Joining variable to strings fails in PHP - php

Question

I'm sorry that this is basic. When I use this PHP code it works fine:
$data = '{"reportID":1092480021}';
However, when I run my URL like this:
http://localhost:8000/new/reportget.php?type=1092480021
and use this PHP code:
$reportref = $_GET['type'];
$data = '{"reportID:".$reportref."}"';
I get the error
Error_description:reportID is required
I think it's an error with how I am joining my variable to the string but I can't understand where I am going wrong.

Answer 1

Your string is improperly quoted. To match the format in your first example use:
$data = '{"reportID":' . $reportref.'}';
Note there are no double quotes on the last curly.
Even better:
$reportref = 1092480021;
$data = [ 'reportId' => $reportref ];
var_dump(json_encode($data));
Output:
string(23) "{"reportId":1092480021}"

Answer 2

For simple view and understanding, can you try out:
$data = "{\"reportID\":$reportref}";
Think that should sort it out

Answer 3

Use it like this
data = '{"reportID:"'.$reportref.'"}"';

Answer 4

It isn't working because you wrap all the value within single quote and when it come to concatenate the $reprtref you put directly .$reportref without closing the first single quote and after putting the value to concatenate you forget to open another single quote
'{"reportID:".$reportref."}"';
the correct value is
'{"reportID:"' . $reportref . '"}"';
and to match the way you specify your $data value It must be like this
'{"reportID":' . $reportref . '}';