Using VBA, I am trying to download images from url, renaming them and saving them to folder.
I have found code that facilitates this, but it seems that all "names" with a "/" in it won't download.
Is this possible? Is there a way around it?
I have tried the code from the link Downloading Images from URL and Renaming
No error messages are provided. The images simply won't download.
Files are not allowed to have a / in their name, that's always been the case. The code works unless you put a symbol that's not allowed, which are / \ : * ? " < > |
Related
I have received a file from a customer. The file is said to be
SQL code (application/sql)
However, this has turned out to be wrong: nothing could open it. It turns out it was secretely a .zip file. By renaming it to '.zip' and manually extracting it I was able to get the files contained in it. I would like to do a similar process in python.
So far I've renamed the file:
file_name_zip = file_name.replace('.sql', '.zip')
os.rename(file_name, file_name_zip)
And I've tried extracting it:
zip_ref = zipfile.ZipFile(file_name_zip, 'r')
zip_ref.extractall(extracted_file)
However, this failed because
zipfile.BadZipFile: File is not a zip file
I've googled, and apparently this can sometimes be fixed using:
zip_file_name_2 = zip_file_name.replace('.zip', '2.zip')
os.system(f'zip -FF {zip_file_name} --out {zip_file_name_2}')
This required me to put in a bunch of settings, which I wasn't able to figure out. There must be a better way to go about this.
Does anybody know how to parse such an .sql file?
I'm sorry if this is basic and I missed something simple. I'm trying to run the code below to iterate through files in a folder and merge all files that start with a specific string, into a dataframe. All files sit in a lake.
file_list=[]
path = "/dbfs/rawdata/2019/01/01/parent/"
files = dbutils.fs.ls(path)
for file in files:
if(file.name.startswith("CW")):
file_list.append(file.name)
df = spark.read.load(path=file_list)
# check point
print("Shape: ", df.count(),"," , len(df.columns))
db.printSchema()
This looks fine to me, but apparently something is wrong here. I'm getting an error on this line:
files = dbutils.fs.ls(path)
Error message reads:
java.io.FileNotFoundException: File/6199764716474501/dbfs/rawdata/2019/01/01/parent does not exist.
The path, the files, and everything else definitely exist. I tried with and without the 'dbfs' part. Could it be a permission issue? Something else? I Googled for a solution. Still can't get traction with this.
Make sure you have a folder named "dbfs" if your parent folder starts from "rawdata" the path should be "/rawdata/2019/01/01/parent" or "rawdata/2019/01/01/parent".
The error is thrown in case of incorrect path.
This is an old thread, but if someone is still looking for a solution:
It does require path to be listed as:
"dbfs:/rawdata/2019/01/01/parent/"
I am uploading a file, that contains spaces in the name, to Amazon S3 using cffile action="upload". The file name is burger+beans n beetroot.jpg.
As you can see, the name contains spaces and a plus sign.
When I read the directory, to list the contents, the file name returned by ColdFusion in the query is: burger+beans+n+beetroot.jpg. However, when viewing the file using Amazon S3 Browser, it is correctly listed as: burger+beans n beetroot.jpg. So it appears ColdFusion is replacing the spaces with + signs.
Does anyone know why this happens and if there is a way to disable this? I tried using both the DirectoryList() method as well as the <cfdirectory action="list"> tag, and both do this.
Please note: I am aware that the file name could be cleaned up before processing - that's a workaround, but not the solution I am looking for. Thanks!
I believe this is not a CF problem, it's a S3 problem. They send out their file names escaped. Which makes this a non-answer.
I created a folder in a S3 bucket. Then I uploaded a file named burger+beans n beetroot.jpg. I can see in AWS' console the file properly named. I select it, then in the Actions menu select Download. I get the modal window. Take a look at the URL in the browser footer - the file name is escaped.
I right-click their link and choose "Save Link As..." - the file name is escaped as well.
So I don't think there is anything you can do once the file is up there. You'll need to clean it before uploading. I know it's not what you want to hear.
Try URL encoding the filename, so the + sign will be converted to an url encoded space (%2b). You could use URLEncodedFormat, but make sure that the path to the file isn't urlencoded as well.
