dps43378 2016-03-07 08:50 采纳率: 0%
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AJAX:为什么Autosuggest(给定)不起作用?

I got the plugin from the net and made some required changes. But I think PHP is not responding.

exp1.html code

<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('');
            });
        }
    });
});
});

Console also not showing any error, which points to some mistake in php code. I have provided complete code. Thank you!

  • 写回答

1条回答 默认 最新

  • dongxuan2577 2016-03-07 09:44
    关注

    I edited your code a bit and it is working fine for me now.

    <?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!";
    }
    ?>
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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