Why the the newline doesn't work with printf() in PHP

Question

IGNORE THIS QUESTION. ASKED IN ERROR:

What's the difference between the following 2 printf statements:

printf($variable, "
");

printf("Hello
");

The newline gets ignored in the 1st printf() statement. But it works fine with the 2nd statement.

The only way I can use the new line is by splitting the 1st statement in to 2 separate statements:

printf($variable);
printf("
");

This sounds like a query from an absolute novice however I feel the newline was supported pretty good in Java but not in PHP.

Answer 1

is being ignored because as stated in documentation first parameter is format which should be printed and rest of parameters are values which this format can use, for instance:

$num = 2.12; 
printf("formatted value = %.1f", $num);
//      ^^^^^^^^^^^^^^^^^^^^^^   ^^^^
//             |                  |
//             format             |
// value which can be put in format in place of `%X` where `X` represents type

will print formatted value = 2.1 because in format %.1f you decided to print only one digit after dot in floating point number.

To make format use string argument you need to use %s placeholder like in case of print("hello %s world","beautiful") which would put beautiful in place of %s and print hello beautiful world.

Now lets back to your code. In printf($variable, " "); $variable represents format, and it most probably doesn't have any %s for string in it which would let you put use " " argument in this format. This means that " " will be ignored (not used in format) so will not be printed.

Code like this

printf("Hello
");

or

printf($variable);
printf("
");

doesn't have this problem because it explicitly use in format which should be printed.

So it seems that you may either want to use

printf("%s
", $value) 

which seems like overkill because you can simply concatenate strings using . operator and print them like

print($value."
")

or

echo $value."
";