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
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 need to delete many folders with an exempt list which has a list of folders and files that I should not delete. So I tried delete activity and tried to use if function of Add dynamic content to check whether the file or folder name is the same of the specified ones. But I do not have what should be the parameters #if(). In other words, to use these functions, how do we get the file name or folder name?
It is difficult to get all files and folders by using GetMetadata activity.
As workarounds, you can try these ways:
1.create a Delete activity and select List of files option. Then create a file filled with those path of files and folders need to be deleted.(relative path to the path configured in the dataset).
2.using blob SDK to do this thing.
is it possible to load module from file with extension other than .lua?
require("grid.txt") results in:
module 'grid.txt' not found:
no field package.preload['grid.txt']
no file './grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt.lua'
no file '/usr/local/share/lua/5.1/grid/txt/init.lua'
no file '/usr/local/lib/lua/5.1/grid/txt.lua'
no file '/usr/local/lib/lua/5.1/grid/txt/init.lua'
no file './grid/txt.so'
no file '/usr/local/lib/lua/5.1/grid/txt.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
no file './grid.so'
no file '/usr/local/lib/lua/5.1/grid.so'
no file '/usr/local/lib/lua/5.1/loadall.so'
I suspect that it's somehow possible to load the script into package.preaload['grid.txt'] (whatever that is) before calling require?
It depends on what you mean by load.
If you want to execute the code in a file named grid.txt in the current directory, then just do dofile"grid.txt". If grid.txt is in a different directory, give a path to it.
If you want to use the path search that require performs, then add a template for .txt in package.path, with the correct path and then do require"grid". Note the absence of suffix: require loads modules identified by names, not by paths.
If you want require("grid.txt") to work should someone try that then yes, you'll need to manually loadfile and run the script and put whatever it returns (or whatever require is documented to return when the module doesn't return anything) into package.loaded["grid.txt"].
Alternatively, you could write your own loader just for entries like this which you set into package.preload["grid.txt"] which finds and loads/runs the file or, more generically, you could write yourself a loader function, insert it into package.loaders, and then let it do its job whenever it sees a "*.txt" module come its way.
I'm attempting to create a process to import data. I created the entire process and it works, but I'm having trouble creating the variable to find the file name of the csv i want to import automatically. Each time a new csv is uploaded to me it has a timestamp on it. I want to be able to grab that file no matter what the name is and do work to it.
So for example this week the file name would be
filename_4-14-2014.csv
And next week
filename_4_21_2014.csv
And so on into eternity. . .
Is there a way to create a variable that picks up the full file name even though its changing?
After doing some poking around, I've discovered the following...
You can use a file system task to perform the copy operation I was referring to. You can set the input file and the output file as variables. This way you can always know that the file you use for import is always named the same, and has the right data.
You just need to add the variables and a File System Task to your package.
Ok so to accomplish what I wanted I created a Foreach Loop Container. Using the foreach loop container I had it look for any files ending with .csv in my specified folder by using a wildcard [denoted by asterisk: *.csv] .
Within the Foreach Loop container is as follows.
Step 1: File System Task - rename file.
Step 2: Data Flow Task - Import data to sql
Step 3: File System Task - Copy the file to another folder, append datetime to filename
Step 4: File System Task - Delete source file.
I used variables to get all the file and folder names plus datetimes.
1) I have a folder called CCBuilds containing couple of files in this path: E:\Testing\Builds\CCBuilds.
2) I have written C# code (Process.Start) to Rar this folder and save it in E:\Testing\Builds\CCBuilds.rar using the following command
"C:\program files\winrar\rar.exe a E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds"
3) The problem is that, though the rar file gets created properly, when I unrar the file to CCBuilds2 folder (both through code using rar.exe x command or using Extract in context menu), the unrared folder contains the full path, ie. extracting E:\Testing\Builds\CCBuilds.rar ->
E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\<<my files>>
Whereas I want it to be something like this: E:\Testing\Builds\CCBuilds2\CCBuilds\<<my files>>
How can I avoid this full path persistence while adding to rar / extracting back from it. Any help is appreciated.
Use the -ep1 switch.
More info:
-ep = Files are added to an archive without including the path information. Could result in multiple files existing in the archive
with same name.
-ep1 = Do not store the path entered at the command line in archive. Exclude base folder from names.
-ep2 = Expand paths to full. Store full file paths (except drive letter and leading backslash) when archiving.
(source: http://www.qa.downappz.com/questions/winrar-command-line-to-add-files-with-relative-path-only.html)
Just in case this helps: I am currently working on an MS Access Database project (customer relations management for a small company), and one of the tasks there is to zip docx-files to be sent to customers, with a certain password encryption used.
In the VBA procedure that triggers the zip-packaging of the docx-files, I call WinRAR as follows:
c:\Programme\WinRAR\winrar.exe a -afzip -ep -pThisIsThePassword "OutputFullName" "InputFullName"
-afzip says: "Create a zip file (as opposed to a rar file)
-ep says: Do not include the paths of the source file, i.e. put the file directly into the zip folder
A full list of such switches is available in the WinRAR Help, section "Command line".
x extracts it as E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\, because you're using full path when declaring the source. Either use -ep1 or set the default working dir to E:\Testing\Builds.
Use of -ep1 is needed but it's a bit tricky.
If you use:
Winrar.exe a output.rar inputpath
Winrar.exe a E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds
it will include the input path declared:
E:\Testing\Builds\CCBuilds -> E:\Testing\Builds\CCBuilds.rar:
Testing\Builds\CCBuilds\file1
Testing\Builds\CCBuilds\file2
Testing\Builds\CCBuilds\folder1\file3
...
which will end up unpacked as you've mentioned:
E:\Testing\Builds\CCBuilds2\Testing\Builds\CCBuilds\
There are two ways of using -ep1.
If you want the simple path:
E:\Testing\Builds\CCBuilds\
to be extracted as:
E:\Testing\Builds\CCBuilds2\CCBuilds\file1
E:\Testing\Builds\CCBuilds2\CCBuilds\file2
E:\Testing\Builds\CCBuilds2\CCBuilds\path1\file3
...
use
Winrar.exe a -ep1 E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds
the files inside the archive will look like:
CCBuilds\file1
CCBuilds\file2
CCBuilds\folder1\file3
...
or you could use ep1 to just add the files and folder structure sans the base folder with the help of recursion and defining the base path as the inner path of the structure:
Winrar.exe a -ep1 -r E:\Testing\Builds\CCBuilds.rar E:\Testing\Builds\CCBuilds\*
The files:
E:\Testing\Builds\CCBuilds\file1
E:\Testing\Builds\CCBuilds\file2
E:\Testing\Builds\CCBuilds\folder1\file3
...
inside the archive will look like:
file1
file2
folder1\file3
...
when extracted will look like:
E:\Testing\Builds\CCBuilds2\file1
E:\Testing\Builds\CCBuilds2\file2
E:\Testing\Builds\CCBuilds2\folder1\file3
...
Anyway, these are two ways -ep1 can be used to exclude base path with or without the folder containing the files (the base folder / or base path).