populating an array with a for loop in php - php

Question

I need to add * in an array. This is how i do it in javaScript.
function makeStarString(grade) {
var starArray = [];
var starString = "";
for (var i = 0; i < grade; i++){
starArray[i] = "*";
starString = starString.concat(starArray[i]);
}
return starString;
}
The javascript version works fine but I cant make it work with php.
This is as far as i got with php.
function makeStarString($varGrade) {
$starArray = array ();
$starString = "";
for ($i = 0; $i < strlen($varGrade); $i++){
$starArray($i) = $star;
$starString = $starString.(starArray(i));
}
return $starString;
}
I get this error "Fatal error: Can't use function return value in write context" because of line $starArray($i) = $star;
The argument I send to the function is an Integer.
So the purpose of the function is that if i send a 5 to the function it will return *****
How can i fix my php function?

Answer 1

Use $starArray[$i] instead of $starArray($i) and starArray(i).

Answer 2

If you really need stars in an array, you don't need to use a loop:
$starArray = array_fill(0, $varGrade, '*');
and if you need to turn the array into a string:
$starString = implode('', $starArray);
But if you don't really ever need to use the array of stars, it would be easier to just use str_repeat:
str_repeat('*', $varGrade);

Answer 3

It would generally be a great idea to use strlen($varGrade) in a variable, because the foreach loop will have to count the length every iteration. This might bring performance issues.
Your $star variable is not defined, so I have no idea what you're trying to put there. Revise your own code.
Finally, you may use the .= operator to add something to existing string.
function makeStarString($varGrade) {
$starArray = array ();
$starString = "";
$length = strlen($varGrade);
for ($i = 0; $i < $length; $i++){
$starArray[$i] = $star;
$starString .= $starArray[$i];
// equivalent to $starString = $starString . $starArray[$];
}
return $starString;
}
UPDATE
If you send an int to the function, you don't need strlen:
function makeStarString($varGrade) {
$starArray = array ();
$starString = "";
for ($i = 0; $i < $varGrade; $i++){
$starArray[$i] = $star;
$starString .= $starArray[$i];
// equivalent to $starString = $starString . $starArray[$];
}
return $starString;
}

Answer 4

This code should work
function makeStarString($varGrade) {
$star = "*"; // star variable wasn't defined to.
$starArray = array ();
$starString = "";
for ($i = 0; $i < $varGrade; $i++){
$starArray[$i] = $star; // [ ] instead of ( )
$starString .= ($starArray[$i]); // $a.=$b instead of $a=$a+$b, added the $ at the i and [ ] instead of ( )
}
return $starString;
}

Answer 5

Is this what you need?
function makeStarString($varGrade) {
$starString = '';
for ($i = 0; $i < $varGrade; $i++) {
$starString .= '*';
}
return $starString;
}
A few notes on my modifications.
I don't see why you need that $starArray array in the first place. So I skipped it.
I revised your indentation. Strange indentation can make very simple code seem much more complicated than it really is. :)
You should ask if $i < $varGrade, not if $i < strlen($varGrade). If you ask by strlen(), then you get the width of the number you enter. For example, strlen(55) is 2, strlen(555) is 3, strlen(5555) is 4 etc. - ignoring the fact that it's a number. You just want the for-loop to give you $varGrade many stars so there is no need for strlen().
As a detail, I've put used single-quotes instead of double-quotes for strings because they're lighter (PHP doesn't parse variables inside them).
I hope it helps.