我正在尝试自动刷新多个PHP变量。到目前为止,我只能自动刷新一个变量而其余变量不起作用(它始终是最后一个变量,所以var2
)。
我正在考虑制作一个JSON脚本并尝试这样做,但是如何在JSON中使用我的PHP页面自动刷新而不刷新整个页面?
的index.php
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<script language="javascript" src="jquery-1.2.6.min.js"></script>
<script language="javascript" src="jquery.timer-1.0.0.js"></script>
<link rel="stylesheet" href="css/bootstrap.css">
<link rel="stylesheet" href="css/bootstrap.min.css">
<script src="jquery-1.9.0.min.js"></script>
<?php $ra = "<div id='c1b'></div>
<script type='text/javascript'>
$(document).ready(function() {
$('#c1b').load('var1.php');
refresh();
});
function refresh(){
setTimeout( function() {
$('#c1b').fadeOut('slow').load('var1.php').fadeIn('slow');
refresh();
}, 2000);
}
</script>";
$rb = "<div id='c2b'></div>
<script type='text/javascript'>
$(document).ready(function() {
$('#c2b').load('var2.php');
refresh();
});
function refresh(){
setTimeout( function() {
$('#c2b').fadeOut('slow').load('var2.php').fadeIn('slow');
refresh();
}, 2000);
}
</script>"; ?>
<?php echo $ra . ' & ' . $rb ?>
var1.php
<?php
include_once 'connect.php';
$d= 'var1';
$q = $handler->prepare("SELECT * FROM var WHERE name = ?");
$q->bindParam(1, $d);
$q->execute();
while($r = $q->fetch()){
$e = $r['names'];
}
echo $e;
?>
var2.php
<?php
include_once 'connect.php';
$d= 'var2';
$q = $handler->prepare("SELECT * FROM var WHERE name = ?");
$q->bindParam(1, $d);
$q->execute();
while($r = $q->fetch()){
$e = $r['names'];
}
echo $e;
?>
答案 0 :(得分:0)
尽管我可以阅读本文,但您将拥有两个名为&#34; refresh&#34;在您的页面上定义。
尝试将第二个脚本更改为唯一名称,例如
$(document).ready(function() {
$('#c2b').load('var2.php');
refreshvar2();
});
function refreshvar2(){
setTimeout( function() {
$('#c2b').fadeOut('slow').load('var2.php').fadeIn('slow');
refreshvar2();
}, 2000);
}