The script receives two variables from a previous web page. From those variables, the code determines which images are desired. It sends those images to a temp folder, zips up that folder and places it in an output folder for pickup. That's where things go south. I'm trying to allow the webpage to provide a button for the user to click on and download the zip file. Because the zip file's name needs to change based on the variables the script receives, I cannot just make a generic link to the zip file.
import arcpy, sys, shutil, os
path = "C:/output/exportedData/raw/"
pathZip = "C:/output/exportedData/zip/"
#First arg is the mxd base filename which is the same as the geodatabase name
geodatabaseName = "C:/output/" + sys.argv[1] + ".gdb"
#this is where the images are determined and sent to a folder
zipFileName = sys.argv[1]
zipFile = shutil.make_archive(path + zipFileName,"zip")
movedZip = os.rename(zipFile, pathZip + zipFileName + ".zip")
shutil.rmtree(path + zipFileName)
print """<h3>Download zip file</h3>""".format(movedZip)
And the last line indicates where the problem comes in. Firebug indicates the link made is
Download zip file
The string substitution isn't working in this case and I'm at a loss as to why. Thank you, in advance for any assistance you can provide.
os.rename() doesn't return anything, which means that movedZip becomes None.
Here's what you probably want to do instead:
movedZip = pathZip + zipFileName + ".zip"
os.rename(zipFile, movedZip)
The os.rename method does not return any value. You could see the official document here. It rename the file or directory src to dst. Some exceptions might be thrown. But does not return anything.
Related
I'm trying to store a path in a variable. see below
target = r"C:\Users\User\CodeProjects\WebSafer"
However, I need it to be dynamic. Not hardcoded to my username, so I get the login username by doing:
val = os.getlogin()
So I need to put the variable val in the path. But every time I tried doing it I always get a truncating/syntax error. Please help me! Below is the code snippet:
print("No copy found...making a copy\n")
val = os.getlogin()
original = r"C:\*******\********\*******\***\****"
target = r"C:\Users\User\CodeProjects\WebSafer"
shutil.copy(original, target)
The "*" are just for privacy reasons, there actually replaced with the right path location to what I'm copying.
What I have tried so far:
target = r"C:\Users\{val}\CodeProjects\WebSafer".format(val = os.getlogin)
target = r"C:\Users\{}\CodeProjects\WebSafer".format(val)
target = rf"C:\Users\{val}\CodeProjects\WebSafer".format(val = os.getlogin)
target = rf"C:\Users\{}\CodeProjects\WebSafer".format(val)
Don't mix f with .format, this is working for me:
import os
val = os.getlogin()
print(rf"C:\Users\{val}\CodeProjects\WebSafer")
And I think better way is:
import os.path
from pathlib import Path
home = str(Path.home())
print(os.path.join(home, "CodeProjects\WebSafer"))
Then if you encounter some error when copying, you need clarify what you want to copy, copy a file, or a folder, if a folder, should it go to within the dest folder, or overwrite dest folder?
You may want try different methods such as
shutil.copy, shutil.copytree, and different parameters.
"r" means the string will be treated as raw string so try removing that and using escaped characters target = "C:\\Users\\{val}\\CodeProjects\\WebSafer".format(val = os.getlogin)
You can use f-string (to directly enter your variable) and r-string (to enter the path without escape characters like \) together.
val = os.getlogin() # Returns username
target = fr"C:\Users\{val}\CodeProjects\WebSafer"
If you're getting a No such file or directory error, it means that the actual file or folder does not exist. Check the actual path to ensure every part (Your Username, CodeProjects, Websafer) exists on your computer.
In case you don't know if your user will have that folder on their system, you can use a try-except block to alert the user or to revert to some default folder instead.
I'm manipulating some svg's using lib-svgpathtools and have made use of the wsvg() function. This directly saves the created file(ex:file1.svg)on the left panel under file section of
Google colaboratory. Under content just like that..now that i have multiple files to be created its difficult to go on downloading each on of them ..instead is there any way to get them directly into a folder and further that folder to my system?? its being difficult since this file creation is happening in this function. Plz help me out!
