PHP doesn't replace variables with defined values in a String

Question

Here is a part of my PHP application:

Stage 1:
I get an array of payment methods from user and assign an ID to it, because it can contain multiple choices.

I filled $paymentMethod variable with some examples.

$paymentMethod = ['PayPal','Visa','Master'];
$paymentMethodNames = array("PayPal", "Visa", "Master", "COD");
$paymentMethodNumerical = array("1", "2", "3", "4");
$paymentMethodCalculated = implode(str_replace($paymentMethodNames, $paymentMethodNumerical, $paymentMethod));

the result of echo $paymentMethodCalculated is 123 which is perfectly correct.

Stage 2:
I need to create a special variable which is needed to work for another part of my application.

for ($x = 0; $x < strlen($paymentMethodCalculated); $x++) {
$paymentFinal .= '$_'.substr($paymentMethodCalculated, $x, 1);
    if(isset($paymentFinal) && $x !== strlen($paymentMethodCalculated) - 1) {
       $paymentFinal .= '.",".';
    }
}

The result of echo $paymentFinal is $_1.",".$_2.",".$_3 which is perfectly correct.

Stage 3:
Now I define:

$_1 = Test;
$_2 = Test2;
$_3 = Test3;

Now, when I echo $paymentFinal
It still shows:
$_1.",".$_2.",".$_3

But my desired result is:
Test.",".Test2.",".Test3

Question: Why PHP does not replace defined variables in $paymentFinal variable?

You would have to define $_1 prior to referring to them in your statement for them to be automatically replaced. — Nigel Ren
– Nigel Ren, Commented Dec 30, 2021 at 19:08
Because those are variable of variable, so you'll have to eval it. Although I would really recommend not doing it, as pointed in the manual: Caution The eval() language construct is very dangerous because it allows execution of arbitrary PHP code. Its use thus is discouraged. If you have carefully verified that there is no other option than to use this construct, pay special attention not to pass any user provided data into it without properly validating it beforehand. — β.εηοιτ.βε
– β.εηοιτ.βε, Commented Dec 30, 2021 at 19:09
Why don't you use an array like you did for $paymentMethodNumerical, in this specific case? — β.εηοιτ.βε
– β.εηοιτ.βε, Commented Dec 30, 2021 at 19:13

bobkorinek · Accepted Answer · 2021-12-30 19:15:44Z

1

The problem is that PHP doesn't alter your string based on your defined variables. You have variable, which contains set of characters which remain the same until you alter the string by yourself (like str_replace or something similar). The "variables" inside $paymentFinal ("$_1.",".$_2.",".$_3") are just set of characters; they don't represent the actual variables you have defined inside your code.

answered Dec 30, 2021 at 19:15

bobkorinek

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3 Comments

Nima Masalegooha Over a year ago

Thanks for your definite answer. Can you explain the logic of this behavior of PHP? Wouldn't be better if it was possible?

bobkorinek Over a year ago

This is a normal behavior of any programming language. The variable is just a pointer to some space inside the memory, where the values of variables are stored. And very simply, the string variables are stored there as well. They are just bytes located inside the memory. Nothing more. If you put "$_1" inside your memory, it will be set of bytes representing this value. The programming language won't try to replace this string with other variables (as you thought), unless you tell it explicitly to do it (like str_replace).

bobkorinek Over a year ago

Of course if we would create a language, where this could be an automatic feature, it would be a really dangerous. Imagine creating a program, where users could put their own input. If the user put the "variable" we're talking about, it could brake or expose the program. If you want similar behavior, it would be better to create some list/map, where you could put those "variables" (for example as a key) and access them there. This way if someone would put something that is not inside the list, it wouldn't allow them to access it. Of course, there can be many other ways to do this thing!

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