PuTTY Plink save files name to text file, and PSFTP only those - sftp

I am trying to take the files that I get with PuTTY's Plink command but save the file names to a text file so that I can only pull those files with PSFTP afterwards. Or can this be done without a temp text file?
The files I get are modified in the last 15 min, and I only want to get those files. I am new to PuTTY and FTP in general. I searched everywhere but cannot find anything that helps.
Any help is appreciated,
Thank you

You have to generate the PSFTP script file dynamically by reading the Plink output file line by line and producing a put command for each line.
See Batch files: How to read a file?
Or use an SFTP client that can directly download only files created in the last 15 minutes.
For example with WinSCP scripting:
winscp.com /command ^
"open sftp://username:password#example.com/ -hostkey=""fingerprint""" ^
"get /path/*>15N c:\path\" ^
"exit"
Read about file masks with a time constraint.
(I'm the author of WinSCP)

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But it is failing with error , No such file .
Is that anything I am missing . Any idea on this is much appreciated .
You need to put either exact name or use a file list with the name of the file and then use indirect file type in the session that is reading the file.
You can use a pre session shell command like this ls -1 Gemstone_*.csv>/infa/home/tmp/Gemstone_filelist.txt. Or you can create a shell script too with this command for better control.
in the session that is reading this file, set the property to indirect file type and mention /infa/home/tmp/Gemstone_filelist.txt as file to be extracted.
Infa will pick files one by one and process them.
Once the file gets processed, delete it using a post session command task rm -f Gemstone_*..

How to convert .xls or .xlxs file to csv file without any plugins or tools using Unix command

I have to convert .xls or .xlxs file to .csv file without using plugins or tools using Unix Command
Is their any way to do this ?
I Tried to do like this below ...But not working
Change the characterSet code from .xls file to UTF-8 encoding
Then create file again with extension change
cp temp.xls temp.csv
It is possible, but you need to realise that an *.xls file is a zipped directory structure (just unzip such a file, using Winzip or 7-zip). The unzipping can also be done using UNIX commands.
But what then? The directory structure is quite complicated to understand, and in order to create a script or a program which can do this (without using any external tools) is a tremendous work, so I'd propose you, either to use external tools anyway, or to make sure the files you receive already are CSV format.

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I am trying to read the files present in a directory in a remote server through SFTP. The catch is that I only had read and write permissions on the directory and not execute. This means any method that requires opening (cding) into the folder would fail. I need to read the file names since they are variable. From what I understand ls does not require execute privs. If I can get a list of files in the directory then reading then would be fine. Here is the directory structure:
Inbox
--file-a.txt
--file_b.txt
...
I have tried libssh but sftp_readdir required a handle of the open directory. I also looked at paramiko for python but that too requires to open the directory to read the file names.
I am able to do this in bash using send "get Inbox/* ${destination_dir}". Is there anyway I can use a similar pattern match but on c++ or python?
Also, I cannot execute bash commands through my binary. Does anyone know of any library in python or c++ (preferred) that would support this?
I have not posted here in a while so please excuse me if I am not following the formatting. I will learn from your suggestions. Thank you!

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I am trying to zip file which is in the format of Amazon*.xls in unix and also remove the source file after compression.Below is the used command
zip -m Amazon`date +%Y-%m-%d:%H:%M:%S`.zip Amazon*.xls
For the above command i am getting below error
zip I/O error: No such file or directory
zip error: Could not create output file Amazon.zip
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It is not the zip, it is how your shell deals with expanding/substituting variables. Two lines solution for bash
export mydate=`date +%Y-%m-%d:%H:%M:%S`
zip -m Amazon_$mydate.zip *matrix*
Execute by hand (few secs difference) or better put in a shell script myzipper.sh and just source it.
Use '-p' instead of '-m', if zip files are to be extracted on Windows OS.
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zip -p Amazon_$mydate.zip matrix

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I have a number of zip files located in a single folder eg:
file1.gz
file2.gz
file3.gz
file4.gz
I'm looking for a way of automatically unzipping these using a batch job to a similarly named folder structure so for example the contents of file1.gz will drop into a folder named file1.
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Any help is greatly appreciated.
Which OS are you using? This is something you'd do using the shell's capabilities, you could write
for A in *.gz ; do gunzip $A ; done
I'm using gunzip here, because .gz is actually gzip, But you can use the 7zip CLI tool as well, of course. If you're on Windows, then I recommend installing a real shell (the standard cmd.exe can not really be considered a shell IMHO).

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