我创建了一个小的PHP脚本。见下面:
$x=5;
if ($x >= 8 ) {
echo "x is ".$x;
} else {
echo 'error';
$errorWindow = 'www.google.com';
echo "<script type='text/javascript'>
popWin('$errorWindow', 'windowname', 'width=400,height=300,scrollbars=yes');
</script>";
}
但是我的popwin无法正常工作,我想知道为什么,浏览器页面上显示了“错误”消息,但没有出现弹出窗口。谁能告诉我我在做什么错?我如何获得popwin?
编辑
我对语法进行了一些更改,这是该页面的完整代码:
$x=5;
if ($x >= 8 ) {
echo "x is ".$x;
} else {
?><script type="text/javascript">
function popWin(url)
{
var thePopCode = window.open(url,'','height=800, width=1000, top=500, left=200, scrollable=yes, menubar=yes, resizable=yes');
if (window.focus)
{
thePopCode.focus();
}
}
</script>
<?php
$errorWindow='google.com';
echo "<script type='text/javascript'>
popWin($errorWindow);
</script>";
}
最佳答案
可以了
<?php
$x=5;
if($x >= 8 ) {
echo "x is ".$x;
} else {
echo "Error";
?><script type="text/javascript">
function popWin(url)
{
var thePopCode = window.open(url,'','height=800, width=1000, top=500, left=200, scrollable=yes, menubar=yes, resizable=yes');
if (window.focus)
{
thePopCode.focus();
}
}
</script>
<?php
$errorWindow='google.com';
echo "<script type='text/javascript'>
popWin('google.com');
</script>";
}
?>
这也正在工作
<?php
$x=5;
if($x >= 8 ) {
echo "x is ".$x;
} else {
echo "Error";
?><script type="text/javascript">
function popWin(url)
{
var thePopCode = window.open(url,'','height=800, width=1000, top=500, left=200, scrollable=yes, menubar=yes, resizable=yes');
if (window.focus)
{
thePopCode.focus();
}
}
</script>
<?php
$errorWindow='google.com';
echo "<script type='text/javascript'>
popWin('".$errorWindow."');
</script>";
}
?>