从弹出窗口中的列表中选择，并在父窗口中设置文本框的值 [英] select from list in pop-up window and set value of text box in parent window

问题描述

如何''s：


我的java脚本现在在我的弹出窗口中打开一个新窗口我有名单列表显示，我想选择名称并将其插入我的PHP页面文本框。谁能告诉我怎么做？


谢谢！

How to''s:

I have java script that opens a new window now inside my pop-up window I have list of names display and I want to select the names and insert it on my PHP page text box. Anybody who can show me how to do this?

thanks!

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推荐答案

你的问题中没有Java;你想让我把你的问题转移到哪里？一个Javascript论坛或一个PHP论坛？


亲切的问候，


Jos（主持人）

There no Java in your question; where do you want me to move your question to? A Javascript forum or a PHP forum?

kind regards,

Jos (moderator)

嗨Josah，


最初我在PHP上发布了这个帖子，但Markus告诉我把它放在JAVA论坛上。


基本上做完之后弹出窗口我只想抓取选中的变量到另一页。我希望我能描述一下我想做得好的过程。现在我正在寻找答案如何做到这一点。


再次感谢

Hi Josah,

Originally I posted this thread on PHP but I was told by Markus to put this on JAVA forum.

Basically after doing the pop-up window I just wanted to grab the variable selected to the other page. I hope I discribe the process I wanted to do well. Now Im looking for answer how to do this.

thanks again

我的php里面有java脚本打开第二页窗口。我想要做的是从弹出窗口中选择名称（例如），并将值发送到PHP文本框上的原始页面。有谁能告诉我怎么做？

I have java script inside my php that opens a 2nd page window. What I wanted to do is to select the name (for example) from the pop-up window and send the value to my original page on PHP text box. Anybody who can show me how to do this?

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