How to count the number of multi-valued attributes in an LDAP entry... or lines in a paragraph, where diff paragraphs have different numbers of lines?

Using Miller (mlr) to read the data as "xtab" input (a format where records are separated by empty lines and fields are separated by newlines, with the field name and a tab at the start of the line), with the tabs in the format replaced by : (colon+space):

$ mlr --xtab --ips ': ' put -q 'print $dn, NF - 1' file
[email protected],ou=test,dc=acme,dc=com 1
[email protected],ou=test,dc=acme,dc=com 2
[email protected],ou=test,dc=acme,dc=com 0
[email protected],ou=test,dc=acme,dc=com 1

This simply outputs the dn field and the count of however many other fields there are in the record.

If you need commas in the output, use print $dn . ", " . string(NF - 1) as the put expression. Wrap that in a conditional that only prints the expression if NF > 1 (if you wish), like so:

$ mlr --xtab --ips ': ' put -q 'NF > 1 { print $dn . ", " . string(NF - 1) }' file
[email protected],ou=test,dc=acme,dc=com, 1
[email protected],ou=test,dc=acme,dc=com, 2
[email protected],ou=test,dc=acme,dc=com, 1

Alternatively, add the count as a new field and then cut out the dn field and your new field (output is on a whitespace-delimited indexed fields format ("nidx")):

$ mlr --x2n --ips ': ' put '$c = NF - 1' then cut -f dn,c file
[email protected],ou=test,dc=acme,dc=com 1
[email protected],ou=test,dc=acme,dc=com 2
[email protected],ou=test,dc=acme,dc=com 0
[email protected],ou=test,dc=acme,dc=com 1

Add --ofs ', ' to the options if you want comma+space as the output field delimiter. Here's how that may look, together with filtering out any record whose calculated c value is zero:

$ mlr --x2n --ips ': ' --ofs ', ' put '$c = NF - 1' then filter -x '$c == 0' then cut -f dn,c file
[email protected],ou=test,dc=acme,dc=com, 1
[email protected],ou=test,dc=acme,dc=com, 2
[email protected],ou=test,dc=acme,dc=com, 1

And here's an awk approach:

$ awk -F'[ ,]' -v RS='\n\n' '{n=split($0,a,"\n"); print $2,n-1}' file 
[email protected]  1
[email protected]  2
[email protected]  0
[email protected]  2   <-- WRONG! See text

Explanation

• -F'[ ,]': set the field separator to space or comma.
• -v RS='\n\n'" set the record separator to be \n\n, so a paragraph.
• n=split($0,a,"\n");: split the current record (paragraph) on \n into the array a. The number returned (n) is the number of elements in this array, so the number of \n characters in this record, and therefore the number of attributes plus one.
• print $2,n-1: print the second field (since we are using space and = as the field separator, on your file this will be the string after the first space and before he first ,), and the value of n minus one, so the number of lines minus one.

Note that this is getting the last record wrong. That's because the file doesn't end with two consecutive newline characters but only with the default trailing newline that all text files require. As a result, this is also counted for the last record. You can get around this by adding an extra newline:

$ printf '%s\n\n' "$(cat file)" | 
   awk -F'[ ,]' -v RS='\n\n' '{n=split($0,a,"\n"); print $2,n-1}' 
[email protected] 1
[email protected] 2
[email protected] 0
[email protected] 1

Or, you print only after moving to the next line, and then print the last one specially:

$ awk -F'[ ,]' -v RS='\n\n' '{ 
   if(last){print last}; n=split($0,a,"\n"); last=$2" "n-1}
    END{print $2,n-2
 }' file 
[email protected] 1
[email protected] 2
[email protected] 0
[email protected] 1

Using awk:

   $ awk '/^accountid: /{c++}
          /^dn: /{ 
          if (dn) print dn,c;c=0;
          sub(/^dn /,"");
          dn=$0
    }END{print dn,c}'

Or awk in paragraph mode:

$ awk -v RS= -F'\n' '{sub(/^dn: /,"");print $1"," NF-1}'