Reputation: 1216
this is a json file:
myjson.php
mylink([
{
"link": "link1",
"url": "http://www.example1.com"
},
{
"link": "link2",
"url": "http://www.example2.com"
},
{
"link": "link3",
"url": "http://www.example3.com"
}
])
And I try call it.
$.ajax({
url: "json/physician.php",
dataType: "text",
success: function(data) {
json = $.parseJSON(data)
alert(json.mylink.link[0]);
}
});
but it do not work.(do not return any alert.) what's my wrong?
Upvotes: 0
Views: 83
Reputation: 51
$.ajax({
url: 'https://app.zencoder.com/api/v2/jobs/1234/progress.js?api_key=asdf1234',
dataType: 'jsonp',
success: function(data){
// write your code
}
});
Upvotes: 0
Reputation: 37
Why you are creating a complex json array , also your json is not a valid json array,while you are creating your json array you can check the valid json in jsonlint
In simple form here is JSON array in javascript and how to retrieve value
var mylink=[
{
"link": "link1",
"url": "http://www.example1.com"
},
{
"link": "link2",
"url": "http://www.example2.com"
},
{
"link": "link3",
"url": "http://www.example3.com"
}
];
after that you can get value from any index of array :-
alert(mylink[0].link);
alert(mylink[0].url);
Upvotes: 0
Reputation: 43156
Your JSON is not valid. Valid format for what you're trying to do will be:
{ //json
"myLink": { //.mylink
"link": [ //.link
{ //[0]
"link": "link1",
"url": "http://www.example1.com"
},
{
"link": "link2",
"url": "http://www.example2.com"
},
{
"link": "link3",
"url": "http://www.example3.com"
}
]
}
}
You can validate JSON using tools like jsonlint.com
Not sure why the file is a .php
file, you can use getJSON method to retrieve json data which will automatically parse the response for you.
Upvotes: 1