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using conditional awk statement to control regex pattern

unix awk begin statementConditional block vs conditional statement (if)AWK print regex patternAwk: if and conditional statement in same blockhelp to rectify awk statementBreaking awk statement while searching a pattern in a filePass month name dynamically in AWK (GNU) with control statementUsing if/else statement in awkUsing variables in awk statementsearch pattern pair using awk

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1

I have code that parses through a csv file and then checks each field against a regex. however there are a few fields that need to be mandatory if another field has data in it, so I essentially need a conditional block to control the flow of data . so for example of sample file see below

 "S","HEY","J","B","0",""

so what I need is a way to say

if $1 == "S"
USE THIS regex ($3~/^("[A-Z0-9]1")$/) print "3RDfield invalid-HEADER-FILE";
else 
USE THIS REGEX ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

I have tried using inline version

$1== "S" && ($3~/^"[A-Z0-9]1"$/) print "3RD field invalid-HEADER- FILE"; 
$1 != "S" && ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

awk

share|improve this question

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why not $1 != "S" && ($5~/^("")$/) for that second block? That's what you initially described.
  
  – Kusalananda♦
  Mar 11 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That should be $1==""S"" && .., $44!=""B"" && .., right? (otherwise it will match S,... (an unquoted S as $1).
  
  – mosvy
  Mar 11 at 10:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kusalananda apologies I copied the wrong segment of code see post for correct version
  
  – jordanb111
  Mar 11 at 10:35
  

  
  
  
  

add a comment | 

1

I have code that parses through a csv file and then checks each field against a regex. however there are a few fields that need to be mandatory if another field has data in it, so I essentially need a conditional block to control the flow of data . so for example of sample file see below

 "S","HEY","J","B","0",""

so what I need is a way to say

if $1 == "S"
USE THIS regex ($3~/^("[A-Z0-9]1")$/) print "3RDfield invalid-HEADER-FILE";
else 
USE THIS REGEX ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

I have tried using inline version

$1== "S" && ($3~/^"[A-Z0-9]1"$/) print "3RD field invalid-HEADER- FILE"; 
$1 != "S" && ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

awk

share|improve this question

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Why not $1 != "S" && ($5~/^("")$/) for that second block? That's what you initially described.
  
  – Kusalananda♦
  Mar 11 at 10:24
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  That should be $1==""S"" && .., $44!=""B"" && .., right? (otherwise it will match S,... (an unquoted S as $1).
  
  – mosvy
  Mar 11 at 10:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Kusalananda apologies I copied the wrong segment of code see post for correct version
  
  – jordanb111
  Mar 11 at 10:35
  

  
  
  
  

add a comment | 

I have code that parses through a csv file and then checks each field against a regex. however there are a few fields that need to be mandatory if another field has data in it, so I essentially need a conditional block to control the flow of data . so for example of sample file see below

 "S","HEY","J","B","0",""

so what I need is a way to say

if $1 == "S"
USE THIS regex ($3~/^("[A-Z0-9]1")$/) print "3RDfield invalid-HEADER-FILE";
else 
USE THIS REGEX ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

I have tried using inline version

$1== "S" && ($3~/^"[A-Z0-9]1"$/) print "3RD field invalid-HEADER- FILE"; 
$1 != "S" && ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

awk

share|improve this question

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

 "S","HEY","J","B","0",""

if $1 == "S"
USE THIS regex ($3~/^("[A-Z0-9]1")$/) print "3RDfield invalid-HEADER-FILE";
else 
USE THIS REGEX ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

$1== "S" && ($3~/^"[A-Z0-9]1"$/) print "3RD field invalid-HEADER- FILE"; 
$1 != "S" && ($3~/^("")$/) print "3RDfield invalid-HEADER-FILE";

share|improve this question

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

share|improve this question

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

edited 2 days ago

Rui F Ribeiro

41.9k1483142

edited 2 days ago

Rui F Ribeiro

41.9k1483142

asked Mar 11 at 10:21

jordanb111jordanb111

83

asked Mar 11 at 10:21

jordanb111jordanb111

83

1 Answer
1

active

oldest

votes

1

 
 if ($1 == ""S"")
 regex = "^"[[:upper:][:digit:]]"$"
 else
 regex = "^""$"
 
 $5 ~ regex print "error"

Or using the ternary operator:

 $5 ~ ($1 == ""S"" ? "^"[[:upper:][:digit:]]"$" : "^""$") 
 print "error"
 

Note that [A-Z], [0-9] could (and in practice sometimes do) match just about anything in locales other than C, while [[:digit:]] matches [0123456789] and [[:upper:]] uppercase letters (all the ones in the locale, not necessarily limited to latin ones without diacritics).

1 is superfluous.

share|improve this answer

edited Mar 11 at 11:01

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately due to the system this will be implemented on it doesn't have posix support so [[:upper:]] and the like wont work on the system hence the older a-z method
  
  – jordanb111
  Mar 11 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  and yes you are correct about 1 I have now removed it
  
  – jordanb111
  Mar 11 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, OK. If it doesn't support POSIX character classes, chances are it won't support 1 either (superfluous anyway). If [A-Z] matches other characters than ABCDEFGHIJKLMNOPQRSTUVWXYZ, you can always replace it with [ABCDEFGHIJKLMNOPQRSTUVWXYZ] or fix the locale to C.
  
