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Usage with PHP generated json
Asked 11 years ago
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has this question
11 years ago James Thompson posted:
I am having issues with php generated json.Any json we create does not load. The only way we found that there was a problem wa using a behavior.
I created an "On Error" behavior and it is being triggered even with this simple PHP generated json
This is the PHP.
<?php header('Content-Type: application/json'); $arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5); echo json_encode($arr); echo ''; ?>
If I take the generated json and put in in a file and use it as a data source the error in not triggered
The URL for the datasource is;
recruitping.com/api/simplejson.php
Can any one help.
Replies
Replied 11 years ago
11 years ago George Petrov replied:
Hi James,
We actually changed the JSON in the latest HTML5 Data Bindings, to use JSONP so it can be executed across domains and also avoid bad caching you IE.
So you just need to adjust your PHP to return also JSONP compliant JSON. It is just an extra prefix.
See:
www.geekality.net/2010/06/27/php-how-to-easily-provide-json-and-jsonp/
Greetings,
George
We actually changed the JSON in the latest HTML5 Data Bindings, to use JSONP so it can be executed across domains and also avoid bad caching you IE.
So you just need to adjust your PHP to return also JSONP compliant JSON. It is just an extra prefix.
See:
www.geekality.net/2010/06/27/php-how-to-easily-provide-json-and-jsonp/
Greetings,
George
Replied 11 years ago
11 years ago Miroslav Zografski replied:
Hello James,
Try returning a JSONP object. How to generate such in your case:
Regards.
Try returning a JSONP object. How to generate such in your case:
<?php header('Content-Type: text/javascript; charset=utf8'); $arr = array('a' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5); $callback = $_GET['callback']; echo $callback.'('. json_encode($arr).')'; ?>
Regards.