wget: downloaded file name

Question

I'm writing a script for Bash and I need to get the name of the downloaded file using wget and put the name into $string.

For example, if I downloading this file below, I want to put its name, mxKL17DdgUhcr.jpg, to $string.

wget http://pics.sitename.com/images/191211/mxKL17DdgUhcr.jpg
45439 (44K) [image/jpeg]
Saving to: «mxKL17DdgUhcr.jpg»

100%[===================================================================================================>] 45 439      --.-K/s   в 0s

2011-12-20 12:25:33 (388 MB/s) - «mxKL17DdgUhcr.jpg» saved [45439/45439]

Extravaganza

maybe wget --content-disposition 'url' – Chalcidice

Answer 1

Use the basename command to extract the filename from the URL. For example:

url=http://pics.sitename.com/images/191211/mxKL17DdgUhcr.jpg
filename=$(basename "$url")
wget "$url"

Taking

Warning: This will not work for urls that contains redirects or dynamic content. Refer to est's answer for correct solution. – Outbid

I like it! But it also won't quite work if there are URL parameters. For example https://github.com/awslabs/aws-well-architected-labs/blob/master/Reliability/300_Testing_for_Resiliency_of_EC2_RDS_and_S3/Code/Python/server.py?raw=1 – Pt

Answer 2

wget --server-response -q -O - "https://very.long/url/here" 2>&1 | 
  grep "Content-Disposition:" | tail -1 | 
  awk 'match($0, /filename=(.+)/, f){ print f[1] }' )

This is the correct version as there are may be several 301/302 redirects and finally a Content-Disposition: header to set the file name

Guessing file name based on URL is not always correct.

Sardella

I like this approach, but unfortunately the awk in Debian derivatives (Ubuntu, e.g.) does not support the 3rd argument in match. – Marquetry

while not always perfectly accurate this is the correct approach. – Locomotion

In Ubuntu, you can use: wget --server-response -q -O - "https://very.long/url/here" 2>&1 | grep "Content-Disposition:" | tail -1 | awk -F"filename=" '{print $2}' – Outbid

Modern easy way to achive it: wget {link} --content-disposition – Materials

@balbelias: this will correctly make wget use the name suggested by the server. But... how to retrieve it to assign it to a variable? – Memberg

Answer 3

You can just specify the filename before downloading, with the -O option to wget:

wget -O myfile.html http://www.example.com/

Lanthanide

While not as "clever" as the other answers, this method actually has the advantage of simplicity and predictability – Pt

Answer 4

As PizzaBeer mentioned, wget says where he's going to save the file. And that's important because it will ensure to not overwrite existing files by adding a number at the end of the filename.

So here's my solution with grep to narrow down the good line (--line-buffered is needed because of how wget works, see here) and sed to extract the filename.

wget --content-disposition $1 2>&1 | grep "Saving to" --line-buffered | sed -r 's/Saving to: ‘(.*)’/\1/'

You can store this in a variable, which will be populated at the end of the download.

Dallis

I found that the quotes on the sed command were formatted incorrectly, but this worked - sed -r "s/Saving to: '(.*)'/\1/" – Astound

Answer 5

You can be explicit about the name like this:

url='http://pics.sitename.com/images/191211/mxKL17DdgUhcr.jpg'
file=`basename "$url"`
wget "$url" -O "$file"

Segal

Answer 6

To handle URL-encoded filenames:

URL="http://www.example.com/ESTAD%C3%8DSTICA(2012).pdf"
BASE=$(basename ${URL})             # ESTAD%C3%8DSTICA(2012).pdf
FILE=$(printf '%b' ${BASE//%/\\x})  # ESTADÍSTICA(2012).pdf
wget ${URL}

Hussy

Answer 7

#!/bin/bash
file=$(wget $1 2>&1 | grep Saving | cut -d ' ' -f 3 | sed -e 's/[^A-Za-z0-9._-]//g')

I like this because wget already tells you the filename it's saving. The sed strips non-filename characters ie. the apostrophes.

Walburga

Answer 8

~ $ URL='http://pics.sitename.com/images/191211/mxKL17DdgUhcr.jpg'
~ $ echo ${URL##*/}
mxKL17DdgUhcr.jpg
~ $ wget $URL -O ${URL##*/}
--18:34:26--  http://pics.sitename.com/images/191211/mxKL17DdgUhcr.jpg
           => `mxKL17DdgUhcr.jpg'

Severn

Answer 9

An alternative to @Gowtham Gopalakrishnan's answer is simply:

wget --server-response -q "https://very.long/url/here" 2>&1 | awk -F"filename=" '{if ($2) print $2}'

Which just outputs the name of the file that is set in the content disposition.

Example

$ wget --server-response -q https://hostname/filename-that-i-liek.zip 2>&1 | awk -F"filename=" '{if ($2) print $2}'
"filename-that-i-liek.zip"

Mosira

Answer 10

I guess you already have the full URL of the file somewhere in a variable. Use Bash parameter expansion to strip the prefix:

echo ${url##*/}

Lefton

Answer 11

So you want to give the file / image name as a parameter.

Try this:

echo -n "Give me the name of file in http://pics.sitename.com/images/191211/ :"

read $string

sudo wget http://pics.sitename.com/images/191211/$string ;;

I think this could help you

Haemal