[Solved] Shell scripting to find the delimiter

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To count the number of columns with awk you can use the NF variable:

$ cat file
ABC|12345|EAR
PQRST|123|TWOEYES
ssdf|fdas,sdfsf
$ awk -F\| 'NF!=3' file
ssdf|fdas,sdfsf

However, this does not seem to cover all the possible ways the data could be corrupted based on the various revisions of the question and the comments.

A better approach would be to define the exact format that the data must follow. For example, assuming that a line is “correct” if it is three columns, with the first and third letters only, and the second numeric, you could write the following script to match all non conforming lines:

awk -F\| '!(NF==3 && $1$3 ~ /^[a-zA-Z]+$/ && $2+0==$2)' file

Test (notice that only the second line (which is conforming) does not get printed):

$ cat file
A,BC|12345|EAR
PQRST|123|TWOEYES
ssdf|fdas,sdfsf
ABC|3983|MAKE,
sf dl lfsdklf |kldsamfklmadkfmask |mfkmadskfmdslafmka
ABC|abs|EWE
sdf|123|123
$ awk -F\| '!(NF==3&&$1$3~/^[a-zA-Z]+$/&&$2+0==$2)' file
A,BC|12345|EAR
ssdf|fdas,sdfsf
ABC|3983|MAKE,
sf dl lfsdklf |kldsamfklmadkfmask |mfkmadskfmdslafmka
ABC|abs|EWE
sdf|123|12

You can adapt the above command to your specific needs, based on what you think is a valid input. For example, if you wanted to also restrict the length of each line to 50 characters, you could do

awk -F\| '!(NF==3 && $1$3 ~ /^[a-zA-Z]+$/ && $2+0==$2 && length($0)<50)' file

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solved Shell scripting to find the delimiter