如何以编程方式为视图分配ID？

Question

In an XML file, we can assign an ID to a view like android:id="@+id/something" and then call findViewById() , but when creating a view programmatically, how do I assign an ID?

I think setId() is not the same as default assignment. setId() is extra.

Can anybody correct me?

Answer 1

Android id overview

An Android id is an integer commonly used to identify views; this id can be assigned via XML (when possible) and via code (programmatically.) The id is most useful for getting references for XML-defined View s generated by an Inflater (such as by using setContentView .)

Assign id via XML

• Add an attribute of android:id="@+id/ somename " to your view.
• When your application is built, the android:id will be assigned a unique int for use in code.
• Reference your android:id 's int value in code using " R.id. somename" (effectively a constant.)
• this int can change from build to build so never copy an id from gen/ package.name/ R.java , just use " R.id. somename".
• (Also, an id assigned to a Preference in XML is not used when the Preference generates its View .)

Assign id via code (programmatically)

• Manually set id s using someView.setId( int );
• The int must be positive, but is otherwise arbitrary- it can be whatever you want (keep reading if this is frightful.)
• For example, if creating and numbering several views representing items, you could use their item number.

Uniqueness of id s

• XML -assigned id s will be unique.
• Code-assigned id s do not have to be unique
• Code-assigned id s can (theoretically) conflict with XML -assigned id s.
• These conflicting id s won't matter if queried correctly (keep reading) .

When (and why) conflicting id s don't matter

• findViewById(int) will iterate depth-first recursively through the view hierarchy from the View you specify and return the first View it finds with a matching id .
• As long as there are no code-assigned id s assigned before an XML-defined id in the hierarchy, findViewById(R.id.somename) will always return the XML-defined View so id 'd.

Dynamically Creating Views and Assigning ID s

• In layout XML, define an empty ViewGroup with id .
• Such as a LinearLayout with android:id="@+id/placeholder" .
• Use code to populate the placeholder ViewGroup with View s.
• If you need or want, assign any id s that are convenient to each view.

• Query these child views using placeholder.findViewById(convenientInt);

• API 17 introduced View.generateViewId() which allows you to generate a unique ID.

If you choose to keep references to your views around , be sure to instantiate them with getApplicationContext() and be sure to set each reference to null in onDestroy . Apparently leaking the Activity (hanging onto it after is is destroyed) is wasteful.. :)

Reserve an XML android:id for use in code

API 17 introduced View.generateViewId() which generates a unique ID. (Thanks to take-chances-make-changes for pointing this out.)*

If your ViewGroup cannot be defined via XML (or you don't want it to be) you can reserve the id via XML to ensure it remains unique:

Here, values/ids.xml defines a custom id :

<?xml version="1.0" encoding="utf-8"?>
<resources>
    <item name="reservedNamedId" type="id"/>
</resources>

Then once the ViewGroup or View has been created, you can attach the custom id

myViewGroup.setId(R.id.reservedNamedId);

Conflicting id example

For clarity by way of obfuscating example, lets examine what happens when there is an id conflict behind the scenes.

layout/mylayout.xml

<?xml version="1.0" encoding="utf-8"?>
<LinearLayout
    xmlns:android="http://schemas.android.com/apk/res/android"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:orientation="vertical" >
    <LinearLayout
        android:id="@+id/placeholder"
        android:layout_width="fill_parent"
        android:layout_height="wrap_content"
        android:orientation="horizontal" >
</LinearLayout>

To simulate a conflict, lets say our latest build assigned R.id.placeholder ( @+id/placeholder ) an int value of 12 ..

Next, MyActivity.java defines some adds views programmatically (via code):

int placeholderId = R.id.placeholder; // placeholderId==12
// returns *placeholder* which has id==12:
ViewGroup placeholder = (ViewGroup)this.findViewById(placeholderId);
for (int i=0; i<20; i++){
    TextView tv = new TextView(this.getApplicationContext());
    // One new TextView will also be assigned an id==12:
    tv.setId(i);
    placeholder.addView(tv);
}

So placeholder and one of our new TextView s both have an id of 12! But this isn't really a problem if we query placeholder's child views:

// Will return a generated TextView:
 placeholder.findViewById(12);

// Whereas this will return the ViewGroup *placeholder*;
// as long as its R.id remains 12: 
Activity.this.findViewById(12);

*Not so bad

Answer 2

You can just use the View.setId(integer) for this. In the XML, even though you're setting a String id, this gets converted into an integer. Due to this, you can use any (positive) Integer for the Views you add programmatically.

According to View documentation

The identifier does not have to be unique in this view's hierarchy. The identifier should be a positive number.

So you can use any positive integer you like, but in this case there can be some views with equivalent id's. If you want to search for some view in hierarchy calling to setTag with some key objects may be handy.

Credits to this answer .

Answer 3

Yes, you can call setId(value) in any view with any (positive) integer value that you like and then find it in the parent container using findViewById(value) . Note that it is valid to call setId() with the same value for different sibling views, but findViewById() will return only the first one.