joao.sauer
Lurker
Hi guys,
Sorry if this is really easy and someone already answered it, but the problem is that I have no idea how to search for it.
Let's say that my Layout is huge and split it in 2-3 different layouts.
So, I have 1.xml, 2.xml and 3.xml.
The 1.xml have a Scrollview that contains 1 linearLayout.
The 2.xml have just 1 LinearLayout with a lot of items inside of it.
The 3.xml have just 1 LinearLayout with a lot of items inside of it.
So, in my onCreate, I need do the:
setContentView(R.layout.1);
Now, let's say that I have a flag that if is true I will need to use 2.xml and false I will use the 3.xml.
My question is:
How can I get a pointer to 2.xml or 3.xml and join it to the LinearLayout inside of the 1.xml?
So, in the end I will have one the screen with 1 ScrollView that contains 1 LinearLayout that will contain 1 more LinearLayout that could be from 2.xml or 3.xml.
It's simple I know, but I have no idea in how to find about it, really sorry about my question.
Thanks a lot in advanced to answer this simple question.
Joao
Sorry if this is really easy and someone already answered it, but the problem is that I have no idea how to search for it.
Let's say that my Layout is huge and split it in 2-3 different layouts.
So, I have 1.xml, 2.xml and 3.xml.
The 1.xml have a Scrollview that contains 1 linearLayout.
The 2.xml have just 1 LinearLayout with a lot of items inside of it.
The 3.xml have just 1 LinearLayout with a lot of items inside of it.
So, in my onCreate, I need do the:
setContentView(R.layout.1);
Now, let's say that I have a flag that if is true I will need to use 2.xml and false I will use the 3.xml.
My question is:
How can I get a pointer to 2.xml or 3.xml and join it to the LinearLayout inside of the 1.xml?
So, in the end I will have one the screen with 1 ScrollView that contains 1 LinearLayout that will contain 1 more LinearLayout that could be from 2.xml or 3.xml.
It's simple I know, but I have no idea in how to find about it, really sorry about my question.
Thanks a lot in advanced to answer this simple question.
Joao