在 Linux 上通过 Python 脚本截取屏幕截图
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Take a screenshot via a Python script on Linux
提问by skyronic
I want to take a screenshot via a python script and unobtrusively save it.
我想通过python脚本截取屏幕截图并不显眼地保存它。
I'm only interested in the Linux solution, and should support any X based environment.
我只对 Linux 解决方案感兴趣，应该支持任何基于 X 的环境。
采纳答案by Rusty
This works without having to use scrot or ImageMagick.
这无需使用 scrot 或 ImageMagick 即可工作。
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."
Borrowed from http://ubuntuforums.org/showpost.php?p=2681009&postcount=5
借自http://ubuntuforums.org/showpost.php?p=2681009&postcount=5
回答by Slava V
import ImageGrab
img = ImageGrab.grab()
img.save('test.jpg','JPEG')
this requires Python Imaging Library
这需要 Python 图像库
回答by Juliano
回答by Snowball
Cross platform solution using wxPython:
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)
回答by Alex Snet
Compile all answers in one class. Outputs PIL image.
将所有答案汇总在一节课中。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()
回答by ponty
I have a wrapper project (pyscreenshot) for scrot, imagemagick, pyqt, wx and pygtk. If you have one of them, you can use it. All solutions are included from this discussion.
我有一个用于 scrot、imagemagick、pyqt、wx 和 pygtk的包装项目 ( pyscreenshot)。如果您有其中之一，则可以使用它。所有解决方案都包含在此讨论中。
Install:
安装：
easy_install pyscreenshot
Example:
例子：
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')
回答by JHolta
Just for completeness: Xlib - But it's somewhat slow when capturing the whole screen:
只是为了完整性：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()
One could try to trow some types in the bottleneck-files in PyXlib, and then compile it using Cython. That could increase the speed a bit.
可以尝试在 PyXlib 的瓶颈文件中添加一些类型，然后使用 Cython 编译它。这样可以稍微提高速度。
Edit:We can write the core of the function in C, and then use it in python from ctypes, here is something I hacked together:
编辑：我们可以在 C 中编写函数的核心，然后从 ctypes 在 python 中使用它，这是我一起破解的东西：
#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);
}
And then the python-file:
然后是python文件：
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()
回答by Anand
Try it:
尝试一下：
#!/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."
回答by ouille
回答by rominf
It's an old question. I would like to answer it using new tools.
这是一个老问题。我想使用新工具来回答它。
Works with python 3 (should work with python 2, but I haven't test it) and PyQt5.
适用于 python 3（应该适用于 python 2，但我还没有测试过）和 PyQt5。
Minimal working example. Copy it to the python shell and get the result.
最小的工作示例。将其复制到python shell并获取结果。
from PyQt5.QtWidgets import QApplication
app = QApplication([])
screen = app.primaryScreen()
screenshot = screen.grabWindow(QApplication.desktop().winId())
screenshot.save('/tmp/screenshot.png')
