03.06.2018, 05:40
You already assigned something to "var" here "new var = dini_Int(filename2,"Int");" , then y are you assigning something else to var here? var=var+raison; . I think thats the issue , you could try replacing
Because variables are containers so it can hold one at a time.
Aren't you trying to save a string? you cant do it like that
PHP код:
var=var+raison;
to
new "variable name"=var+raison;
Aren't you trying to save a string? you cant do it like that
