使用jQuery和AJAX从数据库中显示更多结果

问题描述：

I'm trying to create news block with AJAX in my template. so i wrote a simple code, in HTML Part:

<a href="javascript:ShowPost();">Show Posts</a>
<div id="result"></div>

jQuery Part:

function ShowPost(){
  $.post(dir + 'engine/ajax/posts.php', {action:"showpost"},
    function(data) {
      $("#result").html(data);
});
};

PHP Part:

if ($_POST['action'] == "showpost") {    
 $db->query( "SELECT title FROM post LIMIT 0,5 " );
  while ( $row = $db->get_row() ) {
   echo $row['title']."<br>";
  }    
}

Question is, how i can get more result after first click? for example, after first click on Show Posts link, i can show the 5 news from database. after second click on Show Posts link i need to show 6 to 10 news, in third click get result from 11 to 15 and continue ...

i hope you understand my question.

我正在尝试在模板中使用AJAX创建新闻块。所以我在HTML部分写了一个简单的代码： p>

 ＆lt; a href =“javascript：ShowPost（）;”＆gt; Show Posts＆lt; / a＆gt; 
＆lt; div  id =“result”＆gt;＆lt; / div＆gt; 
  code>  pre> 
 
  jQuery Part： p> 
 
 
  function ShowPost（）  {
 $ .post（dir +'engine / ajax / posts.php'，{action：“showpost”}，
 function（data）{
 $（“＃result”）。html（data）; \  n}）; 
}; 
  code>  pre> 
 
  PHP部分： p> 
 
 
  if（$ _POST ['action'  ] ==“showpost”）{
 $ db-＆gt; query（“SELECT title FROM post LIMIT 0,5”）; 
 while（$ row = $ db-＆gt; get_row（））{
 echo $  row ['title']。“＆lt; br＆gt;”; 
} 
} 
  code>  pre> 
 
 问题是，如何在首次点击后获得更多结果？ 例如，在第一次 strong>点击显示帖子 strong>链接后，我可以显示来自数据库的5条新闻。 在第二次 strong>点击显示帖子链接后，我需要显示6到10个新闻，在第三个 strong>中点击从11到15获取结果并继续......  p> 
  
 
我希望你理解我的问题。 p> 
  div>

答

Based on your implementation you need to keep track how many items you have shown and pass the page number in. For example:

jQuery Part:

var pageNumber = 0; 
function ShowPost() {   
   pageNumber++;   
   $.post(dir+ 'engine/ajax/posts.php', {action:"showpost", pageNum: pageNumber},
      function(data) {
       $("#result").html(data); 
   }); 
};

Disclaimer: I m not a PHP developer so please treat teh code below as pseudo-code.

PHP Part:

if ($_POST['action'] == "showpost") {    
var pageSize = 5;  
var pageNumber = $_POST['pageNumber'];    
var from = (pageNumber - 1) * pageSize;   
var to = pageNumber * pageSize;  
$db->query( "SELECT title FROM post LIMIT " + from + "," + pageSize);   
while ( $row = $db->get_row()) {    echo $row['title']."<br>";   }     
}

答

Take a hidden field in html and update it's value after every success call and when next time you call for next record send that value in post.

<input type="hidden" name="limit" value="5">
<a href="javascript:ShowPost();">Show Posts</a>
<div id="result"></div>

答

Just implement the pagination limit in ur query

var offset = -5;
function ShowPost(){
offset = offset + 5;
  $.post(dir + 'engine/ajax/posts.php', {action:"showpost",offset:offset},
    function(data) {
      $("#result").html(data);
});
};

PHP part

if ($_POST['action'] == "showpost") {

$vOffset = $_POST['offset'];    
 $db->query( "SELECT title FROM post LIMIT $vOffset,5 " );
  while ( $row = $db->get_row() ) {
   echo $row['title']."<br>";
  }    
}

使用jQuery和AJAX从数据库中显示更多结果

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