i have
$age = implode(',', $wage); // which is object return: [1,4],[7,11],[15,11]
$ww = json_encode($age);
and then i retrieve it here
var age = JSON.parse(<?php echo json_encode($ww); ?>);
so if i make
alert(typeof(<?php echo $age; ?>)) // object
alert(typeof(age)) //string
in my case JSON.parse retuned as string.
how can i let json return as object?
EDIT:
var age = JSON.parse(<?php echo $ww; ?>); // didnt work , its something syntax error
implode returns a string, so it is only natural that json_encode encodes it as such. It does not recognize already JSON-like data passed as a string.
If you want to get an object, you have to pass an associative array to json_encode:
$foo = array(
1 => 4,
7 => 11,
15 => 11
);
echo json_encode($foo); // {1:4,7:11,15:11}
With so little info about what $wage looks like before it's imploded, it's hard to tell exactly what you want to get. How is that structure ([1,4],[7,11],[15,11]) an object? Is the first element of each tuple a key? That's what I assumed with my example, but it might be off.
typeof([1,4],[7,11],[15,11]), and typeof on an array returns object (put simply, typeof returns the type of the last comma-separated value it receives as an argument). In your second case, you get typeof("[1,4],[7,11],[15,11]"), which definitely is a string literal (notice the quotes).typeof("[1,4],[7,11],[15,11]") with double quotesvar age = JSON.parse("\"[1,4],[7,11],[15,11]\""); and then typeof(age). (Look at how you call json_encode twice, once on $wage to get $age and then again on $age to put it in your Javascript.)var age = [<?php echo $age; ?>];
a. You get a syntax error because you need to enclose the string within quotes, like so:
var age = JSON.parse("<?php echo $ww; ?>");
b. Moreover, you don't need JSON.parse. You can simply echo the php var after it was already json_encoded in the server side:
var age = <?php echo $ww; ?>;
JSON.parse is there to convert a JavaScript string to an object. In the case of PHP string, once it is built as JSON, echoing it in the right place is equivalent to coding it yourself.
view source in the browser please and paste it here so I can see how it is parsed
$ww=json_encode($age)then you echojson_encode($ww);so reallyvar age=JSON.parse(<?php echo json_encode(json_encode($age)); ?>Is this what you really want to be doing?