Passing data between MySql and Objective C - php

Question

I am working on a small social web application as a final project for my iOS class. I have a profile view controller where all the info about the user from the database is supposed to be displayed on the labels. The problem is that I don't really know the best way to do this. Here is my php script:
<?
// Database credentials
$host = 'localhost';
$db = 'blabla';
$uid = 'blabla';
$pwd = 'blabla';
// Connect to the database server
$link = mysql_connect($host, $uid, $pwd) or die("Could not connect");
//select the json database
mysql_select_db($db) or die("Could not select database");
// Create an array to hold our results
$arr = array();
//Execute the query
$rs = mysql_query("SELECT IdUser, username, fullname, phonenumber, facebook, instagram FROM login");
// Add the rows to the array
while($obj = mysql_fetch_object($rs)) {
$arr[] = $obj;
}
//return the json result.
echo '{"users":'.json_encode($arr).'}';
?>
So here I get the info about all the users in the database. I am sure this is not the right way to go, so I guess I need to change the SQL query to retrieve the data for the current user only. But how can I do this? Should I put the username which I enter on the login page into an extra variable and then pass it with JSON to this php script and add the 'WHERE username = 'blabla' statement to the SQL query then? If so, how can I pass the variable to this script with JSON?
Can you please give me some sample code? Or is there a different way to do this?
Thank you so much!

Answer 1

<?php
// Database credentials
$host = 'localhost';
$db = 'blabla';
$uid = 'blabla';
$pwd = 'blabla';
// Connect to the database server
$link = mysql_connect($host, $uid, $pwd) or die("Could not connect");
//select the json database
mysql_select_db($db) or die("Could not select database");
//Execute the query
$rs = mysql_query("SELECT IdUser, username, fullname, phonenumber, facebook, instagram FROM login");
// Add the rows to the array
$data = mysql_fetch_array($rs);
foreach($data as $rec){
echo "user: $rec<br>";
}
?>