<?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>
我们说我有这个剧本。如果user_online是我放入的文件(如何让这个脚本在内部工作 - 在一个名为url的php文件中),我该如何使它工作?
基本上我想将上述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:
<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>
用我想要的更新了它。谁能告诉我它是否可以?这是index.php的一部分
var removedId = 2;
for (var i = 0; i < data.length; i++) {
for (var j = data[i].child.length - 1; j >= 0; j--) {
if (data[i].child[j].boxId == removedId)
data[i].child.splice(j, 1);
}
}
console.log(data);
答案 0 :(得分:0)
您已使用ajax在user_online.php
中执行某项操作,因此需要在.load()
中再次加载数据。所以
改变这一点,
success: function(data)
{
$('.chat-o').html(data);
}
如果您在user_online.php
中仅使用php,请确保echo
值。