it's work, cool.
On 6月11日, 下午11時58分, Tomáš Kavalek <tomas.kava...@gmail.com> wrote:
> 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
