Question

问题是数据库查询存在一些问题。我首先检查它是否存在于数据库中。如果它存在，那么我只是从数据库中取出后启动一个意图。如果它不存在则从web获取并放入数据库

这就是我称之为数据库功能的地方

public static void main(String[] args){
method();

boolean x = true;
    if(x == y()){
       System.out.println("You're right!");

    }else{
        System.out.println("You're wrong!");
        main(args);
  }
}


public static void method(){
    System.out.println("a+b=5 If a=2 what does b equal?"); 
}
public static boolean y(){
    try{
        Scanner s = new Scanner(System.in);
        String input = s.nextLine();
        int input2 = Integer.parseInt(input);
        int b = 3;
        return input2 == b;
}

        catch( Exception e ) {
            return false;
  }
 }
}

这是我的databaseChecker方法

protected void onListItemClick(ListView l, View v, int position, long id) {
    super.onListItemClick(l, v, position, id);
    urlBlogtitle = mainBlogPost.get(position).title; // getting title of the clicked item
    Log.d(TAG, urlBlogtitle);
    summarygetter=databaseChecker(urlBlogtitle);
    if(summarygetter=="") new DownloadXmlTaskContent(MainListActivity.this).execute(URL);
    else startingIntent(summarygetter);
    // intitating the main object for Content parsing
}

这是我的数据库方法

private String databaseChecker(String title) {

    DatabaseHandler db = new DatabaseHandler(MainListActivity.this);

    String content=" ";
    content = db.getContact(title);
    Log.d(TAG,content);
    return content;
}

这是错误的logcat视图。

String getContact(String title) {
    SQLiteDatabase db = this.getReadableDatabase();
    String content = " ";
    Cursor cursor = db.query(TABLE_CONTACTS, new String[] { KEY_ID,
                    KEY_TITLE, KEY_CONTENT }, KEY_ID + "=?",
            new String[] { title }, null, null, null, null);
    if (cursor != null){
        cursor.moveToFirst();}
    else
    {
        return content;
    }

    Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
            cursor.getString(1), cursor.getString(2));
    content=contact.title;
    // return contact
    return content;
}

Answer 1

问题是你没有检查cursor.moveToFirst()的返回值，如果没有条目，它将返回false。而不是：

if (cursor != null){
    cursor.moveToFirst();}
else
{
    return content;
}

Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
        cursor.getString(1), cursor.getString(2));
content=contact.title;
// return contact
return content;

试试这个：

if (cursor.moveToFirst()){
    Contact contact = new Contact(Integer.parseInt(cursor.getString(0)),
        cursor.getString(1), cursor.getString(2));
    content=contact.title;
    // return contact
    return content; 
} else{ 
   return "Whatever_you_return_when_not_found"
}

Sqllite数据库查询在获取时有一些问题

1 个答案: