Variable question regarding single quotion

[xAdrianRonsten]

This one is working:

$characters = "\s\-\'";

This one is not working:

$characters = '\s\-\'';

How do i make the one below working without using double quotations?

 

[Dave]

$characters = '\s\-\\\'';

Escape backslashes.

 

[xAdrianRonsten]

$characters = '\s\-\\\'';

Escape backslashes.

so this one $name = 'Bob'Bob';

how do so echo $name; outputs Bob'Bob and not only Bob?`

 

[AngraMainyu]

so this one $name = 'Bob'Bob';

how do so echo $name; outputs Bob'Bob and not only Bob?`

Option 1: Alternate quotes (") and ('). Works for simple strings with one type of quote when you don't want to escape ****.
Option 2: Just escape the (') (i.e. 'Bob'Bob' becomes 'Bob\'Bob'). This is normally what you should do.