Parse error: syntax error, unexpected 'curl_setopt' (T_STRING) - php

Question

I wached a youtube video and came across these codes.
my php file index.php
<?php
//initialize session
$ch = curl_init();
//set the URL
$url = "http://localhost:81/data.php"
//set options
curl_setopt($ch, CURLOPT_URL,$url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_HEADER, 0);
//execution
$json=curl_exec($ch);
//close
curl_close($ch);
//decode the json
$json=json_decode($json, true);
//loop through the results
for($i=0;$i<['Metadata']['TotalResults'];$i++){
echo "JSON : <b>First Name = </b>". $json['Result'][$i]["FirstName"]." , <b>Last Name = </b>".$json['Result'][$i]["LastName"];
}
?>
my json file data.php
<?php
//metadata which will contain how many resukts we have
$meta_array['TotalResults'] = 1;
$metadata = '{"Metadata" : ';
$metadata .= json_encode($meta_array). ',';
//the data
$array["FirstName"] = 'Gary';
$array["LastName"] = 'George';
$data[] = $array;
//json encode the array
$json_encoded = utf8_encode(json_encode($data));
echo $metadata . ' "Result" : '.$json_encoded. '}' ;
exit();
?>
I am continuously getting the above mentioned error.
I am new to jason and curl.
please help me to solve my problem.
thanks in advance :)

Answer 1

I can't see any semicolon after the url.
$url = "http://localhost:81/data.php";
^ it is missing
In your code block you are using $i<['Metadata']['TotalResults'] at loop. Here you are missing the variable name that is eventually your array name. An example could be:
$i<$your_array['Metadata']['TotalResults']