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Merge multiple files found in config file

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# 1  
Old 01-26-2012
Merge multiple files found in config file
I have a config file with a bunch of these type of blocks:

Code:
        <concat destfile="${standard.js.file}" append="true">
            <filelist dir="${js.dir}/foo" files="foo.js, foo2.js"/>
            <filelist dir="${js.dir}" files="foo3.js"/>
            <filelist dir="${js.dir}/bar/js" files="bar.js" />
            <!-- misc comment to throw things off a bit -->
            <filelist dir="${js.dir}" files="finalfoo.js"/>
        </concat>

I need to grab all these files, in the order they are here, and only within this section (the unique factor is the destfile value), and merge them all into a single file, newfoo.js

I already know how to take multiple files and concatenate them with cat file1 file2 > concatfile, but I'm struggling to figure out how to get the files out of the config file, then combine those.
# 2  
Old 01-26-2012
You mean you want only the data between the concat tags?

--ahamed
# 3  
Old 01-26-2012
I want the content of the files merged into one, so if the contents of ${js.dir}/foo/foo.js is 'foo' and ${js.dir}/foo/foo2.js contains 'bar', the first part of 'newfoo.js' would be:
Code:
foo
bar

and would be followed by the contents of all the other JS files within that <concat> element

---------- Post updated at 07:14 PM ---------- Previous update was at 06:42 PM ----------

To clarify:

I want to take that bit of code in the OP, come back with essentially this:

Code:
cat $jsdir/foo/foo.js $jsdir/foo/foo2.js $jsdir/foo3.js $jsdir/bar/bar.js $jsdir/finalfoo.js > /finalfoo.js

$jsdir will be defined in the calling script, so it'll look like this:

Code:
configfile="config.xml"
jsdir="./js/"

open $configfile
for every filelist between "${standard.js.file}" and "/concat"
files=/dir=\"${js.dir}(.+)\" files=\"(.+)\"/
for each file, prepend $jsdir/
sed $files s/\,// #(remove commas)
cat $files > /finalfoo.js

# 4  
Old 01-26-2012
Try this...
Code:
jsdir="./js/"

for file in $( awk -F"[=\"]" '/filelist/{gsub(/\./,"",$3); print $3}' infile )
do
        if [ -f "$file" ]
        then
                cat $file >> newfile.js
        fi
done

--ahamed
This User Gave Thanks to ahamed101 For This Post:
Validatorian
# 5  
Old 01-26-2012
Quote:
Originally Posted by ahamed101
Try this...
Code:
jsdir="./js/"

for file in $( awk -F"[=\"]" '/filelist/{gsub(/\./,"",$3); print $3}' infile )
do
        if [ -f "$file" ]
        then
                cat $file >> newfile.js
        fi
done

--ahamed
For testing, I've replaced the check to see if it's a file, and echo'd the file instead, so:

Code:
for file in $( awk -F"[=\"]" '/filelist/{gsub(/\./,"",$3); print $3}' infile )
do
    echo "$file"
done

And this is the output I get:
Quote:
${jsdir}/foo
${jsdir}
${jsdir}/prototip-2101/js #note: This should be /prototip-2.1.0.1 -- it apprears periods are being stripped out, but they shouldn't
${jsdir}
${cssdir} # from a separate <concat> -- needs to stay within the bounds of the standard.js.file concat
${jsdir}
${jsdir}
${jsdir}
${cssdir}
# 6  
Old 01-26-2012
Try this...
Code:
#!/bin/bash

jsdir="./js"

file_list=$(awk -F"[=\"]" '
        /standard.js.file/,/<\/concat>/{
        if(/filelist/){dir=$3;l=split($6,arr,", ")
        for(i=1;i<=l;i++){val=val" "dir"/"arr[i]}}}
        END{print val}' infile )

for file in $file_list
do
        eval echo ${file/./}
done

--ahamed
Last edited by ahamed101; 01-26-2012 at 11:08 PM.. Reason: Updated Code!
This User Gave Thanks to ahamed101 For This Post:
Validatorian
# 7  
Old 01-27-2012
Super close -- only problem is it's ouputting two forward slashes before the filename:

Code:
./path/to/scripts//foo.js

Last edited by Validatorian; 01-27-2012 at 01:02 AM..
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