Intellij taking file path incorrectly: D:\jetty-distribution-9.2.1.v20140609\?D:\images\a.txt
When I give files like
File file = new File("D:\images\1.txt");
System.out.println(file.getAbsolutePath());
The output is
D:\jetty-distribution-9.2.1.v20140609\?D:\images\1.txt
It is considering the file path along with server bin
Where do i need to change the setting to detect the absolute path in intelij.
Please help me.
Use double backslashes \\ to escape the \, then it will work.
File file = new File("D:\\images\\1.txt"); System.out.println(file.getAbsolutePath());
Related
For example, I want to open a PDF file in the browser from the command line (just because it's much faster and I need to open many files at once) and when I use the command start [file name] from its directory it try to open it as a executable, so I need to open the browser and type the full path of the file as an attribute, is there a way to call the full path without typing it?
what I exactly need is I need the full path of a file to convert it to string (for example in the browser)
Using tab completions may help. For example, if your target file is named thisPDFisTotallyBananas.pdf and you have another file in the same folder named thisOtherPDFisNot.pdf, you could type thisP then TAB to complete the file name in the command prompt without needing to type the whole filename.
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 am trying to setup a file watcher for scss files which is working on files with a filename not starting with _.
But if I have a file named _file_name.scss the output of any macros that include the filename will be file.name.scss.
The first _ is removed and following ones are replaced by ..
Even though in the insert macros selection tool I can see that the output when you select a macro is correct.
Like $FilePathRelativeToProjectRoot$ will display mypath/_file_name.scss in the selection tool but then my command from this file watcher will output mypath/file.name.scss.
Am I missing a parameter here ?
Full configuration:
For me, existing file names are not changed when using similar file watcher. But files with names starting the _ are not prettified, the main .scss that includes them is processed instead.
To avoid this, try adding COMPILE_PARTIAL=true variable to your file watcher settings:
Also, make sure that Track only root files is off.
See the comments in https://youtrack.jetbrains.com/issue/WEB-13459
I am trying to compile my styles.less to styles.css. My folder structure is following:
assets->less->styles.less
assets->css->styles.css
I believe my configurations are wrong. In PhpStorm I set less output path to refresh: ../css/$FileNameWithoutExtension$.css
I do have a styles.css file under the less file and it is compiling.
So far I only know regular CSS so I'm not very familiar with Less yet.
Any help?
Your File Watcher setup is incomplete.
Right now it will save the generated file next to the source... but you need it 2 folders up.
You did set up correctly in Output paths to refresh .. but that file tells IDE what file to check when file watcher is finished running. It is not where the generated file will be placed.
You need to alter your Arguments field.
Currently you have ... $FileName$ $FileNameWithoutExtension$.css ...
You need to adjust the path there -- it has to be ... $FileName$ ../../css/$FileNameWithoutExtension$.css ... -- because that's where you specify such path.
(leading and trailing "..." means other parameters that you have got there)
You should have changed the Arguments field accordingly;
Like:
Arguments: --no-color $FileName$ $ProjectFileDir$/themes/elisa/assets/css/$FileDirPathFromParent(less)$$FileNameWithoutExtension$.css
Output paths to refresh: $ProjectFileDir$/themes/elisa/assets/css/$FileDirPathFromParent(less)$$FileNameWithoutExtension$.css
I have a separate folder Input for the text files. i'm trying to read the lines in a certain file 'train.txt'. So how do i do it using this code?
for line in io.lines '???' do
end
io.lines takes an optional argument to represent which file it iterates. Since this file is in a different folder, use an absolute path, or a proper relative path. For instance, in Unix-like system, you can use "/some/path/Input/train.txt".
for line in io.lines("/some/path/Input/train.txt") do
--print(line)
end