javascript - 在同一个 php 文件中使用 js 加载

Question

<script>
 setInterval(function(){
         $.ajax({
               url :'user_online.php',

               success: function()
                   {
                       $('.chat-o').load('user_online.php');
                   }
            });
     },1000); 
</script>

假设我有这个脚本。如果 user_online 是我放入的文件，我该如何让它工作(也就是说，我怎样才能让这个脚本在内部工作——在一个与 url 名称相同的 php 文件中)

基本上我想将与上面相同的内容应用到另一个页面，使其自动显示结果而无需重新加载:

<?php 
   $gameid = fetchinfo("value", "info", "name", "current_game");
$query  = mysql_query("SELECT * FROM `games` WHERE `id` < $gameid ORDER BY `id` DESC LIMIT 10");
while($rowd=mysql_fetch_array($query)):
  //define stuff
  $lastwinner=$rowd["userid"];
  $winnercos =$rowd['cost'];
  $winnerpercent = $rowd['percent'];
  $winneravatar=fetchinfo("avatar", "users", "steamid", $lastwinner);
  $winnername = fetchinfo("name", "users", "steamid", $lastwinner);
  $steamlink = fetchinfo("steamprofile", "users", "steamid", $lastwinner); ?>
  <div class="cont row">
  <div class="col-xs-24 header">
  </div>
  <div class="col-xs-24 body">
  <div class="col-xs-16 col-sm-16">
  <a rel="nofollow" target="_blank" href=<?php echo $steamlink; ?> class="img hidden-xs">
  <img src=<?php echo $winneravatar; ?> > </a>
  <a rel="nofollow" target="_blank" href=<?php echo $steamlink; ?> class="name"> <?php echo $winnername; ?> </a>
  </div>
  <div class="right text-right">
  <span class="win">
  Win: <span>$ <?php echo round($winnercos, 3); ?> </span></span>
  <span class="chance">
  Chance: <span><?php echo round($winnerpercent, 2); ?>% </span> </span> </div> </div>

用我想要的更新它。谁能告诉我是否可以？这是来自index.php的一部分

 <div class="site history col-xs-24 col-sm-18 col-md-14 col-lg-12"> <h2>History</h2> <div id="history">
  <script>
 setInterval(function(){
         $.ajax({
               url :'index.php',

               success: function()
                   {
                       $('.history').html(data);
                   }
            });
     },1000); 
</script> 
   <?php 
   $gameid = fetchinfo("value", "info", "name", "current_game");
$query  = mysql_query("SELECT * FROM `games` WHERE `id` < $gameid ORDER BY `id` DESC LIMIT 10");
while($rowd=mysql_fetch_array($query)):
  //define stuff
  $lastwinner=$rowd["userid"];
  $winnercos =$rowd['cost'];
  $winnerpercent = $rowd['percent'];
  $winneravatar=fetchinfo("avatar", "users", "steamid", $lastwinner);
  $winnername = fetchinfo("name", "users", "steamid", $lastwinner);
  $steamlink = fetchinfo("steamprofile", "users", "steamid", $lastwinner); ?>
  <div class="cont row">
  <div class="col-xs-24 header">
  </div>
  <div class="col-xs-24 body">
  <div class="col-xs-16 col-sm-16">
  <a rel="nofollow" target="_blank" href=<?php echo $steamlink; ?> class="img hidden-xs">
  <img src=<?php echo $winneravatar; ?> > </a>
  <a rel="nofollow" target="_blank" href=<?php echo $steamlink; ?> class="name"> <?php echo $winnername; ?> </a>
  </div>
  <div class="right text-right">
  <span class="win">
  Win: <span>$ <?php echo round($winnercos, 3); ?> </span></span>
  <span class="chance">
  Chance: <span><?php echo round($winnerpercent, 2); ?>% </span> </span> </div> </div>

Answer 1

您已经在user_online.php中使用ajax执行了一些操作，因此需要在.load()中再次加载数据。所以

改变这个，

success: function(data)
     {
         $('.chat-o').html(data);
     }

如果您在 user_online.php 中仅使用 php，请确保您回显这些值。