Regex in IF condition in awk Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) 2019 Community Moderator Election Results Why I closed the “Why is Kali so hard” questionHow to use regrex with AWK for string replacement in this example?Regex for matchng anything between parenthesisawk adds extra comma at several placesawk ifs and variables - cannot pass a variable from one line towards subsequent linesUsing multiple awk commands within single lineBest way to iterate through the following awk commandawk or sed command to match regex at specific line, exit true if success, false otherwiseHow to add a condition within a for loop during substitution if string is nullPrint text before and after match, from a specific beginning and to an ending stringsearch column 2 in csv file for value, if value, then insert “invalid” and shift cells right
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Regex in IF condition in awk
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)
2019 Community Moderator Election Results
Why I closed the “Why is Kali so hard” questionHow to use regrex with AWK for string replacement in this example?Regex for matchng anything between parenthesisawk adds extra comma at several placesawk ifs and variables - cannot pass a variable from one line towards subsequent linesUsing multiple awk commands within single lineBest way to iterate through the following awk commandawk or sed command to match regex at specific line, exit true if success, false otherwiseHow to add a condition within a for loop during substitution if string is nullPrint text before and after match, from a specific beginning and to an ending stringsearch column 2 in csv file for value, if value, then insert “invalid” and shift cells right
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?
Required regex condition: [[ "$1" =~ [A-Za-z] ]]
BEGIN FS = ";"; counter=0
if ((length($1) != 10 && length($1) != 12))
counter++
print counter, $1;
if ($counter -gt 2)
print "Invalid input file";
exit;
I am getting error if I use the same condition which I have posted. How to add the condition?
awk regular-expression
edited Apr 15 at 13:05
muru
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
add a comment |
I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?
Required regex condition: [[ "$1" =~ [A-Za-z] ]]
BEGIN FS = ";"; counter=0
if ((length($1) != 10 && length($1) != 12))
counter++
print counter, $1;
if ($counter -gt 2)
print "Invalid input file";
exit;
I am getting error if I use the same condition which I have posted. How to add the condition?
awk regular-expression
edited Apr 15 at 13:05
muru
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
1
See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just[A-Za-z]. What are you adding to yourawk?
– terdon♦
Apr 15 at 12:34
add a comment |
I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?
Required regex condition: [[ "$1" =~ [A-Za-z] ]]
BEGIN FS = ";"; counter=0
if ((length($1) != 10 && length($1) != 12))
counter++
print counter, $1;
if ($counter -gt 2)
print "Invalid input file";
exit;
I am getting error if I use the same condition which I have posted. How to add the condition?
awk regular-expression
edited Apr 15 at 13:05
muru
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
I have awk script file as below. I need to add another condition in the if statement to check if the string contains atleast one alphabet. How can I add the extra condition to the present if statement?
Required regex condition: [[ "$1" =~ [A-Za-z] ]]
BEGIN FS = ";"; counter=0
if ((length($1) != 10 && length($1) != 12))
counter++
print counter, $1;
if ($counter -gt 2)
print "Invalid input file";
exit;
I am getting error if I use the same condition which I have posted. How to add the condition?
awk regular-expression
awk regular-expression
edited Apr 15 at 13:05
muru
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
edited Apr 15 at 13:05
muru
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
edited Apr 15 at 13:05
muru
38k590166
edited Apr 15 at 13:05
muru
38k590166
edited Apr 15 at 13:05
muru
38k590166
38k590166
asked Apr 15 at 12:20
LaxmanLaxman
254
asked Apr 15 at 12:20
LaxmanLaxman
254
asked Apr 15 at 12:20
LaxmanLaxman
254
254
1
See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just[A-Za-z]. What are you adding to yourawk?
– terdon♦
Apr 15 at 12:34
add a comment |
1
See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just[A-Za-z]. What are you adding to yourawk?
– terdon♦
Apr 15 at 12:34
1
1
See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just
[A-Za-z]. What are you adding to your awk?– terdon♦
Apr 15 at 12:34
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just
[A-Za-z]. What are you adding to your awk?– terdon♦
Apr 15 at 12:34
add a comment |
1 Answer
1
active
oldest
votes
You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.
BEGIN FS = ";"; counter=0
if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
counter++
print counter, $1;
if (counter > 2)
print "Invalid input file";
exit;
Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
Yes. I was using=~format.
– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ...seems more « awkish » to me, avoiding some seriously nested braces
– D. Ben Knoble
Apr 15 at 16:53
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
add a comment |
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You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.
BEGIN FS = ";"; counter=0
if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
counter++
print counter, $1;
if (counter > 2)
print "Invalid input file";
exit;
Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
Yes. I was using=~format.
– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ...seems more « awkish » to me, avoiding some seriously nested braces
– D. Ben Knoble
Apr 15 at 16:53
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
add a comment |
You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.
BEGIN FS = ";"; counter=0
if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
counter++
print counter, $1;
if (counter > 2)
print "Invalid input file";
exit;
Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
Yes. I was using=~format.
– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ...seems more « awkish » to me, avoiding some seriously nested braces
– D. Ben Knoble
Apr 15 at 16:53
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
add a comment |
You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.
BEGIN FS = ";"; counter=0
if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
counter++
print counter, $1;
if (counter > 2)
print "Invalid input file";
exit;
Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
You don't actually show how you add the regex, so I am guessing you are using the same format: =~ [A-Za-z]. That won't work. Each language has its own syntax for regex matching. In awk, the format is $target ~ /$regex/, so $1 ~ /[A-Za-z]/.
BEGIN FS = ";"; counter=0
if (length($1) != 10 && length($1) != 12 && $1 ~ /[A-Za-z]/)
counter++
print counter, $1;
if (counter > 2)
print "Invalid input file";
exit;
Also, in awk, the $ sign is used to mark fields, not variables. So $counter will be evaluated to the field number of counter. If counter is 2, then $counter will be the value of the second field. And the -gt is also not an awk thing. Just use >.
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
edited Apr 15 at 16:58
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
answered Apr 15 at 12:36
terdon♦terdon
134k33270450
134k33270450
Yes. I was using=~format.
– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ...seems more « awkish » to me, avoiding some seriously nested braces
– D. Ben Knoble
Apr 15 at 16:53
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
add a comment |
Yes. I was using=~format.
– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ...seems more « awkish » to me, avoiding some seriously nested braces
– D. Ben Knoble
Apr 15 at 16:53
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
Yes. I was using
=~ format.– Laxman
Apr 15 at 12:44
Yes. I was using
=~ format.– Laxman
Apr 15 at 12:44
Why couldnt you move the condition to be the pattern?
length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ... seems more « awkish » to me, avoiding some seriously nested braces– D. Ben Knoble
Apr 15 at 16:53
Why couldnt you move the condition to be the pattern?
length($1) != 10 && ... && $1 ~ /[A-Za-z]/ counter++ ... seems more « awkish » to me, avoiding some seriously nested braces– D. Ben Knoble
Apr 15 at 16:53
1
1
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
@D.BenKnoble why indeed! I just copy/pasted the OP's code and added the regex. I just didn't realize it was that nested. Thanks!
– terdon♦
Apr 15 at 17:00
add a comment |
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See the examples here: gnu.org/software/gawk/manual/gawk.html#Regexp-Usage
– steeldriver
Apr 15 at 12:24
How do you add it? You are showing us a bash-style if statement. The regex is fine, but the regex is just
[A-Za-z]. What are you adding to yourawk?– terdon♦
Apr 15 at 12:34