将值传递给php是未定义的 [英] Passing value to php is undefined

问题描述

嗨我不知道我在哪里出错了也许有人可以帮助我。


我使用以下函数在我的代码中生成对象（这可行）

Hi I am not sure where I am going wrong here maybe someone can help me.

I generate objects in my code using the following function (This works)

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.post函数如何传递数据值render.php？

.post function pass the data values to render.php?

我不太确定它是如何工作的。我之前发布了一个问题，因为我一次读到一行直到mysql，但我遇到的问题是只有几条随机行在我的表中更新。这是某人提出的解决方案。

I''m not quite sure how it works. I had previously posted a problem as I was reading one line at a time through to mysql but the problem I was having was that only a few random lines where getting updated in my table. This was a solution that someone came up with.

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