如何使用该svn export命令从存储库中获取单个文件?
我试过这个:
svn export e:\repositories\process\test.txt c:\
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但我得到这个错误:
svn:e:\ repositories\process不是工作副本
我怎样才能做到这一点:
Class A : DependencyObject {}
Class B : DependencyObject {}
Class C : A , B {}
Run Code Online (Sandbox Code Playgroud) 有图书馆可以做这个pyes和pyelasticsearch.pyelasticsearch 网站看起来不错,pyes采取其他方法,但也没关系.
另一方面,这段代码有效,而且非常简单.
import urllib2 as urllib
import json
import pprint
query = {
"from":0,
"size":10,
"query":{
"field" : {
"name" : "david"
}
},
"sort":[
{"name":"asc"},
{"lastName":"asc"}
]
}
query = json.dumps(query)
response = urllib.urlopen(
'http://localhost:9200/users/users/_search',
query
)
result = json.loads( response.read() )
pprint.pprint(result)
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所以我正在考虑使用简单的代码而不是学习库的技巧.
我想将树视图绑定到类似这样的类:
public class Folder : Base_FileFolder
{
public Folder()
{
Folders = new ObservableCollection<Folder>();
Files = new ObservableCollection<File>();
}
public ObservableCollection<Folder> Folders { get; set; }
public ObservableCollection<File> Files { get; set; }
}
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其他类是:
public class File : Base_FileFolder
{
}
public class Base_FileFolder : DependencyObject
{
public string Name
{
get { return (string)GetValue(NameProperty); }
set { SetValue(NameProperty, value); }
}
public static readonly DependencyProperty NameProperty = DependencyProperty.Register("Name", typeof(string), typeof(Base_FileFolder), new UIPropertyMetadata(""));
}
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如何创建显示"文件和文件夹"集合的树视图
我想用这样的东西:
<HierarchicalDataTemplate
DataType="{x:Type model:Folder}" …Run Code Online (Sandbox Code Playgroud) 我想读一个文件然后把它写回来.这是我的代码:
file = open( zipname , 'r' )
content = file.read()
file.close()
alt = open('x.zip', 'w')
alt.write(content )
alt.close()
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这不起作用,为什么?????
编辑:
重写的文件已损坏(Windows上的python 2.7.1)
我下载 OpenCV (OpenCV-2.3.1-win-superpack.exe) 并在 python 2.6 上使用它。
我使用find_obj.py目录中的OpenCV\samples\python。
一切都很神奇,但即使我使用 try except 块来包装代码,如果代码引发错误,OpenCV 也会打印错误。
OpenCV错误:未知函数中的错误标志(参数或结构字段)(无法识别或不支持的数组类型),文件......\OpenCV-2.3.1\modules\core\src\array.cpp,第2482行
那么如何禁用 OpenCV 警告消息呢?
编辑:
这是代码:
import numpy as np
import cv2
from functools import partial
help_message = '''SURF image match
USAGE: findobj.py [ <image1> <image2> ]
'''
FLANN_INDEX_KDTREE = 1 # bug: flann enums are missing
flann_params = dict(algorithm = FLANN_INDEX_KDTREE, trees = 4)
def anorm(a):
return np.sqrt( anorm2(a) )
def match_bruteforce(desc1, desc2, r_threshold = 0.75):
res = []
for i in …Run Code Online (Sandbox Code Playgroud) 我有两个网络摄像头,我想拍照.以下代码实现了:
import pygame
import pygame.camera
from datetime import datetime
import Image
import threading
import time
class Camera (threading.Thread):
def __init__(self, camera):
self.camera = pygame.camera.Camera(camera,(2304,1536))
self.stop = False
threading.Thread.__init__(self)
def run(self):
self.camera.start()
srf = self.camera.get_image()
img = pygame.image.tostring(srf, 'RGB')
img = Image.fromstring('RGB', srf.get_size(), img)
img.save('%s.png'%datetime.now(), 'PNG')
s = datetime.now()
pygame.init()
pygame.camera.init()
cam1 = Camera("/dev/video0")
cam2 = Camera("/dev/video1")
cam1.start()
cam2.start()
cam1.join()
cam2.join()
print datetime.now() - s
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但我需要设定焦点.
我找到了这些命令行:
apt-get install uvcdynctrl
uvcdynctrl --device=/dev/video1 --clist
uvcdynctrl --device=/dev/video1 --get='Focus, Auto'
uvcdynctrl --device=/dev/video1 --set='Focus, Auto' …Run Code Online (Sandbox Code Playgroud) 我有成千上万个简单对象的数组,我想处理它.
例:
var array = [
{name:"1"}, {name:"2"}, ..., {name:"32,000"}
]
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我想只处理数组(我在其他地方引用了对象).那么,下一个例子是否足以杀死阵列?
例:
var a1 = {name:"1"}, a2 = {name:"2"}, ... , a32000 = {name:"32,000"};
var array = [ a1, a2, ... , a32000 ];
array = null;
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或者我应该做以下事情:
for(var index in array) array[index] = null;
array = null;
Run Code Online (Sandbox Code Playgroud) MS-DOS命令attrib更改单个文件的属性.如何使用它来更改一组文件的属性?
我有这个伟大的pyodbc lib.我尝试下面的代码,它应该插入一行并返回行ID但它不起作用.顺便说一句,我在服务器上使用sql server 2005,客户端是windows os
...
con = pyodbc.connect('conectionString', autocommit = True)
cur = con.execute(
"insert into sometable values('something');
select scope_identity() as id"
)
for id in cur:
print id
...
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一些想法?
我需要在ini文件中更改一个值,这样val2就是要更改的那个
之前
[section1]
var1=val1
var2=val2
var3=va3
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后:
[section1]
var1=val1
var2=value
var3=va3
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