我想通过python脚本截取屏幕截图并且不引人注意地保存它。
我只对Linux解决方案感兴趣，并且应该支持任何基于X的环境。
答案 0 :(得分：64)
无需使用scrot或ImageMagick即可使用。
import gtk.gdk
w = gtk.gdk.get_default_root_window()
sz = w.get_size()
print "The size of the window is %d x %d" % sz
pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False,8,sz[0],sz[1])
pb = pb.get_from_drawable(w,w.get_colormap(),0,0,0,0,sz[0],sz[1])
if (pb != None):
pb.save("screenshot.png","png")
print "Screenshot saved to screenshot.png."
else:
print "Unable to get the screenshot."
借鉴http://ubuntuforums.org/showpost.php?p=2681009&postcount=5
答案 1 :(得分：46)
在一个班级中编译所有答案。 输出PIL图像。
#!/usr/bin/env python
# encoding: utf-8
"""
screengrab.py
Created by Alex Snet on 2011-10-10.
Copyright (c) 2011 CodeTeam. All rights reserved.
"""
import sys
import os
import Image
class screengrab:
def __init__(self):
try:
import gtk
except ImportError:
pass
else:
self.screen = self.getScreenByGtk
try:
import PyQt4
except ImportError:
pass
else:
self.screen = self.getScreenByQt
try:
import wx
except ImportError:
pass
else:
self.screen = self.getScreenByWx
try:
import ImageGrab
except ImportError:
pass
else:
self.screen = self.getScreenByPIL
def getScreenByGtk(self):
import gtk.gdk
w = gtk.gdk.get_default_root_window()
sz = w.get_size()
pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False,8,sz[0],sz[1])
pb = pb.get_from_drawable(w,w.get_colormap(),0,0,0,0,sz[0],sz[1])
if pb is None:
return False
else:
width,height = pb.get_width(),pb.get_height()
return Image.fromstring("RGB",(width,height),pb.get_pixels() )
def getScreenByQt(self):
from PyQt4.QtGui import QPixmap, QApplication
from PyQt4.Qt import QBuffer, QIODevice
import StringIO
app = QApplication(sys.argv)
buffer = QBuffer()
buffer.open(QIODevice.ReadWrite)
QPixmap.grabWindow(QApplication.desktop().winId()).save(buffer, 'png')
strio = StringIO.StringIO()
strio.write(buffer.data())
buffer.close()
del app
strio.seek(0)
return Image.open(strio)
def getScreenByPIL(self):
import ImageGrab
img = ImageGrab.grab()
return img
def getScreenByWx(self):
import wx
wx.App() # Need to create an App instance before doing anything
screen = wx.ScreenDC()
size = screen.GetSize()
bmp = wx.EmptyBitmap(size[0], size[1])
mem = wx.MemoryDC(bmp)
mem.Blit(0, 0, size[0], size[1], screen, 0, 0)
del mem # Release bitmap
#bmp.SaveFile('screenshot.png', wx.BITMAP_TYPE_PNG)
myWxImage = wx.ImageFromBitmap( myBitmap )
PilImage = Image.new( 'RGB', (myWxImage.GetWidth(), myWxImage.GetHeight()) )
PilImage.fromstring( myWxImage.GetData() )
return PilImage
if __name__ == '__main__':
s = screengrab()
screen = s.screen()
screen.show()
答案 2 :(得分：34)
为了完整性： Xlib - 但是在捕获整个屏幕时它有点慢：
from Xlib import display, X
import Image #PIL
W,H = 200,200
dsp = display.Display()
root = dsp.screen().root
raw = root.get_image(0, 0, W,H, X.ZPixmap, 0xffffffff)
image = Image.fromstring("RGB", (W, H), raw.data, "raw", "BGRX")
image.show()
可以尝试在PyXlib中的瓶颈文件中输入一些类型，然后使用Cython进行编译。这可能会提高速度。
修改 我们可以在C中编写函数的核心，然后在ctypes的python中使用它，这是我一起攻击的东西：
#include <stdio.h>
#include <X11/X.h>
#include <X11/Xlib.h>
//Compile hint: gcc -shared -O3 -lX11 -fPIC -Wl,-soname,prtscn -o prtscn.so prtscn.c
void getScreen(const int, const int, const int, const int, unsigned char *);
void getScreen(const int xx,const int yy,const int W, const int H, /*out*/ unsigned char * data)
{
Display *display = XOpenDisplay(NULL);
