问题描述
问题解释如下：
1-在php页面上我有$ x = 2;
2-有一个mysql表名为布局的字段
3-布局中的数据field = 我的问题编号是 $ x
4-现在我想要的是当我做回声$ layout它shoud显示
我的问题编号为2
但它正在显示
我的问题编号是$ x
你能来帮我吗？简单地解释我的问题非常重要我已经解释过它实际上我有一个非常复杂的场景。
请回复。这是我的3-4问题，但从未收到任何回复。
问候，
Sumaiya
www.phpjavascript .com
The problem is explained below:
1- on a php page i have $x=2;
2- there is a mysql table with a field called "layout"
3- The data in "layout" field = My problem number is $x
4- Now what i want is when i do echo $layout it shoud display
My problem number is 2
but it is displaying
My problem number is $x
Can you help me here. It is very important just to simplify my problem i have explained it like that in reality i have quite a complicated scenario.
Please reply back. This is my 3-4 question but never receive any replies from your side.
Regards,
Sumaiya
www.phpjavascript.com
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