You must supply a resource ID for a TextView

问题

I searched existing tutorials related this question, And i followed that example. But still im getting same error. When i change the android.R.layout.simple_list_item1 line it won't work.

MyCode:

public class Lisearch extends Activity {
    private ListView lv;
    private EditText et;
    private String listview_array[] = { "ONE", "TWO", "THREE", "FOUR", "FIVE",
    "SIX", "SEVEN", "EIGHT", "NINE", "TEN" };
    private ArrayList<String> array_sort= new ArrayList<String>();
    int textlength=0;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_lisearch);
        //TextView tv=(TextView)findViewById(R.id.text);
        lv = (ListView) findViewById(R.id.ListView01);
        et = (EditText) findViewById(R.id.EditText01);
        //lv.setAdapter(new ArrayAdapter<String>(this,
        //R.layout.activity_lisearch, listview_array));

        et.addTextChangedListener(new TextWatcher()
        {
        public void afterTextChanged(Editable s)
        {
                                                                        // Abstract Method of TextWatcher Interface.
        }
        public void beforeTextChanged(CharSequence s,
        int start, int count, int after)
        {
        // Abstract Method of TextWatcher Interface.
        }
        public void onTextChanged(CharSequence s,
        int start, int before, int count)
        {
        textlength = et.getText().length();
        array_sort.clear();
        for (int i = 0; i < listview_array.length; i++)
        {
        if (textlength <= listview_array[i].length())
        {
        if(et.getText().toString().equalsIgnoreCase(
        (String)
        listview_array[i].subSequence(0,
        textlength)))
        {
                                                                                                                        array_sort.add(listview_array[i]);
                                                                                                        }
                                                                                        }
                                                                        }
        lv.setAdapter(new ArrayAdapter<String>
        (Lisearch.this,
        R.layout.activity_lisearch,R.id.text, array_sort));
        }
        });

    }

Xml file:

<?xml version="1.0" encoding="utf-8"?>
    <TextView xmlns:android="http://schemas.android.com/apk/res/android"
        android:id="@+id/text"
        android:layout_width="wrap_content"
android:layout_height="wrap_content" >
<EditText android:id="@+id/EditText01"
android:layout_height="wrap_content"
android:layout_width="fill_parent"
android:hint="Search">
</EditText>

<ListView android:id="@+id/ListView01"
android:layout_width="wrap_content"
android:layout_height="wrap_content" >
</ListView>
</TextView>

回答1:

Try this. Using android.R.layout_simple_list_item_1 shows default listview .

lv.setAdapter(new ArrayAdapter<String>(this,android.R.layout.simple_list_item_1,array_sort));

Set Custom Layout

lv.setAdapter(new ArrayAdapter<String>(this,R.layout.conversation_item,R.id.sometextview_from_the_layout,array_sort));

Here

R.layout.conversation_item is the custom layout and R.id.sometextview_from_the_layout is textview id from custom layout.

conversation_item.xml file

<?xml version="1.0" encoding="utf-8"?>
<TextView xmlns:android="http://schemas.android.com/apk/res/android"
    android:id="@+id/text1"
    android:layout_width="fill_parent"
    android:layout_height="wrap_content"
    android:textAppearance="?android:attr/textAppearanceLarge"
    android:gravity="center_vertical"
    android:paddingLeft="6dip"
    android:minHeight="?android:attr/listPreferredItemHeight"
/>

来源：https://stackoverflow.com/questions/12437196/you-must-supply-a-resource-id-for-a-textview

标签

android

android-listview

android-arrayadapter