Shell script using curl to loop through urls

Question

I've been trying to create a simple script that will take a list of queries from a .txt file, append the main url variable, then scrape the content and output it to a text file.

Here's what I have so far:

#!/bin/bash  url="example.com/?q=" for i in $(cat query.txt); do     content=$(curl -o $url $i)     echo $url $i     echo $content >> output.txt done 

list:

images news stuff other 

error log:

curl: (6) Could not resolve host: other; nodename nor servname provided, or not known example.com/?q= other 

If I use this command straight from the command line I get some output into the file:

curl -L http://example.com/?q=other >> output.txt 

Ultimately I would like the output to be:

fetched:    http://example.com/?q=other content:    the output of the page  followed by the next query in the list. 

Answer 1

Use more quotes !

• http://mywiki.wooledge.org/Quotes
• http://mywiki.wooledge.org/Arguments
• http://wiki.bash-hackers.org/syntax/words

Try this instead :

url="example.com/?q=" for i in $(cat query.txt); do     content="$(curl -s "$url/$i")"     echo "$content" >> output.txt done 

Answer 2

You've got nested quotes, try something like this:

#!/bin/bash  url=https://www.google.fr/?q= while read query do     content=$(curl "{$url}${query}")     echo $query     echo $content >> output.txt done < query.txt