I just downloaded psychopy this morning and have spent the day trying to figure out how to work with builder view. I watched the youtube video "Build your first PsychoPy experiment (Stroop task)" by Jon Pierce. In his video he was explaining how to make a conditions file with excel that would be used in his experiment. I wanted to make a very similar test where images would appear and subjects would be required to give a yes or no answer to them (the correct answer is already predefined). In his conditions file he had the columns 'word' 'colour' and 'corrANS'. I was wondering if instead of a 'word' column, I can have an 'image' column. In this column I would like to upload all my images to them in the same way I would words, and have them correlated to a correct answer of either 'yes' or 'no'. We tried doing this and uploaded images to the conditions file, but we haven't had any success in running the test successfully and were hoping somebody could help us.
Thank you in advance.
P.S. we are not familiar with python, or code in general, so we were hoping to get this running using the builder view.
EDIT: Here is the error message we are receiving when running the program
#### Running: C:\Users\mr00004\Desktop\New folder\1_lastrun.py
4.8397 ERROR Couldn't find image file 'C:/Users/mr00004/Desktop/New folder/PPT Retention 1/ Slide102.JPG'; check path?
Traceback (most recent call last):
File "C:\Users\mr00004\Desktop\New folder\1_lastrun.py", line 174, in
image.setImage(images)
File "C:\Program Files (x86)\PsychoPy2\lib\site-packages\psychopy-1.80.03-py2.7.egg\psychopy\visual\image.py", line 271, in setImage
maskParams=self.maskParams, forcePOW2=False)
File "C:\Program Files (x86)\PsychoPy2\lib\site-packages\psychopy-1.80.03-py2.7.egg\psychopy\visual\basevisual.py", line 652, in createTexture
% (tex, os.path.abspath(tex))#ensure we quit
OSError: Couldn't find image file 'C:/Users/mr00004/Desktop/New folder/PPT Retention 1/ Slide102.JPG'; check path? (tried: C:\Users\mr00004\Desktop\New folder\PPT Retention 1\ Slide102.JPG)
Yes, certainly, that is exactly how PsychoPy is designed to work. Simply place the image names in a column in your conditions file. You can then use the name of that column in the Builder Image component's "Image" field. The appropriate image file for a given trial will be selected.
It is difficult to help you further, though, as you haven't specified what went wrong. "we haven't had any success" doesn't give us much to go on.
Common problems:
(1) Make sure you use full filenames, including extensions (.jpg, .png, etc). These aren't always visible in Windows at least I think, but they are needed by Python.
(2) Have the images in the right place. If you just use a bare filename (e.g. image01.jpg), then PsychoPy will expect that the file is in the same directory as your Builder .psyexp file. If you want to tidy the images away, you could put them in a subfolder. If so, you need to specify a relative path along with the filename (e.g. images/image01.jpg).
(3) Avoid full paths (starting at the root level of your disk): they are prone to errors, and stop the experiment being portable to different locations or computers.
(4) Regardless of platform, use forward slashes (/) not backslashes (\) in your paths.
make a new folder in H drive and fill in the column of image in psychopy as e.g. 'H:\psych\cat.jpg' it works for me
Sure a very simple question but I can't seem to find the terminology to find the answer in a search!
I'm using a file-uploader CGI script. Inside the CGI script is some code that generates some HTML. In the HTML I need to put an email address using the # symbol, however this breaks the script. What is the correct way to escape the # symbol in a CGI script?
The error when using the # symbol is:
"FileChucker: load_external_prefs(): Error processing your prefs file ('filechucker_prefs.cgi'): Global symbol "#email" requires explicit package name at (eval 16) line 1526."
Many thanks for any help
Update..
Hi All, many thanks for the replies - I guess it is perl.. (shows my ignorance of what's going on here perfectly!). The code below shows the problem the # in 'email#domain.com'.
'test$PREF{app_output_template} = qq`
%%%ifelse-onpage_uploader%%%
<div id="fcintro">If you're using a mobile or tablet and have problems uploading, we recommend emailing your CV to: email#domain.com<br><span class"upload_limits">We can accept Adobe PDF, Microsoft Word and all popular image and text file types. (max total upload size: 7MB)</span></div>
%%%else%%%
%%subtitle%%
%%%endelse-onpage_uploader%%%
%%app_output_body%%'
Try # instead of #.
Reference: http://www.w3schools.com/tags/ref_ascii.asp