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如何在弹出窗口中将我的网址重定向到下面?在弹出窗口中打开一个网址
$array[] = array ('id' => "$id", 'title' => "$evento", 'start' => "$dtevento2", 'end' => "$dtfim2", 'url' => "/eventos.php?id=$id", 'allDay' => "false");
我的JavaScript代码来弹出
<script language=javascript>
var win = null;
function NovaJanela(pagina,nome,w,h,scroll){
LeftPosition = (screen.width) ? (screen.width-w)/2 : 0;
TopPosition = (screen.height) ? (screen.height-h)/2 : 0;
settings = 'height='+h+',width='+w+',top='+TopPosition+',left='+LeftPosition+',scrollbars='+scroll+',resizable'
win = window.open(pagina,nome,settings);
}
</script>
和链接使用的代码
<a href="eventos.php?id=<? echo $id; ?> "onclick="NovaJanela(this.href,'nomeJanela','620','500','no');return false">Open</a>
对不起,我不明白,你想要什么。什么是PHP网址?它与普通的url有什么不同?你说什么意思:“重定向我的网址”? – user4035
在这里搜索“AJAX”即可满足您的所有需求。很多问题已经解决了这个困境。这里有一个我回答说,可以帮助你开始:http://stackoverflow.com/questions/7932460/submitting-a-form-with-ajax-and-instantly-displaying-the-results-from-a-php-file/7932555#7932555 –
我希望我的阵列中描述的URL在弹出窗口中打开 –