本文介绍了SQLiteException:临近"" :在编译语法错误(code 1)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有这样的错误在编译时和我不知道为什么,谁能帮助我?
I have this error while compiling and I don´t know why, can anyone help me?
公共静态最后弦乐TABLE_BEERS =cervezas;
public static final String TABLE_BEERS = "cervezas";
// Contacts Table Columns names
public static final String KEY_NAME = "_id";
public static final String KEY_COMPANY = "company";
public static final String KEY_TYPE = "type";
public static final String KEY_ALCOHOL = "alcohol";
public DatabaseHandler(Context context) {
super(context, DATABASE_NAME, null, DATABASE_VERSION);
// TODO Auto-generated constructor stub
}
@Override
public void onCreate(SQLiteDatabase db) {
String query = String.format("CREATE TABLE %s (%s STRING PRIMARY KEY,%s TEXT, %s TEXT, %s TEXT);",
TABLE_BEERS, KEY_NAME, KEY_COMPANY,
KEY_TYPE, KEY_ALCOHOL);
/*
String CREATE_BEER_TABLE = "create table " + TABLE_BEERS + "("
+ KEY_NAME + " STRING PRIMARY KEY,"
+ KEY_COMPANY + " TEXT,"
+ KEY_TYPE + " TEXT,"
+ KEY_ALCOHOL + " TEXT )";*/
db.execSQL(query);
这是创建表
public List<Cervezas> getCompanyCervezas(String compania){
List<Cervezas> cervezasList = new ArrayList<Cervezas>();
// Select All Query
String selectQuery = "SELECT _id, type, alcohol FROM " + TABLE_BEERS + " WHERE company=" + compania;
SQLiteDatabase db = this.getReadableDatabase();
Cursor cursor = db.rawQuery(selectQuery, null);
LogCat中
LogCat
android.database.sqlite.SQLiteException: no such column : Alean (code 1): , while compiling: SELECT _id, type, alcohol FROM cervezas WHERE company=Alean
这是怎么回事?
推荐答案
试试下面code: -
try below code:-
String selectQuery = "SELECT _id, type, alcohol FROM " + TABLE_BEERS + " WHERE company= ' " + compania+" ' ";
您错过的单引号。
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