php variables inside json results in invalid json - php

Question

I have the following json created dynamically using php variables and it gives me:
{"message":"invalid json"}
Code:
$json = '{
"sex": "'.$gender.'",
"age": '.$age.',
"evidence": [
{"id": "'.$symptom_id1.'", "choice_id": "present", "initial": true},
{"id": "'.$symptom_id2.'", "choice_id": "present", "initial": true},
{"id": "p_7", "choice_id": "'.$test1.'"},
{"id": "p_8", "choice_id": "'.$test2.'"},
{"id": "p_9", "choice_id": "'.$test3.'"},
{"id": "p_10", "choice_id": "'.$test4.'"},
{"id": "p_28", "choice_id": "'.$test5.'"},
{"id": "'.$location.'", "choice_id": "present"}
],
"extras": {
"disable_groups": true
}
}';
$ch = curl_init('https://api.infermedica.com/v2/diagnosis');
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, $json);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, TRUE);
curl_setopt($ch, CURLOPT_HTTPHEADER, [
'Content-Type: application/json',
'Content-Length: ' . strlen($json),
'app_id: '. APP_ID,
'app_key: '. APP_KEY,
'Interview-Id: '.$interview_id,
'User-Id: '. $hasheduser_id
]);
$result = curl_exec($ch);
echo '<pre>';
print_r( $result );
echo '</pre>';
Hard coding the values in place of variables works fine. Being a newbie, I am unable to understand the reason why it is not well formed.
Help requested.

Answer 1

You should never try and create valid JSON on your own, it will only lead to security issues and bugs, instead use a data structure and pass it to json_encode.
Example data structure for your use:
$to_json = [
'sex' => $gender,
'age' => $age,
'evidence' => [
['id' => $symptom_id1, 'choice_id' => 'present', 'initial' => true],
['id' => $symptom_id2, 'choice_id' => 'present', 'initial' => true],
['id' => 'p_7', 'choice_id' => $test_1],
['id' => 'p_8', 'choice_id' => $test_2],
['id' => 'p_9', 'choice_id' => $test_3],
['id' => 'p_10', 'choice_id' => $test_4],
['id' => 'p_28', 'choice_id' => $test_5],
['id' => $location, 'choice_id' => 'present'],
],
'extras' => [
'disabled_groups' => true
],
];
Then encode it:
$json = json_encode($jo_json);
Hope this helps.

Answer 2

Your JSON is invalid due to a typo on the 2nd line of your variable.
"age": '.$age.',
should be
"age": "'.$age.'",
Or what would be better
"age": "$age",
But the best way to accomplish what you are attempting to do would be to create a PHP array, and pass this to json_encode.
$array = ['hello, i am a key'=>'i am some data'];
var_dump(json_encode($array));