iOS - JSONString to PHP

Question

I send 4 different values in a JSONString. I need to somehow (decode?) convert these to PHP values to send it to a MySQL database.

This function sends is to the php file:

- (void)myFuntionThatWritesToDatabaseInBackgroundWithLatitude:(NSString *)latitude longitude:(NSString *)longitude date:(NSString *)stringFromDate 
{
    _phonenumber = [[NSUserDefaults standardUserDefaults] objectForKey:@"phoneNumber"];

    NSMutableString *postString = [NSMutableString stringWithString:kPostURL];
    NSString*jsonString = [[NSString alloc] initWithFormat:@{\"id\":\"%@\",\"longitude\":\"%@\",\"latitude\":\"%@\",\"timestamp\":\"%@\"}", _phonenumber, longitude , latitude, stringFromDate];

    [postString appendString:[NSString stringWithFormat:@"?data=%@", jsonString]];
    [postString setString:[postString stringByAddingPercentEscapesUsingEncoding:NSUTF8StringEncoding]];
    NSMutableURLRequest *request = [[NSMutableURLRequest alloc] initWithURL:[NSURL URLWithString:postString ]];
    [request setHTTPMethod:@"POST"];

    [[NSURLConnection alloc] initWithRequest:request delegate:self ];
    NSLog(@"Post String =%@", postString);

    //    LocationTestViewController*locationTestViewController = [[LocationTestViewController alloc]init];
    //    phonenumber = locationTestViewController.telefoonnummer;
    NSLog(@"HERE1 : %@", _phonenumber);

}

And it probably goes wrong in this part:

NSString* jsonString = [[NSString alloc] initWithFormat:@{\"id\":\"%@\",\"longitude\":\"%@\",\"latitude\":\"%@\",\"timestamp\":\"%@\"}", _phonenumber, longitude , latitude, stringFromDate];

This is my log:

2012-10-09 15:32:59.869 MyApp[626:c07] Post String=http://www.yourdomain.com/locatie.php?  data=%7B%22id%22:%220612833397%22,%22longitude%22:%22-143.406417%22,%22latitude%22:%2232.785834%22,%22timestamp%22:%2209-10%2015:05%22%7D

It sends it to this PHP file where the id, longitude, latitude and timestamp needs to be made ready for the MySQL insert

 <?php

    $id = $_POST['id'];
    $longitude = $_POST['longitude'];
    $latitude = $_POST['latitude'];
    $timestamp = $_POST['stringFromDate'];

    $link = mysql_connect('server', 'bla', 'bla') or die('Could not connect: ' . mysql_error());

    mysql_select_db('bla') or die('Could not select database');

    // Performing SQL query
    $query="INSERT INTO locatie (id, longitude, latitude, timestamp) 
    VALUES ($id, $longitude,$latitude,$timestamp)";
    $result = mysql_query($query) or die('Query failed: ' . mysql_error());
    echo "OK";

    // Free resultset
    mysql_free_result($result);

    // Closing connection
    mysql_close($link);
?>

On a side note, please use mysql_real_escape_string. Inserting unescaped data into queries is pretty dangerous! — ndm
– ndm, Commented Oct 9, 2012 at 14:57

alan978 · Accepted Answer · 2012-10-09 14:51:15Z

1

Try this code:

$tmpdata = urldecode($_GET['data']);
$data = json_decode($tmpdata);
$id = $data['id'];
$longitude = $data['longitude'];
$latitude = $data['latitude'];
$timestamp = $data['stringFromDate'];

$link = mysql_connect('server', 'bla', 'bla') or die('Could not connect')

mysql_select_db('bla') or die('Could not select database');

// Performing SQL query
$query="INSERT INTO locatie (id, longitude, latitude, timestamp) 
VALUES ($id, $longitude,$latitude,$timestamp)";
$result = mysql_query($query) or die('Query failed: ' . mysql_error());
echo "OK";

// Free resultset
mysql_free_result($result);

// Closing connection
mysql_close($link);

answered Oct 9, 2012 at 14:51

alan978

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3 Comments

David Raijmakers Over a year ago

Didnt work either, I still get the message that there are no values in $id, $longitude, $latitude and $timestamp

ndm Over a year ago

That't least because json_decode creates stdClass instances by default, ie you'd need to access properties like this $id = $data->id;, or use json_decode($tmpdata, true); in order to decode into an associative array.

alan978 Over a year ago

and add "true" to json_decode($tmpdata, true); as ndm suggested :)

ilmiacs · Accepted Answer · 2012-10-09 14:32:32Z

0

Simple typo. Replace

NSString* jsonString = [[NSString alloc] initWithFormat:@{\"id\":\"%@\",\"longitude\":\"%@\",\"latitude\":\"%@\",\"timestamp\":\"%@\"}", _phonenumber, longitude , latitude, stringFromDate];

with

NSString* jsonString = [[NSString alloc] initWithFormat:@"\"id\":\"%@\",\"longitude\":\"%@\",\"latitude\":\"%@\",\"timestamp\":\"%@\"", _phonenumber, longitude , latitude, stringFromDate];

edited Oct 9, 2012 at 14:32

answered Oct 9, 2012 at 14:27

ilmiacs

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3 Comments

David Raijmakers Over a year ago

Doesnt this needs to be removed too? "}"

ndm Over a year ago

Wouldn't that produce an invalid JSON string? They are ment to be enclosed in {}, aren't they?

ilmiacs Over a year ago

If you use json_decode, like alan978 suggested, then yes and you should put the parentheses back. Not like in original code. There you missed a single opening ".

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