Using html input pattern in awk

Question

echo 8d07\'54.520\"W | awk '{ if ($1 ~ /[-+]?[0-9]*[.]?[0-9]+/) print $1; else print "erro" }'

I'm trying to check if it's a number, but it's no working... I use this same regex in a html input text, and it works.

In this case I was expecting "erro". It's not working.

My final goal is to apply 3 different pattern match to 3 fields $1 $2 $3...

jaypal singh · Accepted Answer · 2013-06-26 06:16:55Z

1

Not 100% sure of the requirement but you probably need to put anchors.

$ echo 8d07\'54.520\"W | awk '{ if ($1 ~ /^[-+]?[0-9]+[.]?[0-9]+/) print $1; else print "erro" }'
erro

answered Jun 26, 2013 at 6:16

jaypal singh

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6 Comments

apinhal Over a year ago

It doesn't validate in the same way... I don't understand.

jaypal singh Over a year ago

Without anchors ^ your regex matches any part of your $1 and prints it. If you need to match, you have to tell the regex to start looking from the start of $1 and not anywhere in between $1

apinhal Over a year ago

OK! I'm trying to understand. In html input this is done by default... This is the correct pattern [-+]?[0-9]*[.]?[0-9]+ because with your if you put one digit it gives erro

apinhal Over a year ago

I'm reading this more carefully en.wikipedia.org/wiki/Regular_expression This a more complete pattern for number.

apinhal Over a year ago

The $ at the end is very important the patter like this worked! ^[-+]?[0-9]*[.]?[0-9]+$ Thanks

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