Привет всем, что я успешно достиг своей автоматической формы, но проблема в том, что результаты не соответствуют, когда мы просто возвращаем ее. это мой код ..
<html> <head> <title>Live Search Ajax Example</title> <script language="javascript" type="text/javascript" src="jquery-2.0.2.js"> </script> </head> <body> <h1>Live Search: Ajax Example</h1> <div class="content"> <input type="text" class="search" id="searchid" placeholder="Search for people" /> Ex:arunkumar, shanmu, vicky<br /> <div id="result"></div> </div> <script type="text/javascript" src="jquery-1.8.0.min.js"></script> <script type="text/javascript"> $(function(){ $(".search").keyup(function() { var searchid = $(this).val(); var dataString = 'search='+ searchid; if(searchid !='') { $.ajax({ type: "POST", url: "search.php", data: dataString, cache: false, success: function(html) { $("#result").html(html).show(); } }); }return false; }); jQuery("#result").live("click",function(e){ var $clicked = $(e.target); var $name = $clicked.find('.name').html(); var decoded = $("<div/>").html($name).text(); $('#searchid').val(decoded); }); jQuery(document).live("click", function(e) { var $clicked = $(e.target); if (! $clicked.hasClass("search")){ jQuery("#result").fadeOut(); } }); $('#searchid').click(function(){ jQuery("#result").fadeIn(); }); }); </script> </body> </html> и моя php-страница похожа на
<?php include('database.php'); if($_POST) { $q=$_POST['search']; $sql_res=mysql_query("select id,name,email from fk_mem where name like '$q%' or email like '$q%' order by id LIMIT 5"); while($row=mysql_fetch_array($sql_res)) { $username=$row['name']; $email=$row['email']; $b_username='<strong>'.$q.'</strong>'; $b_email='<strong>'.$q.'</strong>'; $final_username = str_ireplace($q, $b_username, $username); $final_email = str_ireplace($q, $b_email, $email); ?> <div class="show" align="left"> <span class="name"><?php echo $final_username; ?></span> <br/><?php echo $final_email; ?><br/> </div> <?php } } ?>в<?php include('database.php'); if($_POST) { $q=$_POST['search']; $sql_res=mysql_query("select id,name,email from fk_mem where name like '$q%' or email like '$q%' order by id LIMIT 5"); while($row=mysql_fetch_array($sql_res)) { $username=$row['name']; $email=$row['email']; $b_username='<strong>'.$q.'</strong>'; $b_email='<strong>'.$q.'</strong>'; $final_username = str_ireplace($q, $b_username, $username); $final_email = str_ireplace($q, $b_email, $email); ?> <div class="show" align="left"> <span class="name"><?php echo $final_username; ?></span> <br/><?php echo $final_email; ?><br/> </div> <?php } } ?>в<?php include('database.php'); if($_POST) { $q=$_POST['search']; $sql_res=mysql_query("select id,name,email from fk_mem where name like '$q%' or email like '$q%' order by id LIMIT 5"); while($row=mysql_fetch_array($sql_res)) { $username=$row['name']; $email=$row['email']; $b_username='<strong>'.$q.'</strong>'; $b_email='<strong>'.$q.'</strong>'; $final_username = str_ireplace($q, $b_username, $username); $final_email = str_ireplace($q, $b_email, $email); ?> <div class="show" align="left"> <span class="name"><?php echo $final_username; ?></span> <br/><?php echo $final_email; ?><br/> </div> <?php } } ?>
любая помощь пожалуйста ..
Поместите это между <HEAD> и </HEAD> так:
<script src="jquery-2.0.2.js"></script> <script> $.customPOST = function(data,callback){ $.post('search.php',data,callback,'json'); } $(document).ready(function() { $(".search").keyup(function(){ $.customPOST({search: $.('#searchid').val(),function(response){ if(response.success){ var html_code = '<div class="show" style="text-align:left;">'; html_code += '<span class="name">' + response.final_username + '</span>'; html_code += ' <br/>' + response.final_email + '<br/></div>'; $("#result").text(html_code); $("#result").show(); } }); }); </script>
PHP-скрипт должен возвращать ответ JSON таким образом:
<?php ... your code here and .... $final_username = str_ireplace($q, $b_username, $username); $final_email = str_ireplace($q, $b_email, $email); // here we create and return our JSON response $response = array('final_username' => $final_username, 'final_email' => $final_email); echo json_encode($response); ?>