PHP - Pass named function to another function

Question

I have a function collectRows that accepts a sqli_result object, and returns the query results as a array of rows:

function collectRows($result) {
    ...
}

and have another function that accepts a function to decide what to do with the query results before returning them:

function execQuery($query, $resultsF=null) {
    ...    
}

I want to use execQuery like so:

execQuery("SELECT * FROM...", collectRows);

But I get complaints that it can't find the constant "collectRows".

Obviously, I could do something like:

execQuery("SELECT * FROM...", function($result) {
    return collectRows($result);
});

But it would be nice to be able to write it a more succinctly.

Is there a way to pass a named function without wrapping it in an anonymous function?

u_mulder · Accepted Answer · 2016-02-21 20:59:39Z

2

Function name instead of function itself should be passed to a function:

function execQuery($query, $functionName=null) {
    if (is_callable($functionName)) {
        $functionName();
    }
}

And call it:

execQuery("SELECT * FROM...", 'collectRows');

More info about callable type.

answered Feb 21, 2016 at 20:59

u_mulder

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1 Comment

Oh, that's weird. I can just call a string and it will look up the function? Thanks.

Jim Driscoll · Accepted Answer · 2016-02-21 21:01:06Z

1

Quite simply, you only have to pass the name as a string. PHP doesn't care, it'll deal with it anyway.

answered Feb 21, 2016 at 21:01

Jim Driscoll

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