Script to extract passwd name field data into a variable

[Kurtkarl]

HI,
Just want to know if anyone out there has a readymade script wherein:
It extracts $5 (name description) from the /etc/passwd file and put those values into variables. I wanted to grab the first, middle and last name from $5 and store them into variables so we can use those variables to extract data from our database using sql. My problem here is, there are some login id’s from passwd files that has 2 words on it (first and last names) and some has 3 words (first, middle and last). Once I have those variables I can pass it on my Informix sql scripts to extract the data I needed.

I tried awk but I’m having a hard time figuring out how to handle the part that checks if $5 has 2 or 3 values on it. I tried using (if NF == 2) or (if NF == 3) conditions but still I’m not handling the loop so well.

Any help would be greatly appreciated.

Thanks

[Rasheed Tamton]

Hi,

FIRST=`awk -F: ' {print $5}' /etc/passwd |cut -f1 -d" "`

SEC=`awk -F: ' {print $5}' /etc/passwd |cut -f2 -d" "`

THIRD=`awk -F: ' {print $5}' /etc/passwd |cut -f3 -d" "`


You could extract using awk with split

awk -F: '{split($5,a," "); print a[1]}' /etc/passwd
awk -F: '{split($5,a," "); print a[2]}' /etc/passwd
awk -F: '{split($5,a," "); print a[3]}' /etc/passwd

Pass it to variables.

Regards.

[James R. Ferguson]

Hi:

Use 'split' which will return the number of elements it places in the array. For example:

# awk -F":" '{n=split($5,a,/,/);print n,"=",$5}' /etc/passwd

Regards!

...JRF...

[Kurtkarl]

Hi Rasheed,

Thanks for the reply. But how would I tell the script to check first how many fields that $5 has? Reason behind this is we have some entries on our passwd file $5 that has two entries (ex. James Dean) and some has 3 (ex. John H Cook). Your script will always assumed that field 2 is middle name wherein in some cases it is the last name.

Thanks again

[Peter Nikitka · Accepted solution]

Hi,

this assumes, that a field containing only two words results in an empty middle name:

awk -F: '{n=split($5,nn," ")
if(n>2) {nm=nn[2]; n2=nn[3]}
else {nm="";n2=nn[2]}
printf("user=%s first=%s middle=%s last=%s\n",$1,nn[1],nm,n2)
}' /etc/passwd

If field5 contains more than 3 words: they are ignored.

mfG Peter

[Kurtkarl]

thanks for all your replies. Really appreciate it. Peter, thanks for the script. This is what I needed.