RegEx: Get multiline LDAP entry

user_7619269 0
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I'm trying to capture an entire LDAP entry from dn:.+ to the entry's last line, but stopping at last line before next entry, e.g., \n#entry-id: 8266. My trial and error using egrep is getting absolutely nowhere. NOTE: I'm using exported ldif files where the data resides, fwiw.

Closest I've come is with egrep "dn: cn=name,ou=People,dc=example,dc=com.+.|\n*.+\n" but no output on terminal. I've tested the actual regex on regexr.com. I understand that is a completey different env.

Thanks in advance!

Sample Data:


dn: cn=name,ou=People,dc=example,dc=com \
shadowLastChange: 17492 \
userPassword: password \
sn: Last \
givenName: First \
cn: first \
mail: name@example.com \
displayName: First Last \
o: University \
ou: Dept. \
objectClass: top \
objectClass: person

\# entry-id: 8266
ldap grep regex
3 Answers

If the data is always structured like that, and using awk is an option, you can use a range starting with dn: and ending with entry-id: and only print the lines that do not have entry-id:

awk '/^dn:/,/entry-id/ {
  if(!/entry-id:/){print}
}' file

Awk demo

user_54249880

With your shown samples, please try following awk code.

awk '/entry-id/{found=""} /^dn:/{found=1} found' Input_file

OR in case you want to print only 1 set from dn: before entry-id: then try following code:

awk '/entry-id/{exit} /^dn:/{found=1} found' Input_file
user_58665800

egrep uses extended regexp (equivalent to grep -E). Prefer grep -P (perl regexp) instead.
The -z flag makes your regex multiline:

grep -Pz "dn(\n|.)*?(\n\n|$)"

This matches with dn followed by any number of characters (or new line) until (? makes the preceding expression lazy instead of greedy) the next occurrence of \n\n or end of file ($)

user_135255120
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