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I am running the command 

curl -O "http://www.*site*.com/[1-9].png"

but the download is failing. - the output is as follows

[1/10]: http://www.*site*.com/1http://w --> 1.png
  % Total    % Received % Xferd  Average Speed   Time    Time     Time  Current
                                 Dload  Upload   Total   Spent    Left  Speed
100   531    0   531    0     0   1618      0 --:--:-- --:--:-- --:--:--  1618

and the file 1.png has the following html:

<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<html><head>
<title>404 Not Found</title>
</head><body><script type="text/javascript">//<![CDATA[try{(function(a){var b="http://",c="www.*site*.com",d="/cdn-cgi/cl/",e="img.gif",f=new a;f.src=[b,c,d,e].join("")})(Image)}catch(e){}//]]></script>
<h1>Not Found</h1>
<p>The requested URL /1http://w was not found on this server.</p>
<p>Additionally, a 404 Not Found
error was encountered while trying to use an ErrorDocument to handle the request.</p>
</body></html>

looks like it is repeating the address at the end of the request - not sure why

Running the command for each file one at a time works

curl -O "http://www.*site*.com/1.png"

note I have also tried 

curl "http://www.*site*.com/[1-9].png" -o "#1.ext"

so Its not the -O option.

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2 Answers 2

Reset to default
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you could do with xargs or a simple for loop:

for i in `seq 0 9` ; do curl -O "http://www.*site*.com/$i.png"; done

EDIT: i didn't know you could use such syntax with curl... i tried and it works for me with your syntax, even without the -o parameter. (with curl 7.26.0)

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I was able to do it with wget with your example:

wget http://www.*site*.com/[1-9].png

aka in my real life test:

wget https://raw.githubusercontent.com/jodumont/configFile/master/etc/mysql/conf.d/{docker,mysql,mysqld,mysqld_safe_syslog}.cnf
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