- Java code
loginTest = SQLiteDatabase.openOrCreateDatabase("data/data/com.stephen.sqltest/databases/Login_test_db",null); String sql = "Select pass from user where name="+user; int tep=0; Cursor cursor = loginTest.rawQuery(sql,null); cursor.moveToFirst(); try { int tepm =cursor.getInt(cursor.getColumnIndex("pass")); Toast.makeText(SqlTest.this, tepm+" ", Toast.LENGTH_SHORT).show(); } catch (Exception e) { // TODO Auto-generated catch block e.printStackTrace(); Toast.makeText(SqlTest.this, tep+" "+"下方", Toast.LENGTH_SHORT).show(); }
CREATE TABLE android_metadata (locale TEXT);
CREATE TABLE user(id int,name varchar(20),pass int);
sqlite> insert into user values( 1,'stephen',123456)
insert into user values( 1,'stephen',123456);
sqlite> select * from user;
select * from user;
1|stephen|123456
------解决方案--------------------
把user变量换下名字,如果还不行,user表的名字也换下,好像用user做表名会存在问题,记得再access数据库中是这样的。
另外,把你代码中的sql文件拷贝到终端中或者把db pull到pc上,用工具查看,看看能不能查出来。
"Select pass from user where name="+user;