How to include a php variable in json and pass to ajax - php

Question

My Code
var json = xmlhttp.responseText; //ajax response from my php file
obj = JSON.parse(json);
alert(obj.result);
And in my php code
$result = 'Hello';
echo '{
"result":"$result",
"count":3
}';
The problem is: when I alert obj.result, it shows "$result", instead of showing Hello.
How can I solve this?

Answer 1

The basic problem with your example is that $result is wrapped in single-quotes. So the first solution is to unwrap it, eg:
$result = 'Hello';
echo '{
"result":"'.$result.'",
"count":3
}';
But this is still not "good enough", as it is always possible that $result could contain a " character itself, resulting in, for example, {"result":""","count":3}, which is still invalid json. The solution is to escape the $result before it is inserted into the json.
This is actually very straightforward, using the json_encode() function:
$result = 'Hello';
echo '{
"result":'.json_encode($result).',
"count":3
}';
or, even better, we can have PHP do the entirety of the json encoding itself, by passing in a whole array instead of just $result:
$result = 'Hello';
echo json_encode(array(
'result' => $result,
'count' => 3
));

Answer 2

You should use json_encode to encode the data properly:
$data = array(
"result" => $result,
"count" => 3
);
echo json_encode($data);

Answer 3

You're using single quotes in your echo, therefore no string interpolation is happening
use json_encode()
$arr = array(
"result" => $result,
"count" => 3
);
echo json_encode($arr);
As a bonus, json_encode will properly encode your response!

Answer 4

Try:
$result = 'Hello';
echo '{
"result":"'.$result.'",
"count":3
}';

Answer 5

$result = 'Hello';
$json_array=array(
"result"=>$result,
"count"=>3
)
echo json_encode($json_array);
That's all.