【问题标题】:Logically true but 'if statement' with a method making my app crash逻辑上正确但“if 语句”的方法使我的应用程序崩溃
【发布时间】:2012-10-18 15:54:07
【问题描述】:

我的 MapActivity 中有一个 AsyncTask 类。在 doInBacground() 方法中,我必须使用代码块。

看起来像这样:

        @Override
        protected List<GeoPoint> doInBackground(String... arg0) {


            // check if the query (word and the distance) existing queries and return its id 
            int checkQuery = db.getQueryMatch(arg0[3].toString() , arg0[2].toString());
            Log.v(" CheckQuery int ", " equal? " + checkQuery );

            if (checkQuery  < 1) 
            {
                Block 1
            }
            else
            { 
                 Block 2
            }

getQueryMatch 方法将检查是否存在两个文本字段,然后返回记录 ID。

int getQueryMatch(String word, String meter) {
    SQLiteDatabase db = this.getReadableDatabase();
    int getidd = 0; 
    Cursor cursor = db.query(TABLE_HISTORY, new String[] { KEY_ID, QUERYWORD, DISTANCEQ, KEY_TIMESTAMP }, QUERYWORD + "=? AND " + DISTANCEQ + "=?", new String[] { word, meter  }, null, null, null, null);
    //String selectQuery = "SELECT *  FROM " + TABLE_HISTORY + " WHERE " + QUERYWORD +  "= '" + word + "' AND " + DISTANCEQ + "= '"+ meter +"'"   ;
    if (cursor != null)
        cursor.moveToFirst();
    getidd = Integer.parseInt(cursor.getString(0));

    db.close(); // Closing database connection

    // return History
    return getidd;
}

当我在 AsyncTask 类中不使用 if 语句 对其进行测试时,它工作得很好。如果返回值为 0,它会使应用程序崩溃。 所以我想做的是如果没有从数据库中找到记录(通过 getQueryMatch 检查)运行块 1,否则运行块 2。

以下是错误:

10-19 02:47:50.464: E/AndroidRuntime(11967): FATAL EXCEPTION: AsyncTask #2
10-19 02:47:50.464: E/AndroidRuntime(11967): java.lang.RuntimeException: An error occured while executing doInBackground()
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.os.AsyncTask$3.done(AsyncTask.java:299)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.FutureTask$Sync.innerSetException(FutureTask.java:273)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.FutureTask.setException(FutureTask.java:124)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:307)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.FutureTask.run(FutureTask.java:137)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.os.AsyncTask$SerialExecutor$1.run(AsyncTask.java:230)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1076)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:569)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.lang.Thread.run(Thread.java:856)
10-19 02:47:50.464: E/AndroidRuntime(11967): Caused by: android.database.CursorIndexOutOfBoundsException: Index 0 requested, with a size of 0
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.database.AbstractCursor.checkPosition(AbstractCursor.java:418)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.database.AbstractWindowedCursor.checkPosition(AbstractWindowedCursor.java:136)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.database.AbstractWindowedCursor.getString(AbstractWindowedCursor.java:50)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at com.example.phooogle.DatabaseHandler.getQueryMatch(DatabaseHandler.java:148)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at com.example.phooogle.GoogleMapsAppActivity$InitTask.doInBackground(GoogleMapsAppActivity.java:142)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at com.example.phooogle.GoogleMapsAppActivity$InitTask.doInBackground(GoogleMapsAppActivity.java:1)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at android.os.AsyncTask$2.call(AsyncTask.java:287)
10-19 02:47:50.464: E/AndroidRuntime(11967):    at java.util.concurrent.FutureTask$Sync.innerRun(FutureTask.java:305)

对我来说,这在逻辑上是正确的,但不起作用!

【问题讨论】:

    标签: android sqlite android-asynctask


    【解决方案1】:

    您需要检查 movetofirst 返回的值。如果光标为空,则返回 false

       boolean notEmpty = cursor.moveToFirst();
       if(notEmpty)
            getidd = Integer.parseInt(cursor.getString(0));
    

    编辑:

    getidd = -1;
     if (cursor != null){
       boolean notEmpty = cursor.moveToFirst();
       if(notEmpty)
            getidd = Integer.parseInt(cursor.getString(0));    
    }
     db.close(); // Closing database connection
    // return History
    return getidd;
    

    【讨论】:

    • 在这种情况下,如果它是真的,我如何返回 id?
    • 对不起,我不得不改变其他东西,现在它工作得非常完美。非常感谢你:)
    【解决方案2】:

    显然,cursor.getString(0) 失败,因为没有任何数据。

    你也要检查cursor.moveToFirst()的返回值。

    【讨论】:

    • 我这样做了。在方法中看到这一行: if (cursor != null) cursor.moveToFirst();
    猜你喜欢
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 2015-09-28
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    • 1970-01-01
    相关资源
    最近更新 更多