How to get rid of empty values when adding string in php - php

Question

I have something like this:
public function options()
{
$out = '';
$docs = $this->getAll();;
foreach($docs as $key => $doc) {
$out .= ',{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
}
return $out;
}
It gives me a list of options from the DB, but it also gives me a null value at the top.
if I write it like this:
public function options()
{
//$out = '';
$docs = $this->getAll();;
foreach($docs as $key => $doc) {
$out = '';
$out .= '{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
}
return $out;
}
It doesn't give me the null value but it only returns one value.
$out .= ',{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
In this line if I don't add an , it gives me an error message, This because I have $out = ''; at the top. Now can you guys give me an idea how can I get all the values from the DB without the empty value at the beginning.
I also have another question , why we use ;; (double semicolon) in this code:
$docs = $this->getAll();;

Answer 1

test $out to see if it has any length, if so add the comma and the line, otherwise just set it to be the line:
$out="";
foreach($docs as $key=>$doc){
if(strlen($out)){
$out.=',{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
}else{
$out='{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
}
}
as to your other question, er, you wrote the code, so why did you put a double semi-colon?

Answer 2

This is not the correct way to build JSON. First create an array, and use json_encode() on it.

Answer 3

I'd suggest using an array instead to hold the individual values, and using join to concatenate them together.
public function options()
{
$docs = $this->getAll();
// Create an empty array
$items = array();
foreach($docs as $key => $doc) {
// "Push" an item to the end of the array
$items[] = '{"label" : "' . $doc['name'] . '", "value" : "' . $doc['id'] .'"}';
}
// Join the contents together
$out = join(",", $items);
return $out;
}
Also, the double semi-colon is completely unnecessary.