php build or replace part of regular expression

Question

I have this regular expression and I want to replace the "lb" part with the parameter that's being passed in with what ever weight they want to use so what I have is this

protected function validateCheckWeight($attribute, $value, $parameters) 
{        
    // this always outputs correctly either true or false
    return preg_match('/^(?!0\d|0lb$)\d+(?:\.\d+)?lb$/', $value);
}

and what I want is this but it doesn't work when doing validation

protected function validateCheckWeight($attribute, $value, $parameters) 
{        

    return return preg_match('@/^(?!0\d|0' . $parameters[0] .'$)\d+(?:\.\d+)?' . $parameters[0] . '$/@', $value);    
}



protected function validateCheckWeight($attribute, $value, $parameters) 
{   
    dd('@/^(?!0\d|0' . $parameters[0] .'$)\d+(?:\.\d+)?' . $parameters[0] . '$/@');
    // output "@/^(?!0\d|0kg$)\d+(?:\.\d+)?kg$/@"
    return preg_match('@/^(?!0\d|0' . $parameters[0] .'$)\d+(?:\.\d+)?' . $parameters[0] . '$/@', $value);
    // this always outputs false

}

What exactly does "doesn't work" mean? You need to supply a test case. What values are you passing in for $attribute, $value and $parameters, and what return value are you expecting? — Asaph
– Asaph, Commented Mar 12, 2016 at 22:48
it returns true of false and $parameters[0] is the weight type e.g. kg or lbs — ONYX
– ONYX, Commented Mar 12, 2016 at 22:50

Asaph · Accepted Answer · 2016-03-12 22:58:22Z

1

You have 2 different delimiters in your regex, @ and /. I've just confirmed that if you remove @ from the beginning and end of your regular expression, your function returns true for the inputs mentioned in your comments ($value = "kg", $parameters = array('kg')).

answered Mar 12, 2016 at 22:58

Asaph

163k25 gold badges204 silver badges204 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

ONYX Over a year ago

Thank you that was the problem I learned something new today cheers

Naumov · Accepted Answer · 2016-03-12 23:03:23Z

0

If I correct understand this question you can

return strlen(parameters[0]) == strlen(rtrim(parameters[0],'lb'));

but I think again don't correct understand this.

answered Mar 12, 2016 at 23:03

Naumov

1,1879 silver badges22 bronze badges

Collectives™ on Stack Overflow

php build or replace part of regular expression

2 Answers 2

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related