您好，我正在尝试使用脚本在我的 php 执行时显示加载栏。我从中学到了website但是当我做同样的事情时，我的加载栏仍然无法显示，知道为什么吗？
<html>
<head>
<style type="text/css">
div#content {
display: none;
}
div#loading {
top: 200 px;
margin: auto;
position: absolute;
z-index: 1000;
width: 160px;
height: 24px;
background: url(img/142.gif) no-repeat;
cursor: wait;
}
</style>
<script type="text/javascript">
function preloader(){
document.getElementById("loading").style.display = "none";
document.getElementById("content").style.display = "block";
}//preloader
window.onload = preloader;
</script>
<style type="text/css"></style>
</head>
<body>
<div id="loading"></div>
<div id="content" >
<?php
sleep(10);
echo 'This content has been loaded via an AJAX request';
?>
<br>
</div>
</body>
</html>
最佳答案
只需在任何浏览器中运行这段代码
<!DOCTYPE html>
<?php
@ini_set('zlib.output_compression', 0);
@ini_set('implicit_flush', 1);
@ob_end_clean();
set_time_limit(0);
?>
<html>
<head>
<style type="text/css">
div#content {
display: none;
}
img#loading {
top: 200 px;
margin: auto;
position: absolute;
z-index: 1000;
width: 500px;
height: 24px;
cursor: wait;
height: 500px
}
</style>
<style type="text/css"></style>
</head>
<body>
<?php
for ($i = 0; $i < 10; $i++) {
echo str_repeat(' ', 1024 * 64); // this is for the buffer achieve the minimum size in order to flush data
if ($i == 1)
echo '<img id="loading" src="img/142.gif" />';
}
?>
<div id="content" style="display: block;">
<?php
sleep(5);
echo 'This content has been loaded via an AJAX request';
?>
<br>
</div>
<script type="text/javascript">
function preloader() {
document.getElementById("loading").style.display = "none";
document.getElementById("content").style.display = "block";
}//preloader
window.onload = preloader;
</script>
</body>
</html>
如果您有权访问 php.ini，请在您的 php.ini 中设置以下配置并删除 @ini_set('zlib.output_compression', 0); @ini_set('implicit_flush', 1);从一开始
output_buffering = Off
implicit_flush = on
output_compression = Off
如果您对输出缓冲感到好奇 click here
关于javascript - 在使用脚本执行 php 时加载微调器图像，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/21282790/