Question

我一直在使用google jquery API在PHP中进行即时搜索。在youtube视频中，该程序正在运行（因为我正在从该视频中学习）。我的数据库逻辑是正确的，因为我在简单的html页面中尝试过它。但它不适用于jquery。请给我一些建议来解决这个问题。我正在研究ubuntu 14.04。

这是我的jquery包含的PHP页面。

<!DOCTYPE html>
<html>
    <head>
    <title>WebPage</title>
    <script type=text/javascript src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
    <script type=text/javascript>
    function searchq() {     
            var searchTxt = $("input[name='search']").val();
            $.post('search.php', {search_term: searchTxt}, function(output){
                    $('#output').html(output);
            }); 
    }
    </script>
    </head>
    <body>
    <form action="index.php" method="post">
            <b> First Name: </b> <br/>
            <input type="text" name="search" id="search1" placeholder="Search for members ..." onkeydown="searchq();">
            <br/> <br/> <b> Submit: </b>
            <input type="submit" name="Submit">
    </form>
    <div id="output"></div>
    </body>

我的日期连接页面如下

<!DOCTYPE html>
<html>
    <head>
    <title>WebPage</title>
    </head>
    <body>
    <?php
            $connection = mysqli_connect("localhost", "root", "password");
            $output = "Search Results. <br/> <br/>";
            if(isset($_POST["search_term"])){
                    $name = $_POST["search_term"];
                    $sql = "SELECT cust_id, last_name, first_name, email FROM widget_corp.customer WHERE first_name LIKE '%"
                            . $name . "%' OR last_name LIKE '%" . $name . "%'";
                    $result = mysqli_query($connection, $sql);
                    $count = mysqli_num_rows($result);
                    if($count == 0){
                            echo "There were no search results.";
                    }
                    else {
                            while($row = mysqli_fetch_assoc($result)){
                                    $firstName = $row["first_name"] . "  ";
                                    $lastName =  $row["last_name"] . "  ";
                                    $Email = $row["email"] ;
                                    $output .= '<div>'.$firstName.' '.$lastName.'</div>';
                            }
                    }
            }
            echo ($output);
    ?>
    </body>

使用google API与php

0 个答案: