2012-04-16 38 views
0

当我尝试通过使用'Washing Machine'作为搜索字符串来搜索我的数据库以查找'Washing Machine'的数据库条目时,出现一个错误消息:语法在数据库搜索条目时出错

04-16 21:43:28.951: E/AndroidRuntime(545): FATAL EXCEPTION: main 
04-16 21:43:28.951: E/AndroidRuntime(545): java.lang.RuntimeException: Unable to start activity ComponentInfo{com.lukeorpin.theappliancekeeper/com.lukeorpin.theappliancekeeper.EntryStatis tics}: android.database.sqlite.SQLiteException: near "Machine": syntax error: , while compiling: SELECT _id, appliance_name, appliance_wattage, energy_rates FROM ApplianceDetails WHERE appliance_name=Washing Machine 

,更具体地:

at com.lukeorpin.theappliancekeeper.Database.getWattage(Database.java:122) 
04-16 21:43:28.951: E/AndroidRuntime(545): at com.lukeorpin.theappliancekeeper.EntryStatistics.onCreate(EntryStatistics.java:47) 

这里是数据库的搜索查询代码:

public String getWattage(String spinnerChoice) { 
    // TODO Auto-generated method stub 
    String[] columns = new String[] { KEY_ROWID, KEY_NAME, KEY_WATTAGE, KEY_ENERGY}; 
    Cursor c = ourDatabase.query(DATABASE_TABLE, columns, KEY_NAME + "=" + spinnerChoice, null, null, null, null); 
    if (c != null){ 
     c.moveToFirst(); 
     String wattage = c.getString(2); 
     return wattage; 
    } 
    return null; 
} 

public String getEnergyRate(String spinnerChoice) { 
    // TODO Auto-generated method stub 
    String[] columns = new String[] { KEY_ROWID, KEY_NAME, KEY_WATTAGE, KEY_ENERGY}; 
    Cursor c = ourDatabase.query(DATABASE_TABLE, columns, KEY_NAME + "=" + spinnerChoice, null, null, null, null); 
    if (c != null){ 
     c.moveToFirst(); 
     String energyRate = c.getString(3); 
     return energyRate; 
    } 
    return null; 
} 

而这正是该方法创建原始类:

final String spinnerChoice = getIntent().getStringExtra("Name"); 
    if(spinnerChoice==null){ 
     return; 
    } 

Database data = new Database(this); 
    data.open(); 
    String returnedWattage = data.getWattage(spinnerChoice); 
    String returnedEnergyRate = data.getEnergyRate(spinnerChoice); 
    data.close(); 

任何人都不会有为什么这个错误信息出现的任何想法?由于

编辑:这里是代码的错误指向太行(行122):

Cursor c = ourDatabase.query(DATABASE_TABLE, columns, KEY_NAME + "=" + spinnerChoice, null, null, null, null); 
+0

您能显示导致错误的实际sql命令字符串吗? – 2012-04-16 21:54:18

+0

我已经把它放在原帖的末尾,谢谢 – Sketzii 2012-04-16 21:57:14

+0

我希望你已经做了一些调试,并且可以将原始字符串值拉出来让我们看看:) – 2012-04-16 22:01:44

回答

1

提供的错误消息,好像你没有引号表示为一个字符串appliance_name价值比较。它应该是

WHERE appliance_name = 'Washing Machine' 

我不熟悉的API,但你可以尝试改变线122

Cursor c = ourDatabase.query(DATABASE_TABLE, columns, KEY_NAME + "='" + spinnerChoice + "'", null, null, null, null); 
+0

是的,它现在正在工作!非常感谢,这是它的单引号标记:) – Sketzii 2012-04-16 22:10:56

1

你需要附上您搜索在单引号的字符串(在您的SQL查询) 。