(Android-intent)如何修复字符串R.id.chooser,以便java文件引用它

时间:2015-06-28 05:03:58

标签: android android-intent

我正在开发头脑中的第一个进入android开发的练习 并按照指定的书完成练习,但键入R.id.chooser表示,即使我在字符串XML中声明了选择符号,也无法进行重新分配。这很奇怪......有没有人建议解决这个问题?

java文件:

package com.hfad.messenger;

import android.app.Activity;
import android.os.Bundle;
import android.view.View;`enter code here`
import android.content.Intent;
import android.widget.EditText;

public class CreateMessageActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_create_message);
    }

    //Call onSendMessage() when the button is clicked
    public void onSendMessage(View view) {
        EditText messageView = (EditText)findViewById(R.id.message);
        String messageText = messageView.getText().toString();
        Intent intent = new Intent(Intent.ACTION_SEND);
        intent.setType("text/pain");
        intent.putExtra(Intent.EXTRA_TEXT, messageText);
        String chooserTitle = getString(R.string.chooser);
        Intent chosenIntent = Intent.createChooser(intent, chooserTitle);
        startActivity(chosenIntent);
    }
}

xml文件:

<resources>
    <string name="app_name">Messenger</string>
    <string name="send">Send Message</string>
    <String name="chooser">Send Message...</String>
    <string name="action_settings">Settings</string>
    <string name="title_activity_receive_message">ReceiveMessageActivity</string>
</resources>

1 个答案:

答案 0 :(得分:0)

您认为您已使用大写的第一个字母chooser宣布String,但它应该像string一样:

<resources>
    <string name="app_name">Messenger</string>
    <string name="send">Send Message</string>
    <string name="chooser">Send Message...</string>
    <string name="action_settings">Settings</string>
    <string name="title_activity_receive_message">ReceiveMessageActivity</string>
</resources>

这就是为什么Android Studio无法找到chooser字符串。