how to replace a string at a specific position in a csv file using bash - bash

Question

I have several .csv files and each csv file has lines which look like this.
AA,1,CC,1,EE
AA,FF,6,7,8,9
BB,6,7,8,99,AA
I am reading through each line of each csv file and then trying to replace the 4th position of each line beginning with AA with "ZZ"
Expected output
AA,1,CC,ZZ,EE
EE,FF,6,ZZ,8,9
BB,6,7,8,99,AA
However the variable "y" does contain the 4th variable "1" and "7" respectively, but when I use sed command it replaces the first occurrence of "1" with "ZZ".
How do I modify my code to replace only the 4th position of each line irrespective of what value it holds?
My code looks like this
$file = "name of file which contains list of all csv files"
for i in `cat file`
while IFS = read -r line;
do
if [[ $line == AA* ]] ; then
y=$(echo "$line" | cut -d',' -f 4)
sed -i "s/${y}/ZZ/" $i
fi
done < $i

Answer 1

Using sed, you can also direct that only the 4th field of a comma separated values file be changed to "ZZ" for lines beginning "AA" with:
sed -i '/^AA/s/[^,][^,]*/ZZ/4' file
Explanation
sed -i call sed to edit file in place;
general form /find/s/match/replace/occurrence; where
find is /^AA/ line beginning with "AA";
match [^,][^,]* a character not a comma followed by any number of non-commas;
replace /ZZ/4 the 4th occurrence of match with "ZZ".
Note, both awk and sed provide good solutions in this case so see the answers by #perreal and #RavinderSingh13
Example Input File
$ cat file
AA,1,CC,1,EE
AA,FF,6,7,8,9
BB,6,7,8,99,AA
Example Use/Output
(note: -i not used below so the changes are simply output to stdout)
$ sed '/^AA/s/[^,][^,]*/ZZ/4' file
AA,1,CC,ZZ,EE
AA,FF,6,ZZ,8,9
BB,6,7,8,99,AA

Answer 2

To robustly do this is just:
$ awk 'BEGIN{FS=OFS=","} $1=="AA"{$4="ZZ"} 1' csv
AA,1,CC,ZZ,EE
AA,FF,6,ZZ,8,9
BB,6,7,8,99,AA
Note that the above is doing a literal string comparison and a literal string replacement so unlike the other solutions posted so far it won't fail if the target string (AA in this example) contains regexp metachars like . or *, nor if it can be part of another string like AAX, nor if the replacement string (ZZ in this example) contains backreferences like & or \1.
If you want to map multiple strings in one pass:
$ awk 'BEGIN{FS=OFS=","; m["AA"]="ZZ"; m["BB"]="FOO"} $1 in m{$4=m[$1]} 1' csv
AA,1,CC,ZZ,EE
AA,FF,6,ZZ,8,9
BB,6,7,FOO,99,AA
and just like GNU sed has -i for "inplace" editing, GNU awk has -i inplace, so you can discard the shell loop and just do:
awk -i inplace '
BEGIN { FS=OFS="," }
(NR==FNR) { ARGV[ARGC++]=$0 }
(NR!=FNR) && ($1=="AA") { $4="ZZ" }
{ print }
' file
and it'll operate on all of the files named in file in one call to awk. "file" in that last case is your file containing a list of other CSV file names.

Answer 3

EDIT1: Since OP has changed requirement a bit do adding following now.
awk 'BEGIN{FS=OFS=","} /^AA/||/^BB/{$4="ZZ"} /^CC/||/^DD/{$5="NEW_VALUE"} 1' Input_file > temp_file && mv temp_file Input_file
Could you please try following.
awk -F, '/^AA/{$4="ZZ"} 1' OFS=, Input_file > temp_file && mv temp_file Input_file
OR
awk 'BEGIN{FS=OFS=","} /^AA/{$4="ZZ"} 1' Input_file > temp_file && mv temp_file Input_file
Explanation: Adding explanation to above code too now.
awk '
BEGIN{ ##Starting BEGIN section of awk which will be executed before reading Input_file.
FS=OFS="," ##Setting field separator and output field separator as comma here for all lines of Input_file.
} ##Closing block for BEGIN section of this program.
/^AA/{ ##Checking condition if a line starts from string AA then do following.
$4="ZZ" ##Setting 4th field as ZZ string as per OP.
} ##Closing this condition block here.
1 ##By mentioning 1 we are asking awk to print edited or non-edited line of Input_file.
' Input_file ##Mentioning Input_file name here.

Answer 4

Using sed:
sed -i 's/\(^AA,[^,]*,[^,]*,\)[^,]*/\1ZZ/' input_file