Thanks a lot in advance!!
wsvg(paths = paths, attributes = attributes, svg_attributes=svg_attributes, filename='file' + str(num) + '.svg')
You could zip your svgs and download the zip from the side-panel.
from zipfile import ZipFile
import glob
with ZipFile('./all_of_the_svgs.zip', 'w') as zipObj:
for i in glob.glob('./your_folder/*.svg'):
zipObj.write(i)
If the files go directly into the root folder of the left panel, then it would just be: for i in glob.glob('./*.svg'):
To save the svgs directly into a folder in the left side panel you can probably just change the filename argument to include the name of the folder you want it put into:
Before:
wsvg(paths = paths, attributes = attributes, svg_attributes=svg_attributes, filename='file' + str(num) + '.svg')
To:
wsvg(paths = paths, attributes = attributes, svg_attributes=svg_attributes, filename='./folder_name/' + 'file' + str(num) + '.svg')
You would need to create the folder with "folder_name" in the left panel first though.
I am attempting to create a script that will look at a directory and send all of the files of a certain type (PNG) to my Deepstack face recognition. The python script is Deepstack shows (and works fine) is as follows:
import requests
user_image = open("image1.png","rb").read()
response = requests.post("http://localhost:80/v1/vision/face/register",
files={"image":user_image},data={"userid":"User Name"}).json()
print(response)
If I understand your question correctly, you're asking how to check all the files in a directory and operate only on the files ending in ".png".
To do that, you would want to use the os.listdir function to provide a list of all files in a given directory, then check if each one ends in ".png" using a substring.
For example:
import os
# this will list the contents of the current directory by default
# you can also give os.listdir an argument with a path to a specific directory to list
fileList = os.listdir()
for fileName in contents:
if fileName[-4:] == ".png":
### your code goes here, using fileName
This example isn't perfect because it throws an error for filenames with fewer than 4 characters, but you can get the idea.
Here is the below code we have developed for single directory of files
from os import listdir
with open("/user/results.txt", "w") as f:
for filename in listdir("/user/stream"):
with open('/user/stream/' + filename) as currentFile:
text = currentFile.read()
if 'checksum' in text:
f.write('current word in ' + filename[:-4] + '\n')
else:
f.write('NOT ' + filename[:-4] + '\n')
I want loop for all directories
Thanks in advance
If you're using UNIX you can use grep:
grep "checksum" -R /user/stream
The -R flag allows for a recursive search inside the directory, following the symbolic links if there are any.
My suggestion is to use glob.
The glob module allows you to work with files. In the Unix universe, a directory is / should be a file so it should be able to help you with your task.
More over, you don't have to install anything, glob comes with python.
Note: For the following code, you will need python3.5 or greater
This should help you out.
import os
import glob
for path in glob.glob('/ai2/data/prod/admin/inf/**', recursive=True):
# At some point, `path` will be `/ai2/data/prod/admin/inf/inf_<$APP>_pvt/error`
if not os.path.isdir(path):
# Check the `id` of the file
# Do things with the file
# If there are files inside `/ai2/data/prod/admin/inf/inf_<$APP>_pvt/error` you will be able to access them here
What glob.glob does is, it Return a possibly-empty list of path names that match pathname. In this case, it will match every file (including directories) in /user/stream/. If these files are not directories, you can do whatever you want with them.
I hope this will help you!
Clarification
Regarding your 3 point comment attempting to clarify the question, especially this part we need to put appi dynamically in that path then we need to read all files inside that directory
No, you do not need to do this. Please read my answer carefully and please read glob documentation.
In this case, it will match every file (including directories) in /user/stream/
If you replace /user/stream/ with /ai2/data/prod/admin/inf/, you will have access to every file in /ai2/data/prod/admin/inf/. Assuming your app ids are 1, 2, 3, this means, you will have access to the following files.
/ai2/data/prod/admin/inf/inf_1_pvt/error
/ai2/data/prod/admin/inf/inf_2_pvt/error
/ai2/data/prod/admin/inf/inf_3_pvt/error
You do not have to specify the id, because you will be iterating over all files. If you do need the id, you can just extract it from the path.
If everything looks like this, /ai2/data/prod/admin/inf/inf_<$APP>_pvt/error, you can get the id by removing /ai2/data/prod/admin/inf/ and taking everything until you encounter _.
I'm trying to create a program that duplicates itself to another location and creates a batch file on the desktop. I can make it duplicate itself and I can create the batch file but I need some help with the paths.
I can find the path that my program is currently in. Both the direct path and the path to the directory. My problem lies in the fact that I want to place the file in (let's just say for simplicity) 'C:\Users\Me\Documents'. How would I edit the path? I want to be able to place this on a generic windows computer so I can't hard code the path in because each user will be different. This goes the same for placing the batch file and setting it for the right directory to run the python script in documents.
