[SOLVED] Dealing with spaces in awk print

[comptweb]

I am trying to get my google mini to index my redhat box. It needs to follow links to index a folder that is jsut full of pdf files.

I used:

ls | awk {'print "<a href=\"mydocs/"$1"\">"$1"</a>"'} >> jump.html

to create a page full of links that the mini can find and follow.

80% of my filenames do not have spaces in them, but the ones that do have spaces do not show up in the jump.html correctly. All I get is the characters before the space.

audit_report_399.pdf shows up like:

<a href="/mydocs/audit_report_399.pdf">audit_report_399.pdf</a>

but audit report 400.pdf shows up like:

<a href="/mydocs/audit'>audit</a>

anyway to change the awk {print} command I am using to use &20 for the spaces? or another solution?

Greg

[tpl]

welcome to the forum

one thing would be, to run the "ls" output through "sed"
replacing all blanks with...something else: like this--

ls | sed 's/ /_/g'

using the underscore: now no filenames have any blanks,
proceed as before

[comptweb]

I was think along those lines with Sed, but if I actually rename the files it will break the links to them in my CMS (joomla).

If I can replace the spaces with %20, then the links from the CMS should still work...

Hmm....

[ghostdog74]

then try using $0 instead of $1

[aram535]

If you're trying to build a searchable index than you can't change the name of the file or remove the spaces. Unless you do that on the file system as well.

To keep the non-encoded characters you have to change them to their ord value

so create a encode.sed file with:

s/%/%25/g
s/ /%20/g
s/ /%09/g
s/!/%21/g
s/"/%22/g
s/#/%23/g
s/\$/%24/g
s/\&/%26/g
s/'\''/%27/g
s/(/%28/g
s/)/%29/g
s/\*/%2a/g
s/+/%2b/g
s/,/%2c/g
s/-/%2d/g
s/\./%2e/g
s/\//%2f/g
s/:/%3a/g
s/;/%3b/g
s//%3e/g
s/?/%3f/g
s/@/%40/g
s/\[/%5b/g
s/\\/%5c/g
s/\]/%5d/g
s/\^/%5e/g
s/_/%5f/g
s/`/%60/g
s/{/%7b/g
s/|/%7c/g
s/}/%7d/g
s/~/%7e/g

Than you can do:

ENCODED=$(echo "${FOO}" | sed -f encode.sed)

Now $ENCODED should be the encoded string that you can store and request from the webserver.

[lomcevak]

then try using $0 instead of $1

Which might work if you have the ls output one file per line which can be done with the -1 option. Try this:

ls -1 | awk {'print "<a href=\"mydocs/"$0"\">"$0"</a>"'}

[ghostdog74]

Which might work if you have the ls output one file per line which can be done with the -1 option. Try this:

ls -1 | awk {'print "<a href=\"mydocs/"$0"\">"$0"</a>"'}

try this

ls | awk '{print $0}'

[comptweb]

then try using $0 instead of $1

This was the solution.