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Try
sed -n '/row/ {:L; N; /\/row/ !bL; /<source_name>19<\/source_name>/ p; }' file
When found row, collect following lines until /row, and if the pattern space contains source...19..., print, else...

I would also replace this construction
for i in ${3}/$file ; do
i="${i##*/}"

much better indeed but why use any external utility when the shell can do the same:

for i in $(ls "${3}/${file}") ; do
i="${i##*/}"

The only way this is going to break is if the filename...

sed -e 's/JAN_DAT_(.*\)\_([0-9]\{10\}\).*.csv/\1 \2/g'

The first thing is: sed-patterns are always "greedy", which means they match the longest possible part of the line. A pattern like A.*B will...

In this case: cut off the other parts with what i have showed you:

for file in /some/where/*csv ; do
path="${file%/*}" # remove the filename, leaving the path ...

This looks like a continuation of your earlier thread in September (rather than homework/coursework).
Here is an awk script that does a merge like you want:
#!/bin/sh
awk '
$1=="}" { bl-- }
bl>0...

There is (as is always the case) a number of ways of doing this. Here's one with awk:

awk -F'"' '/EqName/ { sub(" ", "_", $2) }1' file
{
EqName Tan_1
....
....
}

{

EqName ...