How to search a pattern like pattern using grep only? - grep

I am searching list of names with pattern "japconfig".There are many files inside one directory. Those files contain names like ixdf_japconfig_FZ.txt,
ixdf_japconfig_AB.txt, ixdf_japconfig_RK.txt, ixdf_japconfig_DK.txt, ixdf_japconfig_LY.txt. But I don't know what are the names present after japconfig word. I need to list down all such names. Also my files contain ixdf_dbconfig.txt, but I don't want to print ixdf_dbconfig.txt in the output.
Each of my file contains one ixdf_japconfig_*.txt and ixdf_dbconfig.txt where * can be FZ,AB,RK,DK,LY. I can achieve my desired result by using grep and then awk to cut the columns.But I don't want to use AWK or other command. I want to achive using grep only.
I need to print below names.
ixdf_japconfig_FZ.txt
ixdf_japconfig_AB.txt
ixdf_japconfig_RK.txt
ixdf_japconfig_DK.txt
ixdf_japconfig_LY.txt
I don't want to print ixdf_dbconfig.txt.
When I tried using "grep -oh "ixdf_japconfig.*.txt" *.dat" command, I am getting below output.
ixdf_japconfig_FZ.txt ixdf_dbconfig.txt
ixdf_japconfig_AB.txt ixdf_dbconfig.txt
ixdf_japconfig_RK.txt ixdf_dbconfig.txt
ixdf_japconfig_DK.txt ixdf_dbconfig.txt
ixdf_japconfig_LY.txt ixdf_dbconfig.txt
where first column is my desired column. But I don't want to print second column. How can I change my code to print only first column?


grep -oh ixdf_japconfig_...txt *.dat
(Your .*. was matching most of the line.)

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Concatenating the files should work for you then:
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James' answer is way better,
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