Question

suddenly my site show new warning –

The relevant code:

printf ("<input type='text' name='C_Comment' value='" . $myComment . "'  >");

The warning I get:

• Warning: printf() function.printf: Too few arguments

probably because $myComment is null.

1. I know I Can fix it if I first test if the value is null, and only then conctenate it. but is there a simpler way?

2. Why did not I get this warning before?

Thanks,

Atara

EDIT: sorry, wrong title. The problem was that $myComment was not NULL, it contained special character.

Answer 1

No, you get that warning because you don’t give enough arguments to printf; probably $myComment contained printf placeholders like %s.

Use echo instead if you don’t want to use printf‘s formatting. You can also rewrite your printf call:

printf ("<input type='text' name='C_Comment' value='%s'>",
    $myComment);

Make sure you’ve escaped special chars in $myComment (see htmlspecialchars).