PHP - Using a variable in a variable (string)

Question

The task is to create a variable (string) in which there will be variables that will be replaced with a value during the output.

Like that:

$s1 = "it's $s2 !"; $s2 = 'OK'; echo $s1; //не работает echo "it's $s2 !"; //должно работать так

Maybe I'm not right approach to solving the problem. Line $ s1 is a piece of HTML code that is output in a loop and the variable $ s2 is constantly changing. Actually, I know how to solve using a function, but the variant with variables seemed more elegant to me.

The $s2 variable must exist when you create $s1 , since it is used in the contents.
I understand why the error, I do not understand how to fix it)
Can you escape the $ sign when creating the first variable?
This seems to be just what you need, but I tried - it does not work, can you give an example of code?

Visman Visman 16.2k eight 21 52 · Accepted Answer · 2017-10-14T12:27:57

Since you do not change variables in some places, use the strtr () function

 $s1 = "it's \$s2 !"; $s2 = 'OK'; echo strtr($s1, ['$s2' => $s2]);

Http://sandbox.onlinephpfunctions.com/code/cbc725c89ce1a85ccfc58dd401350b34568ca39d test

Thank you, thought about this. Apparently this is the best solution. - wakh.ru

Answer 2 · 2017-10-14T11:45:09

Use variable shielding in a string, followed by searching and replacing with one of the appropriate functions for this purpose. For example, str_replace () :

 $s1 = "it's \$s2 !"; $s2 = 'OK'; echo str_replace('$s2', $s2, $s1); // it's OK !

Brilliant!)) Variables are created in different places and of course the order of creation cannot be changed, otherwise the meaning will just disappear.
@ wakh.ru in order to receive your messages in a timely manner, in front of the username to which you are sending a comment, register the dog @ .
@ wakh.ru edited your answer according to your editing in the PP.

ishidex2 ishidex2 6.212 2 12 40 · Answer 3 · 2017-10-14T11:54:14

you have $s2 not yet initialized

 $s2 = 'OK'; $s1 = "it's $s2 !"; echo $s1; echo "it's $s2 !";

PHP - Using a variable in a variable (string)

3 answers 3

More articles: