Removing all spaces in a scraped block of html (help!)

[theninjamanz 29]

I've got a bot that is scraping pretty happily and placing all the contents in a nicely formatted CSV, but there's a  bug that is making it fall foul of its duty.

 

When the bot encounters a block of HTML where the user has made use of a carriage return the scrape is captured like this:

 

Example from this page:

 

        ipb.vars['highlight_color'] = "#ade57a";
        ipb.vars['charset']                = "UTF-8";
        ipb.vars['time_offset']            = "-7";
        ipb.vars['hour_format']            = "12";
        ipb.vars['seo_enabled']            = 1;
       

What I need to do is take the above and perform some kind of regex or other on this to get the following result.

 

ipb.vars['highlight_color'] = "#ade57a";ipb.vars['charset']                = "UTF-8"; ipb.vars['time_offset']            = "-7"; ipb.vars['hour_format']            = "12"; ipb.vars['seo_enabled']            = 1;

 

IE that all the hard breaks get removed and the output gets put into one line.

 

Does anyone have any idea how to accomplish this.

 

My ubot can then return to duties.

 

Cheers.

 

       

[UBotDev 276]

You mean something line $replace(#text,$new line,"") ?

[LazyBotter 188]

Something like that?

set(#Content, "  ipb.vars[\'highlight_color\'] = \"#ade57a\";
        ipb.vars[\'charset\']                = \"UTF-8\";
        ipb.vars[\'time_offset\']            = \"-7\";
        ipb.vars[\'hour_format\']            = \"12\";
        ipb.vars[\'seo_enabled\']            = 1;", "Global")
set(#Pos, 0, "Global")
clear list(%List)
add list to list(%List, $list from text(#Content, "
"), "Delete", "Global")
set(#Output, "", "Global")
loop($list total(%List)) {
    set(#Output, "{#Output}{$list item(%List, #Pos)}", "Global")
    increment(#Pos)
}

[deliter 203]

replace regular expression search text \s+ replace text blank