  – Stéphane Chazelas
  Mar 11 at 10:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  in my testing it hasn't matched any other characters but I will continue and see if it matches anything else
  
  – jordanb111
  Mar 11 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, in a locale using the UTF-8 charset, you can do perl -XCO -le 'print chr($_) for 0x1..0xd7ff,0xe000..0x10ffff' | awk '/^[A-Z]$/' to find the single-character collating elements that [A-Z] matches (it may also match multi-character collating elements like Hungarian Dzs). For /usr/xpg4/bin/awk on Solaris, I get 1205 characters, including b-z
  
  – Stéphane Chazelas
  Mar 11 at 10:59
  
  

  
  
  
  

 | 
show 3 more comments

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1

 
 if ($1 == ""S"")
 regex = "^"[[:upper:][:digit:]]"$"
 else
 regex = "^""$"
 
 $5 ~ regex print "error"

Or using the ternary operator:

 $5 ~ ($1 == ""S"" ? "^"[[:upper:][:digit:]]"$" : "^""$") 
 print "error"
 

Note that [A-Z], [0-9] could (and in practice sometimes do) match just about anything in locales other than C, while [[:digit:]] matches [0123456789] and [[:upper:]] uppercase letters (all the ones in the locale, not necessarily limited to latin ones without diacritics).

1 is superfluous.

share|improve this answer

edited Mar 11 at 11:01

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately due to the system this will be implemented on it doesn't have posix support so [[:upper:]] and the like wont work on the system hence the older a-z method
  
  – jordanb111
  Mar 11 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  and yes you are correct about 1 I have now removed it
  
  – jordanb111
  Mar 11 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, OK. If it doesn't support POSIX character classes, chances are it won't support 1 either (superfluous anyway). If [A-Z] matches other characters than ABCDEFGHIJKLMNOPQRSTUVWXYZ, you can always replace it with [ABCDEFGHIJKLMNOPQRSTUVWXYZ] or fix the locale to C.
  
  – Stéphane Chazelas
  Mar 11 at 10:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  in my testing it hasn't matched any other characters but I will continue and see if it matches anything else
  
  – jordanb111
  Mar 11 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, in a locale using the UTF-8 charset, you can do perl -XCO -le 'print chr($_) for 0x1..0xd7ff,0xe000..0x10ffff' | awk '/^[A-Z]$/' to find the single-character collating elements that [A-Z] matches (it may also match multi-character collating elements like Hungarian Dzs). For /usr/xpg4/bin/awk on Solaris, I get 1205 characters, including b-z
  
  – Stéphane Chazelas
  Mar 11 at 10:59
  
  

  
  
  
  

 | 
show 3 more comments

1

 
 if ($1 == ""S"")
 regex = "^"[[:upper:][:digit:]]"$"
 else
 regex = "^""$"
 
 $5 ~ regex print "error"

Or using the ternary operator:

 $5 ~ ($1 == ""S"" ? "^"[[:upper:][:digit:]]"$" : "^""$") 
 print "error"
 

Note that [A-Z], [0-9] could (and in practice sometimes do) match just about anything in locales other than C, while [[:digit:]] matches [0123456789] and [[:upper:]] uppercase letters (all the ones in the locale, not necessarily limited to latin ones without diacritics).

1 is superfluous.

share|improve this answer

edited Mar 11 at 11:01

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  unfortunately due to the system this will be implemented on it doesn't have posix support so [[:upper:]] and the like wont work on the system hence the older a-z method
  
  – jordanb111
  Mar 11 at 10:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  and yes you are correct about 1 I have now removed it
  
  – jordanb111
  Mar 11 at 10:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, OK. If it doesn't support POSIX character classes, chances are it won't support 1 either (superfluous anyway). If [A-Z] matches other characters than ABCDEFGHIJKLMNOPQRSTUVWXYZ, you can always replace it with [ABCDEFGHIJKLMNOPQRSTUVWXYZ] or fix the locale to C.
  
  – Stéphane Chazelas
  Mar 11 at 10:39
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  in my testing it hasn't matched any other characters but I will continue and see if it matches anything else
  
  – jordanb111
  Mar 11 at 10:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @jordanb111, in a locale using the UTF-8 charset, you can do perl -XCO -le 'print chr($_) for 0x1..0xd7ff,0xe000..0x10ffff' | awk '/^[A-Z]$/' to find the single-character collating elements that [A-Z] matches (it may also match multi-character collating elements like Hungarian Dzs). For /usr/xpg4/bin/awk on Solaris, I get 1205 characters, including b-z
  
  – Stéphane Chazelas
  Mar 11 at 10:59
  
  

  
  
  
  

 | 
show 3 more comments

 
 if ($1 == ""S"")
 regex = "^"[[:upper:][:digit:]]"$"
 else
 regex = "^""$"
 
 $5 ~ regex print "error"

Or using the ternary operator:

 $5 ~ ($1 == ""S"" ? "^"[[:upper:][:digit:]]"$" : "^""$") 
 print "error"
 

Note that [A-Z], [0-9] could (and in practice sometimes do) match just about anything in locales other than C, while [[:digit:]] matches [0123456789] and [[:upper:]] uppercase letters (all the ones in the locale, not necessarily limited to latin ones without diacritics).

1 is superfluous.

share|improve this answer

edited Mar 11 at 11:01

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

 
 if ($1 == ""S"")
 regex = "^"[[:upper:][:digit:]]"$"
 else
 regex = "^""$"
 
 $5 ~ regex print "error"

 $5 ~ ($1 == ""S"" ? "^"[[:upper:][:digit:]]"$" : "^""$") 
 print "error"
 

share|improve this answer

edited Mar 11 at 11:01

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948

answered Mar 11 at 10:32

Stéphane ChazelasStéphane Chazelas

313k57592948