小编Shi*_*ade的帖子

语法错误行继续符后的意外字符python

我的代码是

import MySQLdb
import collections
from pymongo import Connection

from config.MySQLdb import *
from config.MongoDB import *
from config.SyncTables import *

db = MySQLdb.connect(DB_HOST,DB_USR,DB_PWD,DB_NAME)
cursor = db.cursor()
query = "SELECT * FROM table_name WHERE userid = 1"
cursor.execute(query)
row = cursor.fetchall()

mconn = Connection(MONGODB_HOST,MONGODB_PORT)
mdb = mconn[MONGODB_DBNAME]
#col = mdb[trngl_advertiser_agegroup]

db.trngl_advertiser_agegroup .insert({'id': row[0],'userid':row[1], 'ageid':row[2],       'age_value':row[3], \                                             'created':row[4],  'modified':row[5]})
print "after update"
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但是我遇到这个错误,

triongle@triongle.com [~/download/DataInsertionScript]# python IngestDataToMongo.py

  File "IngestDataToMongo.py", line 21
    db.table_name.insert({'id': row[0],'userid':row[1], 'ageid':row[2],  'age_value':row[3],\                                                       'created':row[4], 'modified':row[5]})
                                                                                                                                                                                               ^

SyntaxError: unexpected character …
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python character syntax-error

2
推荐指数
1
解决办法
2万
查看次数

'module'对象在pymongo中没有属性'OrderedDict'错误

我对pymongo很新.在以下代码中,

db = MySQLdb.connect(DB_HOST,DB_USR,DB_PWD,DB_NAME)
cursor = db.cursor()
query = "SELECT * FROM %s WHERE userid = \"%s\"" % (table, userID) 
cursor.execute(query)
colNames = [i[0] for i in cursor.description]
rows = cursor.fetchall()
objects_list = []
# The below logic makes JSON objet based on fetch MySQL rows.
for row in rows:
    d = collections.OrderedDict()
    index = 0
    for col in colNames:
        d[col] = row[index]
        index = index + 1
    objects_list.append(d)
return objects_list
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我收到了错误,

trngl_advertise_perfm
trngl_advertise_activity
trngl_user_fblike
Traceback (most recent call last): …
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python collections runtime-error mongodb pymongo

2
推荐指数
1
解决办法
8002
查看次数

getAllNetworkInfo()已弃用

我写了这段代码.

ConnectivityManager connectivity = (ConnectivityManager)
            context.getSystemService(Context.CONNECTIVITY_SERVICE); 

    if (connectivity != null) {
        NetworkInfo[] info = connectivity.getAllNetworkInfo();
        if (info != null) {
            for (int i = 0; i < info.length; i++) {
                if (info[i].getState() == NetworkInfo.State.CONNECTED) {}
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但是不推荐使用getAllNetworkInfo().请帮忙.谢谢.

connection networking android android-6.0-marshmallow

1
推荐指数
1
解决办法
4763
查看次数

AttributeError:'module'对象在访问命令行参数python时没有属性'arvg'

我访问命令行参数的代码:

length = len(sys.argv)
print length
str = sys.arvg[0]
print str
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给出输出:

triongle@triongle.com [~/download/DataInsertionScript]# python IngestDataToMongo.py python_sucks
2
Traceback (most recent call last):
File "IngestDataToMongo.py", line 83, in <module>
str = sys.arvg[0]
AttributeError: 'module' object has no attribute 'arvg'
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所以,第一次len(sys.argv)工作,但奇怪的是下次在sys.argv [0]上给出错误.那为什么呢?

python attributes runtime-error argv command-line-arguments

0
推荐指数
1
解决办法
8616
查看次数

列表视图项单击未在android中调用

我正在展示一份清单.在onCreate我有代码

listView = (ListView) findViewById(R.id.list_view);

    listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
        @Override
        public void onItemClick(AdapterView<?> arg0, View arg1, int position, long arg3) {

            Log.v("list clicked", "list item no: " + position);
            Toast.makeText(DisplayImagesActivity.this, "Btn delete position: " + position, Toast.LENGTH_LONG).show();

            Intent intentImageDetails = new Intent(DisplayImagesActivity.this, ImageDetailActivity.class);

            intentImageDetails.putExtra("image", imageBitmap);
            intentImageDetails.putExtra("name", listImageInfo.get(position).getImage_name());
            intentImageDetails.putExtra("amount", listImageInfo.get(position).getAmount());
            intentImageDetails.putExtra("desc", listImageInfo.get(position).getImage_description());
            intentImageDetails.putExtra("id", listImageInfo.get(position).getImage_id());
            intentImageDetails.putExtra("date", listImageInfo.get(position).getDate_created());

            startActivity(intentImageDetails);
        }
    });
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和适配器代码是

public class MyImageListAdapter extends BaseAdapter {

    private LayoutInflater inflater = null;

    public MyImageListAdapter() {
        inflater = (LayoutInflater) DisplayImagesActivity.this.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
    }

    @Override
    public int …
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events android click onclick android-listview

0
推荐指数
1
解决办法
1655
查看次数

如何在不使用Intent的情况下从任何其他Activity打开Home Activity

在一次采访中有人问我这个问题.假设申请中有4项活动.他希望后退按钮能够在前三个活动中正常工作(即按后退按钮进入上一个屏幕); 除了最后一个.

他想要的是,当他按下第四个活动(屏幕)上的按钮时,用户应该转到主页活动(开始屏幕或第一个屏幕).但是当用户从第一个屏幕进入第二个屏幕时,Home Activity上的内容应该相同.所以我认为,我不能使用Intent,因为它将创建Starting Activity的新实例.

android launching button back android-activity

0
推荐指数
1
解决办法
861
查看次数

为什么我要旋转图像?

在我的应用程序中,我需要从相机加载图像。

这是我使用的代码:

private void openCamera()
{
    mMediaUri =getOutputMediaFileUri(MEDIA_TYPE_IMAGE);

    Intent photoIntent = new Intent(MediaStore.ACTION_IMAGE_CAPTURE);
    photoIntent.putExtra(MediaStore.EXTRA_OUTPUT, mMediaUri);
    startActivityForResult(photoIntent, REQUEST_TAKE_PHOTO);


}


private Uri getOutputMediaFileUri(int mediaTypeImage)
{
    //check for external storage
    if(isExternalStorageAvaiable())
    {
        File mediaStorageDir = getActivity().getExternalFilesDir(Environment.DIRECTORY_PICTURES);

        String fileName = "";
        String fileType = "";
        String timeStamp = new SimpleDateFormat("yyyyMMdd_HHmmss").format(new java.util.Date());

        fileName = "IMG_"+timeStamp;
        fileType = ".jpg";

        File mediaFile;
        try
        {
            mediaFile = File.createTempFile(fileName,fileType,mediaStorageDir);
            Log.i("st","File: "+Uri.fromFile(mediaFile));
        }
        catch (IOException e)
        {
            e.printStackTrace();
            Log.i("St","Error creating file: " + mediaStorageDir.getAbsolutePath() +fileName +fileType);
            return null;
        }
        return …
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android picasso

-1
推荐指数
1
解决办法
537
查看次数