How to add a TextView to LinearLayout in Android

Question

I am trying to add TextViews to my xml-defined layout in code. I have a xml-sheet, where a lot of Views are defined. But I have to add some views in code, so a create a LinearLayout in the xml-sheet:

<LinearLayout
xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="wrap_content"
android:id="@+id/info"
android:layout_height="wrap_content" 
android:orientation="vertical">
</LinearLayout>

And in this layout, I like to add my TextView:

    View linearLayout =  findViewById(R.id.info);
    //LinearLayout layout = (LinearLayout) findViewById(R.id.info);


    TextView valueTV = new TextView(this);
    valueTV.setText("hallo hallo");
    valueTV.setId(5);
    valueTV.setLayoutParams(new LayoutParams(
            LayoutParams.FILL_PARENT,
            LayoutParams.WRAP_CONTENT));

    ((LinearLayout) linearLayout).addView(valueTV);

But I only get the following error message:

: java.lang.ClassCastException: android.widget.TextView

How can I do it?

Thanks for you help. Martin

Which line is that exception for? It must be from the LinearLayout cast, are you sure the linearLayout variable is a LinearLayout and not a TextView? Also you shouldn't be specifying the Id since you can't guarantee it will be unique. — Robby Pond, Jul 08 '10 at 15:06
You are right, linearLayout is a TextView, but why? I have defined it in the xml-file as a LinearLayout ... — Martin, Jul 08 '10 at 15:14
Make sure you are really operating on the xml shown above. Is `setContentView(R.layout.your_xml_layout);` really loading the right xml? Do you have other xml layouts where you use `android:id="@+id/info"` which happen to be a TextView? — Rodja, Dec 01 '11 at 10:47
Is this issue resolved? Kindly accept as answer or post one. — Talha, Oct 13 '16 at 06:46

score 111 · Answer 1 · edited Feb 22 '14 at 21:37

111

try using

LinearLayout linearLayout = (LinearLayout)findViewById(R.id.info);
...
linearLayout.addView(valueTV);

also make sure that the layout params you're creating are LinearLayout.LayoutParams...

edited Feb 22 '14 at 21:37

Lucas

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answered Jul 08 '10 at 15:36

Ben

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122

there is again an exception, because findViewById returns an TextView. But why? I fetch it with the LinearLayout id.... what do you mean with: also make sure that the layout params you're creating are LinearLayout.LayoutParams... ??? – Martin Jul 08 '10 at 15:59
Why would that change anything? All it does is cast sooner rather than later, which should have the same effect. – yesennes May 19 '16 at 13:06

score 77 · Answer 2 · edited Jun 14 '16 at 09:29

Hey i have checked your code, there is no serious error in your code. this is complete code:

main.xml:-

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

this is Stackoverflow.java

import android.app.Activity;
import android.os.Bundle;
import android.view.View;
import android.view.ViewGroup.LayoutParams;
import android.widget.LinearLayout;
import android.widget.TextView;

public class Stackoverflow extends Activity {
    /** Called when the activity is first created. */
    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
        View linearLayout =  findViewById(R.id.info);
        //LinearLayout layout = (LinearLayout) findViewById(R.id.info);

        TextView valueTV = new TextView(this);
        valueTV.setText("hallo hallo");
        valueTV.setId(5);
        valueTV.setLayoutParams(new LayoutParams(LayoutParams.MATCH_PARENT,LayoutParams.WRAP_CONTENT));

        ((LinearLayout) linearLayout).addView(valueTV);
    }
}

copy this code, and run it. it is completely error free. take care...

+1. This answer is much better than @Ben's as it contains an example of LayoutParams and how the rest of the properties of the `TextView` is initialised. This should be marked as an answer. — Subby, Jun 30 '14 at 09:31

score 23 · Answer 3 · edited Nov 04 '16 at 11:00

for(int j=0;j<30;j++) {
    LinearLayout childLayout = new LinearLayout(MainActivity.this);
    LinearLayout.LayoutParams linearParams = new LinearLayout.LayoutParams(
        LayoutParams.WRAP_CONTENT,
        LayoutParams.WRAP_CONTENT);
    childLayout.setLayoutParams(linearParams);

