表单的JavaScript弹出结果

问题描述

这是我第一次尝试使弹出窗口正常工作.我已经通过互联网进行了研究，但看不到我做错了什么也许你们在这里看到错了.我将通过javascript链接表单，以避免冗长的脚本.

This is the first time I am trying to make a popup working.I have done research over the internet but I could not see what I do wrongPerhaps you guys can see something wrong here.I will just link the form en the javascript for avoiding long script.

<script type="text/javascript">
$('.popup').popupWindow({
height:500,
width:800,
top:50,
left:50
});
</script>

和表格:

echo "<form action='getlist.php' method='post'>
                    <input type='text' name='exte' class='exte' value=".$value['ext'].">
                    <input type='submit' name='aanvragen' class='popup' id='aanvragen' value='aanvragen'></form>";
                    echo "</td>";

我希望它在弹出屏幕中打开getlist.php;)谢谢

I want it to open getlist.php in a popup screen ;)Thanks

当我单击按钮时，它会全屏显示在getlist.php中我希望它出现在弹出窗口中

When I click on the button it just go to getlist.php in a full screenI want this to be in a popup

推荐答案

我没有发现您的代码有什么问题，您的jQuery popupWindow可能在回显未在页面上放置任何内容之前就已被调用，因此侦听器从不会记录自己.

I don't see anything wrong with your code, your jQuery popupWindow is probably being called before the echo ever puts anything on the page so the listener never registers its self.

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