小编Der*_*ter的帖子

如果我在shell中运行:

mongo ds0219xx.mlab.com:219xx/dbname -u user -p pass

它工作,并允许我连接到数据库并提取信息.但是如果我在我的python应用程序(Flask)中并运行它:

import pymongo

client = pymongo.MongoClient("mongodb://user:pass@ds0219xx.mlab.com:219xx/dbname")

db = client["dbname"]

db.users.insert_one({
  "user1": "hello"
})

它给了我一个:

pymongo.errors.OperationFailure: Authentication failed.

我很确定它在进入insert_one()调用之前就失败了,但我并不完全确定.

谢谢!

编辑:根据请求,这是完整的回调:

Traceback (most recent call last):
File "run.py", line 1, in <module>
from app import app
File "/Users/Derek/Documents/programming/shenalum/app/__init__.py", line 6, in <module>
from app import views
File "/Users/Derek/Documents/programming/shenalum/app/views.py", line 4, in <module>
from data import get_posts, get_user_info
File "/Users/Derek/Documents/programming/shenalum/app/data.py", line 9, in <module>
"user1": "hello"
File "/usr/local/lib/python2.7/site-packages/pymongo/collection.py", line 622, in insert_one
with self._socket_for_writes() as …

我用代码行得到'无法解决方法'错误:

FileInputStream fis = openFileInput(fileName);

我不是一个活动,我在一个单独的班级,所以我假设这就是为什么它不起作用.我尝试过这样的事情,但他们仍然给我'无法解决方法错误:

FileInputStream fis = getApplicationContext.openFileInput(fileName);

感谢您的帮助,我是Android开发的新手

我正在尝试绘制背景,然后将按钮放到面板上.如果没有绘制方法,按钮将正确放置在屏幕上,但是当绘制方法存在时,按钮不会显示,直到鼠标悬停在它们上面.我无法弄清楚为什么会这样.谢谢

这是在构造函数中:

setBorder(new EmptyBorder(40, 40, 40, 40));
setSize(1600, 1000);
setLayout(new GridLayout(4, 0, 40, 40));

for(int r = 0; r < rows; r++){
        for(int c = 0; c < cols; c++){
            levels[r][c] = new JButton(String.valueOf(levelNum));
            levels[r][c].setMargin(new Insets(50, 50, 50, 50));
            levels[r][c].addActionListener(e);
            levels[r][c].setBackground(Color.MAGENTA);
            this.add(levels[r][c]);
            levelNum++;
        }
}

然后是:

@Override
public void paint(Graphics g){

    g.setColor(Color.CYAN);
    g.fillRect(0, 0, this.getWidth(), this.getHeight());

    ... (just some basic fillRect()'s and things)
}