax, Jquery and PHP</title>
<link rel="stylesheet" href="exp1.css">
</head>
<body>
<div class="serch-wrap">
    <div class="form-wrapper cf">
        <input type="text" class="abhijitscript" name="search_word">
        <input type="submit" value="Search">
    </div>
    <div class="autodropdown">
        <ul class="suggestresult"></ul>
    </div>
</div>
</body>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript" src="exp1.js"></script>
</html>

exp1.php

<?php
$host="localhost";
$user="root";
$password="jatin";
$link=mysqli_connect($host,$user,$password);
if(@mysqli_select_db($link,'books')){
// echo "Database connected..<br>";
}
else
echo mysqli_connect_error();

if(isset($_POST['name']))
{
$name = mysqli_real_escape_string($link,trim($_POST['name']));
$sql = "SELECT `name` FROM `books` WHERE `name` LIKE 'l%'";
$myquery = mysqli_query($link,$sql) or die(mysql_error());
if(mysqli_num_rows($myquery) !=0)
{
    while(($row = mysqli_fetch_array($myquery)) !== false)
    {
        echo '<li>'.$row['name'].'</li>';
    }
}
else
{
    echo '<li>Not Found</li>';
}
} else{
echo "not set!";
}
?>

exp1.js

$(document).ready(function(){
$('.abhijitscript').keyup(function(){
    var query_string = $(this).val();
    $.ajax({
        type: "POST",
        url: "exp1.php",
        data: { name:query_string },
        success: function(data)
        {   console.log(data);
            $('.suggestresult').html(data);
            $('.suggestresult li').click(function(){
                var return_value = $(this).text();
                $('.abhijitscript').attr('value', return_value); 
                $('.abhijitscript').val(return_value);
                $('.suggestresult').html('');
            });
        }
    });
});
});

控制台也没有显示任何错误，这表明 php 代码中存在一些错误。我已经提供了完整的代码。谢谢！

我稍微编辑了一下你的代码，现在对我来说工作正常。

<?php
$host="localhost";
$user="root";
$password="jatin";
$link=@mysqli_connect($host,$user,$password);
if( !$link || !@mysqli_select_db($link,'books')){
    echo mysqli_connect_error();
    exit ;
}
if(isset($_POST['name'])) {
    $name = mysqli_real_escape_string($link,trim($_POST['name']));
    $sql = "SELECT DISTINCT(`name`) FROM `books` WHERE `name` LIKE '%{$name}%'";
    $myquery = mysqli_query($link,$sql) or die(mysql_error());
    if(mysqli_num_rows($myquery) !=0) {
        while(($row = mysqli_fetch_array($myquery)) !== NULL) {
            echo '<li>'.$row['name'].'</li>';
        }
    } else {
        echo '<li>Not Found</li>';
    }
} else {
    echo "not set!";
}
?>
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