How to send query from iOS to PHP code - mysql

Question

I am kinda new to Obj-C and Stack Overflow, so bear with me here (:D). I newly found out how to send queries from PHP to the MySQL database, but I'm not really sure how to send the query from my iOS app to PHP. Any help would be much appreciated. Thanks.

Answer 1

We don't pass queries from objective C to PHP. What we do is pass data to the server and let the server process the data and respond accordingly.
JSON is popular method to exchange data between server and a mobile device.
For example: You want to get all the fields of the database which has column name, address , contact of table person. You can't do direct database query from the device. like SELECT * from person;.
What we do is send HTTP request to an API like e.g getPersonDetails.php.
Now it's server's responsibility to verify an authorized app is sending HTTP request to it. And after verification it will respond back the data to the application in JSON format or could be XML whatever you feel comfortable. JSON is the preferred way though.
The server response would be this in JSON format:
[
{
name:John,
address:US,
contact:1234
},
{
name:David,
address:UK,
contact:1234
}
]
Now after app receives this JSON object and it needs to decode the JSON into the the suitable form.
Refer to documentation of AFNetworking here:
https://github.com/AFNetworking/AFNetworking
AFNetworking is a popular library for making HTTP request to the server for Objective-C.
Addition 1:
HTTP POST Request from iOS application in Objective C using AFNetworking:
AFHTTPRequestOperationManager *manager = [AFHTTPRequestOperationManager manager];
NSDictionary *parameters = #{#"name": #"Andy",
#"address": #"UK",
#"contact": #"1234",};
[manager POST:#"http://example.com/addToDataBase.php" parameters:parameters success:^(AFHTTPRequestOperation *operation, id responseObject) {
NSLog(#"JSON: %#", responseObject);
} failure:^(AFHTTPRequestOperation *operation, NSError *error) {
NSLog(#"Error: %#", error);
}];
Server side code to handle the POST request
http://example.com/addToDataBase.php
<?php
header('Content-type: application/JSON');
$name = $_POST['name'];
$address = $_POST['address'];
$contact = $_POST['contact'];
$mysqli = new mysqli('localhost', 'my_user', 'my_password', 'world');
/* check connection */
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$stmt = $mysqli->prepare("INSERT INTO person VALUES (?, ?, ?)");
$stmt->bind_param('sssd', $name, $address, $contact);
$stmt->execute();
$stmt->close();
$mysqli->close();
$result = array( "result" => "Successfully Inserted Data");
$output= json_encode($result);
echo $output;
?>

Answer 2

have IOS output something like:
$run_var = '$varName="variable value"';
then run it through eval like so
eval($run_var);
see for referencehttp://php.net/manual/en/function.eval.php