Question

的index.php

  <div class="input-group">
                <input type="text" class="form-control autosuggest" placeholder="Search for">
                <span class="input-group-btn">
                    <button class="btn btn-info"><span class="glyphicon glyphicon-search"></span></button>
                </span>
            </div>
             <div class="dropdown">
                    <ul class="result">

                    </ul>
                </div>

jquery / ajax文件

$(document).ready(function() {
$(".autosuggest").keyup(function() {
    var search_term  = $(this).attr('search');
    var dataString = 'search_term='+ search_term;
    $.ajax({
        type: 'post',
        url : 'search.php',
        data: dataString,
        success: function(data) {
            alert(data);
        }
    });
});

});

search.php文件

<?php
include 'includes/db.php';

if(isset($_POST['search_term']) && !empty($_POST['search_term'])) {
    $search_term = mysqli_real_escape_string($conn, $_POST['search_term']);
    $sql = "SELECT name FROM names WHERE name LIKE $search_term%";
    $run_sql = mysqli_query($conn, $sql);
    while($rows = mysqli_fetch_assoc($run_sql)) {
        echo "<li>$rows[name]</li>";
    }
  }
?>

我只是在php mysql和ajax中制作一个简单的自动填充建议，我的php版本是5，jquery版本是2，我想在autosuggest容器中填充我的mysql数据，首先我试图提醒页面上的数据javascript警报功能的帮助，但是有些错误，任何人都可以查看我的代码

谢谢

Answer 1

使用引号arround搜索变量

 $sql = "SELECT name FROM names WHERE name LIKE '$search_term%' ";

在jquery ajax和php5中自动完成

1 个答案: