Reputation: 36205
I am currently working on a php/javascript project which retrieves information from a database and json encodes the data. It is supposed to show the values from the database inside a combo box on the web page.
In the PHP script that encodes the mysql data I have the following code:
$query = "SELECT pla_platformName as `platform` FROM platforms";
$result = mysql_query($query);
if ($result)
{
while ($myrow = mysql_fetch_array($result))
{
$output[] = $myrow;
}
print json_encode($output);
die();
}
In the javascript code I have the following:
<script>
$(document).ready(function()
{
getPlatforms();
function getPlatforms()
{
$.post("phpHandler/get-platforms.php", function(json)
{
alert(json);
alert(json.platform);
}
);
}
});
</script>
I have alert(json); which shows the entire json data which looks like the following:
[{"0":"hello","platform":"hello"},{"0":"android","platform":"world"}]
The next alert(json.platform) I am expecting it to show either hello or world or both but on this line it keeps on saying undefined. Why isn't this working and how do I get a specific platform, i.e. either hello, or world.
Thanks for any help you can provide.
Upvotes: 0
Views: 2006
Reputation: 4448
You need to first convert your JSON
string into an object
var data = $.parseJSON(json);
In this case, the object returned is an array. Try
alert(data[0].platform);
You can skip the first step if you set the dataType
option to json
in your ajax call.
$.post("phpHandler/get-platforms.php", function(data) {
alert(data[0].platform);
},
'json'
);
See jQuery.post() documentation
Upvotes: 4
Reputation: 7892
Your platform member is defined on each item, you'll have to specify which array item you want the platform for. This should do the trick:
alert(json[0].platform);
I'm assuming that your json parameter actually holds a javascript object, and not simply a string. If it is a string, you should define contentType 'application/json' on your php page. (I'm not sure how you do that in php since I do .NET for server side myself.)
To test it quickly however, you can do a $.parseJSON(json) to get the object.
Upvotes: 4