ios - Accessing JSON elements in PHP -


php noob here. building iphone app sending json blobs web server. on server side receiving json , trying access 3 fields object contains.

two weird things happening me not figure out how fix:

  1. i not getting when printing out decoded object($post_data below).
  2. i able print full object in xcode when printing object without decoding ($content below) have no idea how access various fields in object.

my php code:

$con = mysql_connect("127.0.0.1","root", "") or die('could not connect: ' . mysql_error()); $content = file_get_contents('php://input'); $post_data = json_decode($content , true); echo $content; --> prints object  echo $post_data; --> not print echo $content->lat; --> not print 

my json object:

{   "lat" : 37.33233141,   "long" : -122.0312186,   "speed" : 0 } 

any appreciated.

it's because json_decode($content, true) returning array, has problems displaying if echo container.

try doing echo $post_data['lat'];

you can try using print_r($post_data); have output actual contents of variable can see if isn't working properly


Comments

Popular posts from this blog

c# - Validate object ID from GET to POST -

node.js - Custom Model Validator SailsJS -

php - Find a regex to take part of Email -