I am trying to perform a script on all opened files in DM. For now, I have built a Taglist of all these files, and I am planning to use a for statement to get the path of each file and return it to the existing script. However, I do not know how to extract the path of these opened files from the Taglist. Thanks.
The property of a saving location is not a property of the image object, but of the imageDocument object.
The command you're seeking is
String ImageDocumentGetCurrentFile( ImageDocument img_doc )
Returns the file location to which the image document is currently linked.
And an example script is:
Result("->" + GetFrontImage().ImageGetOrCreateImageDocument().ImageDocumentGetCurrentFile() )
Related
I have a folder structure like this as a source
Source/2021/01/01/*.xlsx files
Source/2021/03/02/*.xlsx files
Source/2021/04/03/.*xlsx files
Source/2021/05/04/.*xlsx files
I want to drop all these excel files into a different folder called Output.
Method 1:
When I am trying this, I used copy activity and I am able to get Files with folder structure( not a requirement) in Output folder. I used Binary file format.
Method 2:
Also, I am able to get files as some random id .xlsx in my output folder. I used Flatten Hierrachy.
My requirement is to get files with the same name as source.
This is what i suggest and I have implemented something in the past and i am pretty confident this should work .
Steps
Use getmetada activity and try to loop through all the folder inside Source/2021/
Use a FE loop and pass the ItemType as folder ( so that you get folder only and NO files , I know at this time you dont; have file )
Inside the IF , add a Execute pipeline activity , they should point to a new pipeline which will take a parameter like
Source/2021/01/01/
Source/2021/03/02/
The new pipeline should have a getmetadata activity and FE loop and this time we will look for files only .
Inside the FE loop add a copy activity and now will have to use the full file name as source name .
I have to load file with csv extension from one particular folder to data base in ssis. file name is not known but folder and extension is fixed.
To load the content of a file, the file name with folder path is required else the connection manager can not be validated and configured.
The easiest way is to get file name is using a For Each Loop container:
Select the option [Foreach File Enumerator]
Provide the Folder path and extension (like *.csv) you already have.
Get the File Name in a variable and use it within the Source of the data flow task within the For each Loop container.
Refer
Is it possible to get the full path of the current file within a live template in IntelliJ? I've tried using groovyScript("new File('.').absolutePath") function, but that returns /Applications/IntelliJ IDEA.app/Contents/bin/. and not the file path as I was hoping for.
Thanks!
According to the docs (emphasis mine):
You can use groovyScript macro with multiple arguments. The first argument is a script text that is executed or a path to the file that contains a script. The next arguments are bound to _1, _2, _3, ..._n variables that are available inside your script. Also, _editor variable is available inside the script. This variable is bound to the current editor.
The _editor is an instance of EditorImpl which holds a reference to the VirtualFile that represents the currently opened file.
Therefore, the following script gets the full path of currently opened file.
groovyScript("_editor.getVirtualFile().getPath()")
Or if you want to get the path relative to the project's root:
groovyScript("_editor.getVirtualFile().getPath().replace(_editor.getProject().getBaseDir().getPath(), \"\")")
Since IntelliJ IDEA 2019.3 the Live Template macros filePath() and fileRelativePath() are available. A complicated Groovy script macro is no longer required.
I'm trying to load group of folders files in one time with when
i set
sourceURI = 'gs://ybbi/bi_landing_zone/files_to_load/app/reports/app_network_analytics_report/201409011*'
all the folders that i'm want to load start with 20140911
but i get the error:
ERROR: Invalid path: gs://ybbi/bi_landing_zone/files_to_load/apn/reports/appnexus_network_analytics_report/20140901191111_3bab8ec0_092a_43de_a157_db35d1555ea0/
20140901191111_3bab8ec0_092a_43de_a157_db35d1555ea0 is one of these folders(don't know why it's print the all folder name of this specific folder)
in some other folder tree cases it's works, but in this specific folder tree it's return the same error .
i know that cloud storage don't have real folders and it's part of the name of the object, but you understand what i mean.
is it bug?
Without more information, what it looks like is that you have a object file called gs://ybbi/bi_landing_zone/files_to_load/apn/reports/appnexus_network_analytics_report/20140901191111_3bab8ec0_092a_43de_a157_db35d1555ea0/ that is not a csv/json file. Some tools may create these dummy files in order to simulate directories. BigQuery requires all objects that match the input glob path to be importable files.
One solution would be to change the glob path to include a narrower set of files. You can pass multiple paths if that makes things easier. For example, you could pass
gs://ybbi/bi_landing_zone/files_to_load/apn/reports/appnexus_network_analytics_report/20140901191111_3bab8ec0_092a_43de_a157_db35d1555ea0/*
and
gs://ybbi/bi_landing_zone/files_to_load/apn/reports/appnexus_network_analytics_report/20140901191111_some_other_path/*
I have about 200 folders with X images in each of them.
I have a master script in the root folder that does some stuff to the images.
Each folder has some variables specific to it and its contents.
I want my master script, when it parses folder Y, load some sort of a config file from within folder Y to get those variables, then when folder Z is to be parsed, load the config file from that one.
I know of #include "config.jsx" that I use at the moment to load it but its at the beginning of the script, I need something dynamic and doesn't need to be a jsx at all.
I store all my parameters in xml format and read that in using the XML objects in extendscript. As long as your parameters file is always named something like 'config.xml' it is easily located.
var file = new File( /c/folder/file.xml );
file.open("r");
var str = file.read();
var xml = new XML(str);