I have tried both
import os
print os.path.dirname(os.path.abspath(__file__))
and
import os
print os.path.abspath(__file__)
but am clueless as to how to edit the path. When I try googling for it and searching this wonderful site, all I get is stuff about configuring the Python path on windows and other stuff that I can't quite understand at my current level of Python.
Now I turn to you, can you help? Any input would be appreciated, if you could explain how it worked that would be even better!
<>
Due to some questions about my code (and a specific one to post it) Here it is
from sys import argv # Imports module
import os
script, create = argv # Gets script name and desired amount of copies
data = open(script) # Creates a variable to store the script
indata = copy.read() # Creates the data to be copied from the script
batData = """
echo off
%s
""" % # This is not finished, creating that batch file
createT = int(create) + 1
for i in range(1, createT): # Runs a set amount of times
copyName = "%s.py" % str(i) # Creates the name for the file
copy = open(copyName, 'w+') # Opens/creates the file for editing
copy.write(indata) # Writies the indata to the file opened
copy.close # Closes that file
batName = "%s.bat" % str(i)
bat = open(batName, 'w+')
It is not finished but hopefully you get the gist. The argv at the beginning is so I can change the amount of copies made, that will be deleted later as I evolve the code but for now I like it there.
I have currently tried the following to find the path:
import os
print os.path.abspath(__file__)
print os.path.dirname(os.path.abspath(__file__))
print os.path.dirname(__file__)
test = os.path.dirname(__file__)
a, b, c, d, e, f, g, h, i = test.split("\\")
print c
What I want to happen (or think I want to happen) is for the path to be found, then split into pieces (either each directory or break off everything after the username). Then I want to append the document folder tag to the end. For the batch instead of the document tag it will be for the desktop.
among a few others that people have posted. I hope this helps!
Your code snippet returns a string. Just take that string and edit to make the path you want. I'm on a mac so I can't test with an actual windows directory but I'll try to make it look Windows-ish. For instance, lets say this code:
directory_path = os.path.dirname(os.path.abspath(__file__))
print(directory_path)
gives you:
C:\Users\username\AppData
You can use the split function to break the path into pieces (docs).
stuff = path_string.split('\')
print(stuff)
Code output:
['C:', 'Users', 'username', 'AppData']
You can use the pieces create the path you want and then use it to write the file. So, if you want the username folder just loop until you find it. Some example code is below (just an example to get you started - read up on Python if you need help understanding the code).
username = ""
for i in range(0, len(stuff)):
if stuff[i] == "Users":
username = stuff[i + 1]
Not sure if that answers your question but hope it helps.
Am I correct that you are trying to figure out how to make a file path to some location on the user's directory without knowing who the user is going to be that is executing the program?
You may be able to take advantage of environment variables for this. For instance, I can get a file path to the C://Users/username/ directory of whoever is executing the code with:
my_path = os.path.join("C:\\", "Users", os.getenv("USERNAME"))
Where os.getenv("USERNAME") returns the value of the USERNAME environment variable (should be the name of the user that is currently logged on).
You can list all of the available environment variables and their values in Python with:
for key, val in os.environ.items():
print("{} \t {}\n".format(key, val)) # or however you want to prettify it
You may get lucky and find an environment variable that gives you most of the path to begin with.
In general, os.path.join(), os.path.relpath(), and os.path.abspath() combined with some environment variables might be able to help you. Check out the os documentation and os.path documentation.
Many times when modifying a path, I am looking to add/remove folders. Here is my simple method for adding a path, e.g. if I want to move the path of an object into a folder added_path='/train/.
Since my paths are usually uniform, I check the last split characters in the first file location. Usually, my experience is that windows have \\ at the end while Mac and Linux have `/', which makes this work across operating systems. (note: if the paths are not uniform, you obviously place the if-else in the for-loop.)
if '\\' in data[0].img_file:
split_char = '\\'
else:
split_char = '/'
for img in data:
img_path = img.img_file.split(split_char)
label_path = img.label_file.split(split_char)
img.img_file = '/'.join(img_path[:-1]) + added_path + img_path[-1]
img.label_file = '/'.join(label_path[:-1]) + added_path + label_path[-1]
So, this for loop uses all the folders up until the file name, where we insert the extra folder, and then, at last, add the file name.
Example input path: 'data/combined/18.png'
Example output path: 'data/combined/train/18.png'