Maybe you should use json_decode($json, true); in case of you have a problem with json_decode($json);
On 11 čvn, 14:17, Chris Chen <cdcc...@gmail.com> wrote: > $json = $_POST['json']; > $user = json_decode($json); > echo $user['name']; > > 2009/6/11 Val Cartei <val.car...@gmail.com> > > > > > > > > > with data.name (data is the json object you pass to your success > > function). Like: > > > success: function(data) { > > alert(data.name); > > } > > > Val > > > On Thu, Jun 11, 2009 at 12:40 PM, David .Wu<chan1...@gmail.com> wrote: > > > > If I send a json format to php, how to get the value from php? for > > > example > > > > front page > > > <script> > > > var jsonStr = '{"name": "David", "age", "23"}'; > > > $.ajax({ > > > url: 'json.php', > > > type: 'POST', > > > cache: false, > > > data: {json: jsonStr}, > > > success: function(data) { > > > alert(data); > > > } > > > }); > > > </script> > > > > php page > > > <?php > > > $json = $_POST['json']; > > > > // it's get {"name":"David","age": "23"}, but how to gete name? > > > ?> > > > -- > > Valentina Cartei > > Telephone Numbers: > > University +44 (0) 1273 877560 > > Work +44 (0) 1273 206306 > > Mobile +44 (0)796 6882820 > > -- > Chris