问题
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?
回答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 uniqueint
for use in code. - Reference your
android:id
'sint
value in code using "R.id.
somename" (effectively a constant.) - this
int
can change from build to build so never copy an id fromgen/
package.name/R.java
, just use "R.id.
somename". - (Also, an
id
assigned to aPreference
in XML is not used when thePreference
generates itsView
.)
Assign id
via code (programmatically)
- Manually set
id
s usingsomeView.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
-assignedid
s will be unique.- Code-assigned
id
s do not have to be unique - Code-assigned
id
s can (theoretically) conflict withXML
-assignedid
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 firstView
it finds with a matchingid
.- As long as there are no code-assigned
id
s assigned before an XML-definedid
in the hierarchy,findViewById(R.id.somename)
will always return the XML-defined View soid
'd.
Dynamically Creating Views and Assigning ID
s
- In layout XML, define an empty
ViewGroup
withid
. - Such as a
LinearLayout
withandroid:id="@+id/placeholder"
. - Use code to populate the placeholder
ViewGroup
withView
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
回答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
documentationThe 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.
回答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.
来源:https://stackoverflow.com/questions/8460680/how-can-i-assign-an-id-to-a-view-programmatically