Replace certain chars in a string based on sub_str location

Alright, Say I have

$string = "000000000000000000000";

And I need to replace the 3rd 0 in with a 1, resulting in "001000000000000000000"

How would I do this?

str_replace won't work because it replaces all 0's obviously, and using sub_str with str_replace like:

str_replace(substr($string, $click2, $click3), "1", $string);

Replaces it with all 1's, so it doesn't work either

Thanks for the help!

 

Is it always the 3rd character? You can use 2 substr's and rejoin them with the new character.

$string = "000000000000000000000";

$string1 = substr($string,0,2);

$string2 = substr($string,4);

$new_string = $string1 . '1' . $string2;

Or combined: $new_string = substr($string,0,2) . '1' . substr($string,4);

 

Not always third, the number will be from user input, I see what you mean though, I cant set the numbers as vars. I'll mess with it a bit.

Thanks.

 

You can also try something like this. You can reference a string's character position as an array.

Basically:

$string = "000000000000000000000";

echo $string; //000000000000000000000

$string[2] = 'p';

echo $string; //00p000000000000000000

The first character is [0], and so on. Probably easier than substr anyway.

 

Try this function:

function replace_character($string, $index, $newchar){
	$before = substr($string, 0, $index - 1);
	$after = substr($string, $index, strlen($string));
	
	return $before.$newchar.$after;
	
}


echo replace_character("000000000000000000000", 3, "1");

PHP:

This Prints:
001000000000000000000

 

Got it

substr_replace();

 

or if you want to use jestop method:

function replace_character($string, $index, $newchar){
 $string[$index-1] = $newchar;
 return $string;
	
}


echo replace_character("000000000000000000000", 3, "1");

PHP:

 

I think you can only replace at the beginning or end, or add in the middle with substr_replace. As far as I know, you can't replace in the middle.

Disregard, was thinking of something else. Just checked the manual and it should work fine.