    TextView mType = new TextView(MainActivity.this);
    TextView mValue = new TextView(MainActivity.this);

    mType.setLayoutParams(new TableLayout.LayoutParams(
        LayoutParams.WRAP_CONTENT,
        LayoutParams.WRAP_CONTENT, 1f));
    mValue.setLayoutParams(new TableLayout.LayoutParams(
        LayoutParams.WRAP_CONTENT,
        LayoutParams.WRAP_CONTENT, 1f));

    mType.setTextSize(17);
    mType.setPadding(5, 3, 0, 3);
    mType.setTypeface(Typeface.DEFAULT_BOLD);
    mType.setGravity(Gravity.LEFT | Gravity.CENTER);

    mValue.setTextSize(16);
    mValue.setPadding(5, 3, 0, 3);
    mValue.setTypeface(null, Typeface.ITALIC);
    mValue.setGravity(Gravity.LEFT | Gravity.CENTER);

    mType.setText("111");
    mValue.setText("111");

    childLayout.addView(mValue, 0);
    childLayout.addView(mType, 0);

    linear.addView(childLayout);
}

has to define, linear.addView(childLayout);, linear as Linear Layout at first — Noor Hossain, Nov 12 '20 at 13:37

score 23 · Answer 4 · edited Aug 18 '17 at 17:41

23

You can add a TextView to your linear layout programmatically like this:

LinearLayout linearLayout = (LinearLayout) findViewById(R.id.mylayout);
TextView txt1 = new TextView(MyClass.this);
linearLayout.setBackgroundColor(Color.TRANSPARENT);
linearLayout.addView(txt1);

edited Aug 18 '17 at 17:41

wbk727

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answered Feb 09 '12 at 14:01

Deepak

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What is MyClass.this? I'm new to Android development; am I supposed to substitute the name of my fragment class in for "MyClass"? – drusepth Jul 29 '12 at 19:38
1

MyClass.this is a context and a text view takes in a context – Mihai Bratulescu Aug 27 '13 at 07:03
2

Just for clarification: `MyClass.this`, in most cases, is the same as `this`. You need to specify, however, the name of the class, if you are in a nested class and want to access the instance of the "outer" class, which is very common when defining callbacks for the events in android. – drigoangelo Nov 19 '14 at 11:56
I guess some android developers got used to putting the name of the class where it's needed and started putting it everywhere. Also, `MyClass.this` is an instance of `MyClass`, and it will only be a context if MyClass implements Context (e.g. extends Activity) – drigoangelo Nov 19 '14 at 12:01

Salman Nazir · Answer 5 · 2021-01-19T15:06:43.743

In Kotlin you can add Textview as follows.

 val textView = TextView(activity).apply {
            layoutParams = LinearLayout.LayoutParams(
                LinearLayout.LayoutParams.MATCH_PARENT,
                LinearLayout.LayoutParams.WRAP_CONTENT
            ).apply {
                setMargins(0, 20, 0, 0)
                setPadding(10, 10, 0, 10)
            }
            text = "SOME TEXT"
            setBackgroundColor(ContextCompat.getColor(this@MainActivity, R.color.colorPrimary))
            setTextColor(ContextCompat.getColor(this@MainActivity, R.color.colorPrimaryDark))
            textSize = 16.0f
            typeface = Typeface.defaultFromStyle(Typeface.BOLD)
        }
 linearLayoutContainer.addView(textView)

score 11 · Answer 6 · answered Mar 07 '18 at 11:56

You should use something similar to this for adding TextView to LinearLayout dynamically:

LinearLayout linearLayout = getActivity().findViewById(R.id.infoLayout);

TextView valueTV = new TextView(context);
valueTV.setText("hallo hallo");
valueTV.setId(Integer.parseInt("5"));
valueTV.setLayoutParams(new LinearLayout.LayoutParams(
       LinearLayout.LayoutParams.FILL_PARENT,
       LinearLayout.LayoutParams.WRAP_CONTENT));

linearLayout.addView(valueTV);

getActivity() is used for inside Fragments, you can use context or anything similar per each instance you are inside.

score 3 · Answer 7 · answered Apr 09 '11 at 10:56

3

You need to access the layout via it's layout resource, not an id resource which is not guaranteed unique. The resource reference should look like R.layout.my_cool_layout where your above XML layout is stored in res/layout/my_cool_layout.xml.

answered Apr 09 '11 at 10:56

penguin359

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score 0 · Answer 8 · answered Aug 23 '13 at 22:42

0

LinearLayout.LayoutParams layoutParams ;
layoutParams= new LinearLayout.LayoutParams(LinearLayout.LayoutParams.FILL_PARENT, LinearLayout.LayoutParams.WRAP_CONTENT);

answered Aug 23 '13 at 22:42

Zapnologica

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score -1 · Answer 9 · edited Nov 29 '15 at 00:15

-1

Here's where the exception occurs

((LinearLayout) linearLayout).addView(valueTV);

addView method takes in a parameter of type View, not TextView. Therefore, typecast the valueTv object into a View object, explicitly.

Therefore, the corrected code would be :

((LinearLayout) linearLayout).addView((TextView)valueTV);

edited Nov 29 '15 at 00:15

sfmirtalebi

370
7
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answered Jan 10 '15 at 05:30

user3509153

343
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Have you tested this? – mrres1 Jan 10 '15 at 05:46
Yes, I have. And it works perfectly! Have you tried it? – user3509153 Jan 10 '15 at 20:21

How to add a TextView to LinearLayout in Android

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