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POiOS - JSONString to PHP
 

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  1. POiOS - JSONString to PHP
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    <p>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.</p> <p>This function sends is to the php file:</p> <pre><code>- (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); } </code></pre> <p>And it probably goes wrong in this part:</p> <pre><code>NSString* jsonString = [[NSString alloc] initWithFormat:@{\"id\":\"%@\",\"longitude\":\"%@\",\"latitude\":\"%@\",\"timestamp\":\"%@\"}", _phonenumber, longitude , latitude, stringFromDate]; </code></pre> <p>This is my log:</p> <pre><code>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 </code></pre> <p>It sends it to this PHP file where the id, longitude, latitude and timestamp needs to be made ready for the MySQL insert</p> <pre><code> &lt;?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); ?&gt; </code></pre>
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    1. USRavi
    1. USDavid Raijmakers
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    1. COWhat's the error?
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      1. POiOS - JSONString to PHP
      1. USilmiacs
    2. COOn a side note, please use [mysql_real_escape_string](http://php.net/mysql_real_escape_string). **Inserting unescaped data into queries is pretty dangerous!**
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      1. USndm
 

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