Using argument text in PHP Function

Question

I am a little new to the PHP world and despite my googling and manual reading, I can't

I am currently (ethically!) scraping a website that has a number of different categories (let us call them 'alpha', 'beta', 'gamma').

Currently, I go through these steps:

Load the file
Locate the correct table (php_siomple_dom)

Then go through the elements to grab the information.

$alpha_one = $element -> find etc;
$alpha_two = $element -> find etc;
$alpha_n   = $element -> find etc;

$beta_one  = $element -> find etc;
$beta _two = $element -> find etc;
$beta_n    = $element -> find etc;

All of the tables are the same for alpha, beta, gamma; thus I would like to write a function, but I am struggling with how to include the argument name in the variables I create. This was my idea, but it does not work.
```
function grab($argument) {

    $argument . "_one" = $element -> find etc;
    $argument . "_two" = $element -> find etc;
    $argument . "_n"   = $element -> find etc;

}
```

Thus, I could use:

grab('alpha');

grab('beta');

grab('gamma');

without having to write out the code for each table

I've looked into 'magic methods', but cannot swing it into working. I would really appreciate if some PHP wizard could shed some light upon this. Thank youu!

If you have a situation where you're numbering a series of variables, then you have a situation that calls for the use of an array instead — Mark Baker
– Mark Baker, Commented Jan 7, 2015 at 20:23
@KimAlexander yes, I did. I will update the question. Thank you. — tsdryt
– tsdryt, Commented Jan 7, 2015 at 20:27
@MarkBaker they actually aren't numbers, they are categories, but the hope of simplicity, I used numbers instead of my confusing category names. Thanks for the reply, though! — tsdryt
– tsdryt, Commented Jan 7, 2015 at 20:28
just simple note: you can do : $myvarname='anyVarNemIWant'; $$myvarname=5; echo $anyVarNameIWant; output: 5 That allows you to use value of string var as a new (or existing) var name — Alex
– Alex, Commented Jan 7, 2015 at 20:29

Marc B · Accepted Answer · 2015-01-07 20:17:42Z

1

That code is rather pointless. the $argument . "_one" variable would exist ONLY within the function, and be destroyed when the function returns. You probably want something more like this:

function grab($argument) {
    $data = array();
    $data[$argument . "_one"] = ...
    $data[$argument . "_two"] = ...
    etc...
    return ($data);
}

Essentially: build an array with dynamic key names, based on your $argument, then return the entire array to the calling context.

answered Jan 7, 2015 at 20:17

Marc B

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1 Comment

AbraCadaver Over a year ago

Then $data = grab('alpha'); echo $data['alpha_one'];

Waldson Patricio · Accepted Answer · 2015-01-07 20:19:20Z

0

$argument only exists inside the function scope. It will not have any effect outside this function. If you're using a class you can create all possible variables (not recommended) and call:

$this->{$argument . "_one"} = ...

or store them in an array variable (best way):

$this->myArray[$argument . '_one'] = ...

answered Jan 7, 2015 at 20:19

Waldson Patricio

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Comments

motanelu · Accepted Answer · 2016-10-16 00:25:28Z

0

Just do this:

function grab($argument) {
    ${$argument . "_one"} = $element -> find etc;
    ${$argument . "_two"} = $element -> find etc;
    ${$argument . "_n"}   = $element -> find etc;
}

edited Oct 16, 2016 at 0:25

answered Jan 7, 2015 at 20:15

motanelu

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1 Comment

Alex Over a year ago

${$argument . "_one"} :-) fix

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