Shell Programming and Scripting

Hi All,

Im bit stuck s in writing shell script which downloads the file from the SFTP server

Script first list the available files over server and check the today's file in the list.
if file available then transfer function gets called and file downloading starts.
iam able to list out the available files found the today's file in below format.

Code:

available_file_variable=`grep $source_file $listing_file | sort -u`

Code:

+ echo -rw-rw---- 1 vgif_VGL vgif_VGL 32298503 May 06 19:12 UAT.pVV.SID.VGL.D170506.T110027.AMA.MAS.FTP.DATA.gz

now I need to pass the actual filename ( ) to the variable by which transfer function gets initated.

so here i have the value in available_file_variable =

Code:

-rw-rw---- 1 vgif_VGL vgif_VGL 32298503 May 06 19:12 UAT.pVV.SID.VGL.D170506.T110027.AMA.BVS.FTP.DATA.gz

i need to cut only filename from the available file variable not file permission ,timestamp etc and set to the transfer_file means available file variable >>transfer_file.

how i can do it by cut or awk to set transfer_file = UAT.pVV.SID.VGL.D170506.T110027.AMA.BVS.FTP.DATA.gz

You could correct the variable, by using it like this:

Code:

${var##* }

But that will only work if the files on the ftp server never contain spaces..
Correctly parsing ls -l output can sometimes be surprisingly difficult, because it may vary because if it is younger or older than 6 months or maybe even the locale setting..

I think best would be to adjust the script so that ls -1 is used instead of ls -l (ls minus one vs. ls minus el).

Hi ,
Sorry didnt get you , my simple requirement is i have value in variable this

Code:

-rw-rw---- 1 vgif_VGL vgif_VGL 32298503 May 06 19:12 UAT.pVV.SID.VGL.D170506.T110027.AMA.MAS.FTP.DATA.gz

i need to fetch only file name

Code:

UAT.pVV.SID.VGL.D170506.T110027.AMA.MAS.FTP.DATA.gz

please advise.

What I meant was you could pass:

Code:

${available_file_variable##* }

instead of

Code:

${available_file_variable}

What I was trying to say is that this should work, if the file name does not contain spaces.

--
You could also change the command substitution to something like this:

Code:

available_file_variable=$(awk -v f="$source_file" "$listing_file" '$0~f {print $NF}' | sort -u)

or

Code:

available_file_variable=$(awk -v f="$source_file" "$listing_file" '$0~f && !A[$NF]++{print $NF}')

Also, providing the file name does not contain space characters...

--
But best would be to adjust the script so that it only stores the file names in the file $listing_file, my suggestion was to use ls -1 rather than ls -l to produce $listing_file

Well all you need once you got your ls -l output is a pipe with awk to get the 9th field: e.g. with ls -l ->

Code:

 ls -l|awk '{print $9}'

That will sometimes work with the right locale and if a time six months ago results in the same number of fields and if filenames do not contain spaces.
Otherwise it may fail.

It will also sometimes fail if owner or group name are too long, causing owner and group fields to "blend", which results in fewer fields than provisioned.

Parsing ls -l output is unreliable, unless the variability of the output is known precisely.

Thank you to both, could manage to do it via awk.