Send JSON to PHP and get a response
Question
I'm just learning how to use jQuery and PHP together. This is my first attempt, and I feel like I'm almost getting the concept. However, there's an issue I failed to address. When I post a JSON object to PHP script and try to return one of the parameters, I get the following error : "Trying to get property of non-object in ..."
index.html:
<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-git2.js"></script>
<meta charset=utf-8 />
<title>JS Bin</title>
<style id="jsbin-css"></style>
</head>
<body>
<button onClick="postData();">Submit me!</button>
<script>
function postData() {
var myData = {
'firstName' : 'John',
'lastName' : 'Doe'
};
$.ajax( {
type: "POST",
url: "postData.php",
contentType: "application/json",
data: myData,
success: function(msg){
alert(msg);
},
error: function(err) {
alert('error!' + err);
}
});
}
</script>
</body>
</html>
postData.php:
<?php
$input = file_get_contents('php://input');
$jsonData = json_decode($input);
$output = $jsonData->{'firstName'};
echo $output;
?>
4 answers
solution1
1 2013-08-13 18:09:51
With a bit more work, you can achieve this using a REST client that will automatically handle data-type conversion and URL parsing among other things.
To name some of the advantages of using a REST architecture:
Simple.
You can easily scale your solution using caching, loading-balancing etc.
Allows you to logically separate your URL-endpoints.
It gives you the flexibility to change implementation easily without changing clients.
Try reading, A Brief Introduction to REST to get a better idea about the design pattern and it's uses. Of course you won't need to write a framework from scratch if you don't want to, since there are already several open-source PHP based implementations out there such as Recess PHP Rest Framework .
Hope this helps!
solution2
0 2013-08-13 17:34:38
json_decode (depending on PHP version) defaults to returning an array, not an object. The proper way to access it would be:
$output = $jsonData['firstname'];
You'll also want it to return an associative array, so pass true as the second argument of json_decode.
$jsonData = json_decode($input, true);
What could alternately be happening is that the JSON is invalid, in which case PHP returns null . You can check for that:
if ($jsonData = json_decode($input, true) === null) {
// Do stuff!
} else {
// Invalid JSON :(
}
solution3
0 2013-08-13 17:50:57
I put it a little bit more simple with the function post of jquery. I hope you found it useful:
first your html and js:
<!DOCTYPE html>
<html>
<head>
<script src="http://code.jquery.com/jquery-git2.js"></script>
<meta charset=utf-8 />
<title>JS Bin</title>
<style id="jsbin-css">
</style>
</head>
<body>
<button onClick="postData();">Submit me!</button>
<script>
function postData() {
$.post(
"postData.php",
{
firstName : 'John',
lastName : 'Doe'
},
function(msg)
{
alert(msg);
}
);
}
</script>
</body>
</html>
then your php:
<?php
echo $_REQUEST['firstName']." - ".$_REQUEST['lastName'];
?>
solution4
0 2013-08-13 21:34:52
I finally figured it out:
js:
function postData() {
var myData = {
firstName: 'John',
lastName: 'Doe'
};
$.ajax({
type: "POST",
url: "postData.php",
data: JSON.stringify(myData),
success: function(msg){
alert(msg);
},
error: function(err){
alert('error!' + JSON.stringify(err));
}
});
}
php:
<?php
$input = file_get_contents('php://input');
$jsonData = json_decode($input);
$output = $jsonData->{'firstName'};
echo $output;
?>
When decoding you DON'T want to put "true" as the second argument, because then, it's not JSON but an associative array (or so I've read). If I put json_decode($input, true), then it won't work.
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.