Window root = DefaultRootWindow(display);
XImage *image = XGetImage(display,root, xx,yy, W,H, AllPlanes, ZPixmap);
unsigned long red_mask = image->red_mask;
unsigned long green_mask = image->green_mask;
unsigned long blue_mask = image->blue_mask;
int x, y;
int ii = 0;
for (y = 0; y < H; y++) {
for (x = 0; x < W; x++) {
unsigned long pixel = XGetPixel(image,x,y);
unsigned char blue = (pixel & blue_mask);
unsigned char green = (pixel & green_mask) >> 8;
unsigned char red = (pixel & red_mask) >> 16;
data[ii + 2] = blue;
data[ii + 1] = green;
data[ii + 0] = red;
ii += 3;
}
}
XDestroyImage(image);
XDestroyWindow(display, root);
XCloseDisplay(display);
}
然后是python-file：
import ctypes
import os
from PIL import Image
LibName = 'prtscn.so'
AbsLibPath = os.path.dirname(os.path.abspath(__file__)) + os.path.sep + LibName
grab = ctypes.CDLL(AbsLibPath)
def grab_screen(x1,y1,x2,y2):
w, h = x2-x1, y2-y1
size = w * h
objlength = size * 3
grab.getScreen.argtypes = []
result = (ctypes.c_ubyte*objlength)()
grab.getScreen(x1,y1, w, h, result)
return Image.frombuffer('RGB', (w, h), result, 'raw', 'RGB', 0, 1)
if __name__ == '__main__':
im = grab_screen(0,0,1440,900)
im.show()
答案 3 :(得分：18)
这个适用于X11，也许适用于Windows（有人，请检查）。需要PyQt4：
import sys
from PyQt4.QtGui import QPixmap, QApplication
app = QApplication(sys.argv)
QPixmap.grabWindow(QApplication.desktop().winId()).save('test.png', 'png')
答案 4 :(得分：15)
我有一个用于scrot，imagemagick，pyqt，wx和pygtk的包装器项目（pyscreenshot）。 如果你有其中一个，你可以使用它。 所有解决方案都包含在本次讨论中。
安装：
easy_install pyscreenshot
示例：
import pyscreenshot as ImageGrab
# fullscreen
im=ImageGrab.grab()
im.show()
# part of the screen
im=ImageGrab.grab(bbox=(10,10,500,500))
im.show()
# to file
ImageGrab.grab_to_file('im.png')
答案 5 :(得分：8)
使用wxPython的跨平台解决方案：
import wx
wx.App() # Need to create an App instance before doing anything
screen = wx.ScreenDC()
size = screen.GetSize()
bmp = wx.EmptyBitmap(size[0], size[1])
mem = wx.MemoryDC(bmp)
mem.Blit(0, 0, size[0], size[1], screen, 0, 0)
del mem # Release bitmap
bmp.SaveFile('screenshot.png', wx.BITMAP_TYPE_PNG)
答案 6 :(得分：7)
import ImageGrab
img = ImageGrab.grab()
img.save('test.jpg','JPEG')
这需要Python Imaging Library
答案 7 :(得分：3)
短暂的搜索gtkShots看起来可能对你有所帮助，因为它是一个GPL的python截图程序，所以应该有你需要的内容。
答案 8 :(得分：3)
您可以使用此
import os
os.system("import -window root screen_shot.png")
答案 9 :(得分：2)
这个Autopy
有一个python包位图模块可以屏幕抓取（bitmap.capture_screen） 它是多平台（Windows，Linux，Osx）。
答案 10 :(得分：2)
有点迟了但是从来没有轻易的是
import autopy
import time
time.sleep(2)
b = autopy.bitmap.capture_screen()
b.save("C:/Users/mak/Desktop/m.png")
答案 11 :(得分：1)
对于ubuntu来说，我可以用以下方法为选择窗口截图：
import gi
gi.require_version('Gtk', '3.0')
from gi.repository import Gdk
from gi.repository import GdkPixbuf
import numpy as np
from Xlib.display import Display
#define the window name
window_name = 'Spotify'
#define xid of your select 'window'
def locate_window(stack,window):
disp = Display()
NET_WM_NAME = disp.intern_atom('_NET_WM_NAME')
WM_NAME = disp.intern_atom('WM_NAME')
name= []
for i, w in enumerate(stack):
win_id =w.get_xid()
window_obj = disp.create_resource_object('window', win_id)
for atom in (NET_WM_NAME, WM_NAME):
window_name=window_obj.get_full_property(atom, 0)
name.append(window_name.value)
for l in range(len(stack)):
if(name[2*l]==window):
return stack[l]
window = Gdk.get_default_root_window()
screen = window.get_screen()
