Passing JSON from PHP to a JavaScript variable
Question
I'm making an aplication with phonegap and I'm stuck trying to send JSON data from the PHP on the server to JavaScript on the device. I want to do something like:
var JSON = '{ "Table" : ' + "http://www.hel.net/LoadDB.php=?table=exhibitions" + '}';
the php works fine and returns somethig like:
"[{"id":"1","name":"Art for everyone","image":null,"name2":"June 29, 2013: 11:00am.","description":"With reservation\r\nFree entrance","description2":null}]"
I want that result in a javascript variable to work later with:
var obj = eval ("(" + JSON + ")");
document.getElementById("rName").innerHTML=obj.Table[1].name;
document.getElementById("lname").innerHTML=obj.Table[1].name2;
What I want to do is something like:
var JSON = '{ "Table" : ' + "http://www.hel.net/LoadDB.php=?table=exhibitions" + '}';
var obj = eval ("(" + JSON + ")");
document.getElementById("rName").innerHTML=obj.Table[1].name;
document.getElementById("lname").innerHTML=obj.Table[1].name2;
How can I make the first line work? is it possible to make the first line work? PS. I do not have much experience with JSON arrays.
Ok I tried ajax and works, I used:
console.log("before");
var jqxhr = $.ajax( "http://www.hel.com/LoadDB.php?table=exhibitions" )
.done(function(data) { console.log(data); })
.fail(function() { console.log("error"); })
.always(function() { console.log("complete"); });
console.log("after");
more info in:
3 answers
solution1
1 2013-06-27 01:47:14
I think all you need is var obj = <?php echo $myjsonencodedvar; ?> var obj = <?php echo $myjsonencodedvar; ?>
or
var obj = <?php echo json_encode($myarray_or_object); ?>
Since I said "I think..." I decided to test it out. I found the following dump() function here on SO.
$arr=array(1,'biscuit'=>'gravy',3,4,5);
$json=json_encode($arr);
?>
<script>
j=<?php echo $json; ?>;
document.write(dump(j));
function dump(obj) {
var out = '';
for (var i in obj) {
out += i + ": " + obj[i] + "\n";
}
return out;
}
</script>
output:
0: 1 biscuit: gravy 1: 3 2: 4 3: 5
solution2
0 2013-06-27 01:49:59
使用JSONP（无回调），并在客户端使用$ .getJSON（），它将从json字符串解析为js对象。
solution3
0 ACCPTED 2013-06-27 02:34:42
Try this:
PHP: (json.php)
<?php
header("Content-Type: text/json");
//the data format your question mentioned
$data = array("Table"=>array(array("id"=>"1","name"=>"Art for everyone","image"=>null,"name2"=>"June 29, 2013","description"=>"With reservation\r\nFree entrance","description2"=>null)));
echo json_encode($data);
?>
Front-end:
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<script>
$.get("json.php",function(json){
alert(json.Table[0].name);
});
</script>
</body>
</html>
Hope this is helpful for you.
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