PHP using 'use' in anonymous function throws exception

Question

I'm working in the latest version of PHP, within a class, and writing a sorting function which includes the following:

public static function sort_alphabetically($data, $sortBy=null)
{
    ...

    if($sortBy)
    {
        function sortBy($a, $b) use ($sortBy)
        {
            if ($a->$sortBy == $b->$sortBy){return 0;}
            if ($a->$sortBy > $b->$sortBy){return 1;}
            else{return -1;}
        };

        usort($data, "sortBy");
    }
    return $data;
}

yet I'm consistently getting

Parse error: syntax error, unexpected 'use' (T_USE), expecting '{'...

It's difficult to google the word use and get useful results, but I have seen a mention that it can't be used within a class or namespace. However, the PHP documentation does exactly that - http://php.net/manual/en/functions.anonymous.php - so I am not sure whether the author of that bit of the Internet was correct.

Can anyone shed some light on this? Is there an alternative method I could use?

If a function has a name, it's not anonymous

Sam Dufel
– Sam Dufel

2014-10-20 14:45:49 +00:00
Commented Oct 20, 2014 at 14:45 — Sam Dufel
– Sam Dufel, Commented Oct 20, 2014 at 14:45
true. I had overlooked that, thanks

BIU
– BIU

2014-10-20 15:03:08 +00:00
Commented Oct 20, 2014 at 15:03 — BIU
– BIU, Commented Oct 20, 2014 at 15:03
.. and it's a parse error, not an exception.

Karoly Horvath
– Karoly Horvath

2014-10-20 15:03:31 +00:00
Commented Oct 20, 2014 at 15:03 — Karoly Horvath
– Karoly Horvath, Commented Oct 20, 2014 at 15:03
yeah yeah...technicalities... ;)

BIU
– BIU

2014-10-20 15:04:48 +00:00
Commented Oct 20, 2014 at 15:04 — BIU
– BIU, Commented Oct 20, 2014 at 15:04
Possible duplicate of Use php namespace inside function

Sunny Patel
– Sunny Patel

2018-09-18 20:21:18 +00:00
Commented Sep 18, 2018 at 20:21 — Sunny Patel
– Sunny Patel, Commented Sep 18, 2018 at 20:21

thpl · Accepted Answer · 2014-11-28 06:44:13Z

The use keyword is (not the only use case) used to import variables into closures or anonymous functions (other uses are the import of namespaces or traits).

That does not apply to conventional functions that have a name (like in your case). What you did is just nesting a function into another function. In order to be able to import variables into a closure or an anonymous function you need to declare one first.

You have the following options:

public static function sort_alphabetically($data, $sortBy=null)
{
    ...

    if($sortBy)
    {
        usort($data, function($a, $b) use ($sortBy){
            if ($a->$sortBy == $b->$sortBy){return 0;}
            if ($a->$sortBy > $b->$sortBy){return 1;}
            else{return -1;}        
        });
    }
    return $data;
}

Here you would make use of a closure and import the $sortBy variable into it.

public static function sort_alphabetically($data, $sortBy=null)
{
    ...

    if($sortBy)
    {
        $callback = function($a, $b) use ($sortBy){
            if ($a->$sortBy == $b->$sortBy){return 0;}
            if ($a->$sortBy > $b->$sortBy){return 1;}
            else{return -1;}        
        };

        usort($data, $callback);
    }
    return $data;
}

In this case an anonymous function is being used.

I wrestled a bear once. · Accepted Answer · 2014-10-20 14:53:03Z

0

The use language construct is valid for anonymous functions only. Your function is not anonymous because you gave it a name.

public static function sort_alphabetically($data, $sortBy=null)
{
    ...

    if($sortBy)
    {
        $f = function($a, $b) use ($sortBy)
        {
            if ($a->$sortBy == $b->$sortBy){return 0;}
            if ($a->$sortBy > $b->$sortBy){return 1;}
            else{return -1;}
        };

        usort($data, $f);
    }
    return $data;
}

Remove the name, you should be fine.

answered Oct 20, 2014 at 14:53

I wrestled a bear once.

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2 Comments

Ben Swinburne Over a year ago

This won't work because the call to usort in your example has no access to the now unammed function. sortBy is referenced by name as the second parameter

I wrestled a bear once. Over a year ago

Woops, forgot about that.. fixed

Ben Swinburne · Accepted Answer · 2014-10-20 14:53:17Z

0

You can't use use unless it's an anonymous function or closure. Assuming you're not reusing the function elsewhere you can easily adjust your code to suit, as below. Similarly you could assign the function to a variable and use the variable as the second argument of usort

public static function sort_alphabetically($data, $sortBy=null)
{
    if($sortBy)
    {
        usort($data, function ($a, $b) use ($sortBy)
        {
            if ($a->{$sortBy} == $b->{$sortBy}){return 0;}
            if ($a->{$sortBy} > $b->{$sortBy}){return 1;}
            else{return -1;}
        });
    }

    return $data;
}

answered Oct 20, 2014 at 14:53

Ben Swinburne

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Comments

Mark Shust at M.academy · Accepted Answer · 2014-10-20 14:48:33Z

-1

move the function outside of the other function. nesting functions is really hard to read, especially the way you are using the names within this function. it's most likely causing the compiler issue.

answered Oct 20, 2014 at 14:48

Mark Shust at M.academy

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