I am trying to look for a string in the first line and if this string is valid I will going down one line and get another validation, if this validation is also ok, i will going down again and print a parameter of if line, here is what a have: 140036.log 102.117.135.00 140036.log 02/27/2009 FAILURE 183422.log 102.117.135.00 183422.log 02/27/2009 SUCCESS 183422.log /CSPGM02_full_refresh//CED/SCED_20090227183439 183321.log 102.117.135.00 183321.log 02/27/2009 SUCCESS 183321.log /physical/switch/CGV/SCGV20090227183341.VM.DAT and here is the awk that I want to use but is not working awk '{ if ($2 = 102.117.135.00) a=$2 getline if ($3 = SUCCESS) getline print p" "$2}'

A single equal sign assigns a value.  To test for a value, you need a double equal sign. String values need to be in double-quotes, otherwise they are assumed to be a variable. Following an if-condition, awk assumes only one statement.    if (a==1)       statement1       statement2 In the above, statement1 is optional based on value of a.  statement2 is not associated with the if-test, and thus will always be executed.  If you want both statements to be based on the if-test, then do:    if (a==1)       {statement1        statement2} In your original code, p" "$2 actually means the variable p followed by a space followed by word2. But p is unassigned, so I am guessing that you wanted a literal p, so I put it inside the quotes. awk '{ if ($2 == "102.117.135.00")    {a=$2     getline     if ($3 == "SUCCESS")        {getline         print "p "$2        }    } }' file.in And just for variety, here is a completely different way to code it: awk '{ if ($2 != "102.117.135.00") next a=$2 getline if ($3 != "SUCCESS") next getline print "p "$2 }' file.in