stack = screen.get_window_stack()
myselectwindow = locate_window(stack,window_name)
img_pixbuf = Gdk.pixbuf_get_from_window(myselectwindow,*myselectwindow.get_geometry())
将pixbuf转换为数组
def pixbuf_to_array(p):
w,h,c,r=(p.get_width(), p.get_height(), p.get_n_channels(), p.get_rowstride())
assert p.get_colorspace() == GdkPixbuf.Colorspace.RGB
assert p.get_bits_per_sample() == 8
if p.get_has_alpha():
assert c == 4
else:
assert c == 3
assert r >= w * c
a=np.frombuffer(p.get_pixels(),dtype=np.uint8)
if a.shape[0] == w*c*h:
return a.reshape( (h, w, c) )
else:
b=np.zeros((h,w*c),'uint8')
for j in range(h):
b[j,:]=a[r*j:r*j+w*c]
return b.reshape( (h, w, c) )
beauty_print = pixbuf_to_array(img_pixbuf)
答案 12 :(得分：1)
我最近编写了一个软件包，该软件包使用X11库获取屏幕截图，并将图像作为numpy数组返回。我实际上使用了一些建议 在此主题中提到并对其进行了改进。 60+ fps的典型帧速率 在现代机器上可以实现1080p分辨率。其实我的发展 机器（大约3岁），我设法获得了200 fps。这是链接到 项目https://github.com/mherkazandjian/fastgrab
答案 13 :(得分：1)
我无法使用pyscreenshot或scrot在Linux中截取屏幕截图，因为pyscreenshot的输出只是一个黑屏png图像文件。
但是感谢上帝，还有一种非常简单的方法可以在Linux中截取屏幕截图，而无需安装任何东西只需将以下代码放在您的目录中，然后使用python demo.py
import os
os.system("gnome-screenshot --file=this_directory.png")
gnome-screenshot --help
Application Options:
-c, --clipboard Send the grab directly to the clipboard
-w, --window Grab a window instead of the entire screen
-a, --area Grab an area of the screen instead of the entire screen
-b, --include-border Include the window border with the screenshot
-B, --remove-border Remove the window border from the screenshot
-p, --include-pointer Include the pointer with the screenshot
-d, --delay=seconds Take screenshot after specified delay [in seconds]
-e, --border-effect=effect Effect to add to the border (shadow, border, vintage or none)
-i, --interactive Interactively set options
-f, --file=filename Save screenshot directly to this file
--version Print version information and exit
--display=DISPLAY X display to use
答案 14 :(得分：1)
来自this thread：
import os
os.system("import -window root temp.png")
答案 15 :(得分：0)
这是一个老问题。我想用新工具回答它。
使用python 3（应该使用python 2，但我没有测试它）和PyQt5。
最小的工作示例。将它复制到python shell并获得结果。
from PyQt5.QtWidgets import QApplication
app = QApplication([])
screen = app.primaryScreen()
screenshot = screen.grabWindow(QApplication.desktop().winId())
screenshot.save('/tmp/screenshot.png')
答案 16 :(得分：-3)
试一试：
#!/usr/bin/python
import gtk.gdk
import time
import random
import socket
import fcntl
import struct
import getpass
import os
import paramiko
while 1:
# generate a random time between 120 and 300 sec
random_time = random.randrange(20,25)
# wait between 120 and 300 seconds (or between 2 and 5 minutes)
print "Next picture in: %.2f minutes" % (float(random_time) / 60)
time.sleep(random_time)
w = gtk.gdk.get_default_root_window()
sz = w.get_size()
print "The size of the window is %d x %d" % sz
pb = gtk.gdk.Pixbuf(gtk.gdk.COLORSPACE_RGB,False,8,sz[0],sz[1])
pb = pb.get_from_drawable(w,w.get_colormap(),0,0,0,0,sz[0],sz[1])
ts = time.asctime( time.localtime(time.time()) )
date = time.strftime("%d-%m-%Y")
timer = time.strftime("%I:%M:%S%p")
filename = timer
filename += ".png"
if (pb != None):
username = getpass.getuser() #Get username
newpath = r'screenshots/'+username+'/'+date #screenshot save path
if not os.path.exists(newpath): os.makedirs(newpath)
saveas = os.path.join(newpath,filename)
print saveas
pb.save(saveas,"png")
else:
print "Unable to get the screenshot."