在jQuery Ajax和php5中自動完成

Question

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中提出一個簡單的自動完成建議，我的php版本是5，jquery版本是2，我想將我的mysql數據填充到autosuggest容器中，首先，我想借助來提醒頁面上的數據javascript警報功能，但是有一些錯誤，任何人都可以查看我的代碼

謝謝

Answer 1

使用引號arround搜索變量

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