get array out of loop - php

Question

I need get array out of loop (I need used array, and not, last value)
loop
for ($x = 1; $x < $numero; $x++) {
$frase = $frase_script[$x];
$distrito1 = (explode(',',$frase));
echo $distrito1[0]}
Variable out
$ultimo_nome = $distrito1[0];
I need used array, and not, last value
echo "<br> I need print array, and not, last value".$ultimo_nome;
error: prints the last value and not an array.
Example
all code
$ch = curl_init();
curl_setopt($ch, CURLOPT_URL, "http://api.geonames.org/searchJSON?username=country=pt&lang=pt&q=lisbon&fcode=ADM2&adminCode1=14&style=SHORT&maxRows=1000");
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($ch, CURLOPT_HEADER, FALSE);
curl_setopt($ch, CURLOPT_HTTPHEADER, array(
"Accept: application/json"
));
$response = curl_exec($ch);
curl_close($ch);
$frase_script = (explode(',',$response));
$frase = $frase_script[0];
$palavra = (explode(':',$frase));
$numero = $palavra[1];
$frase_script = (explode('"name":',$response));
echo '[';
for ($x = 1; $x < $numero; $x++) {
$frase = $frase_script[$x];
$distrito1 = (explode(',',$frase));
echo $distrito1[0]; }
$ultimo_nome = $distrito1[0];
echo $ultimo_nome;
echo ']';
echo "<br> I need print array, and not, last value".$ultimo_nome;

Answer 1

if you need print all the elements but not the lasy you could use
$distrito1 = (explode(',',$frase));
$numElem = count($distrito1);
foreach ($distrito1 as $key => $value){
if ( $key < $numElem-1 ){
echo $value;
} else {
break;
}
}