PHP function missing argument error

Question

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?

Call like validate($lname, 0,'name')

Rikesh
– Rikesh

2011-11-09 12:27:28 +00:00
Commented Nov 9, 2011 at 12:27 — Rikesh
– Rikesh, Commented Nov 9, 2011 at 12:27

hakre · Accepted Answer · 2022-07-02 09:28:00Z

22

Missing argument 3 ($type) for validate() ^[1]

Always list optional arguments as the last arguments, never before non-optional arguments.

Since PHP doesn't have named parameters¹ nor "overloading ala Java", that's the only way:

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

¹ Error with severity E_WARNING until PHP 7.0 (including); Uncaught ArgumentCountError starting with PHP 7.1^rfc (and starting with PHP 8.0 as well for internal functions^rfc).
² before PHP 8.0, see Named Arguments

edited Jul 2, 2022 at 9:28

hakre

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answered Nov 9, 2011 at 12:27

Berry Langerak

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Comments

Aurelio De Rosa · Accepted Answer · 2011-11-09 12:27:14Z

6

You should at least set the $type in this line:

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

at NULL or '' as you can see here:

function validate($data, $data2 = 0, $type = null)

PHP let you to set a value for the parameters, but you can't define a parameter WITHOUT a preset value AFTER parameter(s) which HAVE a preset value. So if you need to always specify the third param, you have to switch the second and the third like this:

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

answered Nov 9, 2011 at 12:27

Aurelio De Rosa

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Comments

Aziz · Accepted Answer · 2011-11-09 12:28:39Z

0

From http://php.net/manual/en/functions.arguments.php

Note that when using default arguments, any defaults should be on the right side of any non-default arguments; otherwise, things will not work as expected

Your should switch the second and third arguments of the function, making the optional argument the last one. So it becomes:

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

answered Nov 9, 2011 at 12:28

Aziz

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Comments

ahmed · Accepted Answer · 2018-03-22 07:30:48Z

0

function validate($data, $data2, $data3, $data4, $data5)

im a beginner but i think that you can use a thousand arguments as long as you call like that

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

answered Mar 22, 2018 at 7:30

ahmed

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Comments

user832146 · Accepted Answer · 2019-03-12 22:37:50Z

0

Notice that starting with PHP 7.1 this will throw a PHP Fatal error, not just a warning:

PHP Fatal error:  Uncaught ArgumentCountError: Too few arguments to function validate(), 2 passed in /path/to/file.php on line X and exactly 3 expected

More info: http://php.net/manual/en/migration71.incompatible.php

answered Mar 12, 2019 at 22:37

user832146

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