dei*_*988 1 database sqlite android cursor
我正在尝试使用此查询从Android中的SQLite数据库中获取数据.
Cursor cursor = db.rawQuery("SELECT * FROM Task where taskId = 1 ", null);
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但是,如果光标返回空,则存在值为1的taskId.
如果我这样做
Cursor cursor = db.rawQuery("SELECT * FROM Task", null);
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我的光标包含所有值 - 包括值为1的taskId.
我也尝试了以下所有命令,它们都没有工作:
Cursor cursor = db.rawQuery("SELECT * FROM Task where taskId = " + 1, null);
Cursor cursor = db.rawQuery("SELECT * FROM Task where taskId = " + "'1'", null);
Cursor cursor = db.rawQuery("SELECT * FROM Task where taskId = ?", new String[]{"1"});
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taskId是Integer类型,也是用Text试过的,也没用.
有没有我没考虑过的事情?帮助将非常感谢!
编辑:创建数据库的代码:
CREATE TABLE Task + " (" +
_id INTEGER PRIMARY KEY," +
"taskDescription TEXT, +
"taskId INTEGER," +
"taskName TEXT" +
"PreviousTaskID " + "REFERENCES " + "PreviousTasks " +
"(" + PreviousTasks._ID + "))"
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Cursor cursor = null;
String Query ="SELECT * FROM Task where taskId = 1 ";
cursor = sqldb.rawQuery(Query, null);
if (cursor != null && cursor.moveToFirst()) {
do {
// ...
} while (cursor.moveToNext());
cursor.close();
}
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我希望它对你有用..
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