从下拉表单加载选择页面 [英] Load page on selection from dropdown form

问题描述

我正在尝试制作一个下拉菜单，让我可以访问各种网页。现在它已经设置完毕，所以你做出选择然后你点击一个Go按钮，然后它会把你带到适当的链接。我试图找出如何做到这一点，当你做出选择时，它会自动进行，你不需要按Go按钮。

I am trying to make a drop down menu that takes me to various webpages. Right now it is set up so you make your selection and then you hit a "Go" button and then it takes you to the appropriate link. I am trying to figure out how to make it so when you make your selection it automatically goes and you don't need to push the "Go" button.

这里是我有什么：

Here is what I have:

<p align="center">
<form name="jump" class="center">
<select name="menu">
<option value="#">Select an option</option>
<option value="/link1.shtml">Link 1</option>
<option value="/link2.shtml">Link 2</option>
<option value="/link3.shtml">Link 3</option>
</select>
<input type="button" onClick="location=document.jump.menu.options[document.jump.menu.selectedIndex].value;" value="GO">
</form>
</p>

任何帮助或解释链接都会很好。谢谢！

Any help or links to explanations would be great. Thank you!

推荐答案

请尝试以下操作：

Try the following:

<select onchange="location = this.options[this.selectedIndex].value;">
    <option>Please select</option>
    <option value="http://www.apple.com/">Apple</option>
    <option value="http://www.bbc.com">BBC</option>
    <option value="http://www.facebook.com">Facebook</option>
</select>​

您正在寻找的是'onchange'，而不是'onsumbit'

What you were looking for was 'onchange' instead of 'onsumbit'

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