我正在编写一个PHP程序从文件夹中获取图像并显示它。我想刷新一下。我不能把javascript代码放在PHP里吗?
<?php
$action = $_REQUEST['action'];
if ($action == "view"){
$entry =$_GET['al'];
refreah
代码 echo '<script type="text/javascript">';
function refresh(){
document.images("img1").src="/latimage.php?device=$al&ref=" + new Date().getTime();
var e = document.getElementById("blinker");
e.style.visibility = ( e.style.visibility == "visible" )? "hidden" : "visible";
setTimeout("refresh()", 3500);}
</script>;
您可以使用echo
输出JavaScript代码:
echo '<script type="text/javascript">',
'function refresh() {',
"document.images('img1').src='/latimage.php?device=$al&ref=' + new Date().getTime();",
'var e = document.getElementById("blinker");',
'e.style.visibility = ( e.style.visibility == "visible" )? "hidden" : "visible";',
'setTimeout(refresh, 3500)',
'}',
'</script>';
Try Like ?> <script> location.reload(); </script> <?php
尝试点赞
?>
<script>
setTimeout(
document.images("img1").src="/latimage.php?device=$al&ref=" + new Date().getTime();
var e = document.getElementById("blinker");
e.style.visibility = ( e.style.visibility == "visible" )? "hidden" : "visible";
, 3500);
</script>
<?php