Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
486 views
in Technique[技术] by (71.8m points)

validation - PHP function missing argument error

My validate function looks like that

function validate($data, $data2 = 0, $type)
{
...

Function call example

if ($result = validate($lname, 'name') !== true)
        response(0, $result, 'lname');

As you see, my validate function has 3 input vars. I'm not using second var - $data2 often, that's why set it to 0 by default. But when I'm calling this function as given example (as far as I know it means $data=$lname, $data2=0, $type='name') getting error message

Missing argument 3 ($type) for validate() 

How can I fix that?

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

Missing argument 3 ($type) for validate()

Always list optional arguments as the last arguments. Since PHP doesn't have named parameters nor "overloading ala Java", that's the only way:

function validate($data, $type, $data2 = 0) {
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...