get a field from a record

Shell Programming and Scripting
1

I have a file as:
A,B,C,D,E
G,H,I,J,K

I need to find if fourth field is blank or has a space and print that line to other file.

I tried using awk but am not getting the desired result.
Pls help.

2 
awk -F, '$4 ~ /^ *$/' FILE > OTHERFILE
3

Thank you.

4

can I do something like this to add one more condition of checking if the 4th field is number or space or blank also:

awk -F, '$4 /^[]*||[0-9]*/' MYFILE >> OTHERFILE

I also want the other part i.e. I need to exclude all lines whose 4th field is space or blank or number:

MYFILE
a,b,c,d,e
a,b,c,2,r
c,d,f,,g
t,y,u, ,i

The new file should contain only first line. I tried using the below but I also get 3rd line

awk -F, '$4 /^ [A-Z]*/' MYFILE >> OTHERFILE

I need OTHERFILE to have only first record.

5 
 
awk -F, '$4 !~ /[0-9]/  && $4 !~ /[ ]/ && $4!=""' filename
6

I have a similar question. I file a file with lines that look like this

02","00:52:19","knoxtnds1xpdn01","1","200","166.181.191.1","94728","1","174.156.215.192","7072804600","000007076943722","32313530343035333332","1","0","0","0A61E074","0","1235955139","68.28.81.76","10.167.21.20","042300014D51","0","15","13","33","1","1","0","3","3","2","0","","","0","0","0","0","0","","0","31OEqfgO","0","0","","","","","","","","","","","","","","","","",""

As you can see each field is separated by a comma. I'm trying to find all the lines that has "1" on the fourth field.

Any help would be appreciated

7 
awk -F, '$4 == "\"1\"" {print}' filename