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android - 如何从 userdictionary content provider 检索单词

转载 作者:行者123 更新时间:2023-11-30 04:02:04 25 4
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我在使用内容解析器从 UserDictionary.words 内容提供程序获取数据时遇到问题。

这是我的要求:1. 我想找出与给定单词匹配的 UserDictionary 单词。

我的实现;1.我的应用有两个屏幕( Activity )2. 我在第一个屏幕上输入一个词并将其作为 Intent 传递给第二个屏幕。在第二个屏幕上,我想显示与第一个屏幕上给定单词匹配的 Userdictionary 单词列表。

我的问题是...我能够将单词传递给第二个 Activity ,但错过了检索 userDicitnary 记录的机会。我给它的任何词都是返回记录。这是我的两个 Activity 的代码。请帮我解决这个问题。

第一个 Activity :

public void onCreate(Bundle savedInstanceState) {

Log.d(TAG,"onCreate");
super.onCreate(savedInstanceState);
setContentView(R.layout.activity_content_user_demo);
}

public void SendWord(View view){
Log.d(TAG,"SendWord");
EditText SearchWord = (EditText) findViewById(R.id.mSearchWord);
String mSearchString = SearchWord.getText().toString();
Intent intent = new Intent(this, WordListActivity.class);
Log.d(TAG,"EXTRA_MESSAGE");
intent.putExtra(EXTRA_MESSAGE, mSearchString);
startActivity(intent);
}

第二个 Activity :

public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
Log.d(TAG,"onCreate");
setContentView(R.layout.activity_word_list);
Intent intent = getIntent();
Log.d(TAG,"getIntent() EXTRA_MESSAGE");

String searchWord = intent.getStringExtra(ContentUserDemo.EXTRA_MESSAGE);
textView = new TextView(this);
Log.d(TAG,"searchWord " +searchWord);
textView.setTextSize(20);
setContentView(textView);
Log.d(TAG,"textSize " +textView.length());
String[] mProjection =
{
UserDictionary.Words._ID,
UserDictionary.Words.WORD,
UserDictionary.Words.LOCALE
};
String mSelectionClause = null;
String[] mSelectionArgs = {""};
String mSortOrder=null;
ContentResolver cr = getContentResolver();
if (TextUtils.isEmpty(searchWord)) {
mSelectionClause = null;
mSelectionArgs[0] = "";
}
else {
mSelectionClause = UserDictionary.Words.WORD + " = ?";
mSelectionArgs[0] = searchWord;
}
Cursor mCursor = cr.query(UserDictionary.Words.CONTENT_URI,
mProjection,
mSelectionClause,
mSelectionArgs,
mSortOrder);
startManagingCursor(mCursor);


int count = mCursor.getCount();


if (null == mCursor) {
Log.d(TAG, "cursor.getCount()=" + count);
String message="No word matches with "+ searchWord;
textView.setText(message);
}
else if (mCursor.getCount() < 1) {
Log.d(TAG, "cursor.getCount()=" + count);
String message="getCount is less than 1 " + "No word matches with "+ searchWord;
textView.setText(message);

}
else {
String message="Else part of the code";
textView.setText(message);
String[] mWordListColumns ={UserDictionary.Words.WORD,UserDictionary.Words.LOCALE};
int[] mWordListItems = { R.id.dictWord, R.id.locale};
SimpleCursorAdapter mCursorAdapter = new SimpleCursorAdapter(
getApplicationContext(),
R.layout.wordlistrow,
mCursor,
mWordListColumns,
mWordListItems,
0);
ListView mWordList = (ListView) findViewById(R.id.mWordList);
mWordList.setAdapter(mCursorAdapter);

}
}

最佳答案

TextUtils.isEmpty() {} 子句中,我建议更改:

mSelectionArgs[0] = "";mSelectionArgs = null;

因为可能存在非法参数异常:

java.lang.IllegalArgumentException: 
Cannot bind argument at index 1 because the index is out of range.
The statement has 0 parameters.

关于android - 如何从 userdictionary content provider 检索单词,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12338943/

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