php get json item return wrong

Question

I have the following code with my php(updated):

$sql = "INSERT INTO uac_user (user_name, user_password, create_time, lastupdateTime) VALUES ('$usernmae', '$password', NOW(), NOW())";

if ($conn->query($sql) === TRUE) {
    $last_id = $conn->insert_id; //get new id
    $records = array();  //select role
    if($result = $conn->query("SELECT role_id FROM uac_role WHERE `role_name` = '$role';")){
        if($result->num_rows){
            while ($row = $result->fetch_object()){
                $records = $row;
            }
            $dateResult=(string)json_encode($records);
            echo ($dateResult);
            echo ($dateResult['role_id']);
            // $sql2 = "INSERT INTO uac_mapping (role_id, user_id) VALUES ('$dateResult['role_id']', '$last_id')"; //insert mapping
            // if ($conn->query($sql2) === TRUE) {
                // echo "success"; 
            // }
            // else {
                // echo "Error: " . $sql . "<br>" . $conn->error;
            // }
        }
        else $records = 'no data';
    }
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}
die();

the first echo return [{"role_id":"4"}]

but the second return [

what I need for the second one is 4

what is the problem about my code?

Add the $records array exactly as it is. It's easier to pinpoint the problem that way. — Andrei
– Andrei, Commented Sep 27, 2017 at 8:27

IMSoP · Accepted Answer · 2017-09-27 08:28:19Z

1

I think you have made a common mistake of confusing what JSON is and is not. JSON is a way of representing some data in a string, for transfer to another system, where it will be turned back into something else for use. JSON is not a type of object that you can use directly to extract data.

Look carefully at this line:

$dateResult=(string)json_encode($records);

The result of this is that $dateResult is a string - you even have an extra (string), but even without that, the manual page for json_encode makes clear that you will always have a string:

Returns a string containing the JSON representation of the supplied value.

You then run this:

$dateResult[role_id]

You are trying to look up role_id inside a string. But a string is just a bunch of symbols, it doesn't know what "role_id" is. You had structured data in $records, it's there that you want to look up role_id.

If $records is an array, what you want is $records['role_id'] (note the quotes around 'role_id'; role_id would be the name of a constant, not a string).

If $records is an object, what you want is $records->role_id (no quotes this time, because ->role_id is more like a variable name than a string).

answered Sep 27, 2017 at 8:28

IMSoP

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3 Comments

Wai Hung Over a year ago

I just updated the code and the $records is an array I have tried both of 'role_id'; role_id but it doesn't work

IMSoP Over a year ago

@WaiHung Read my answer again. You do not need json_encode here, you do not need $dateResult, you do not need anything to do with JSON. Also, the function name "fetch_object" should give you a clue that what it will return is in fact an object.

Wai Hung Over a year ago

I just use echo ($records->role_id); and it works. Thank you so much

Node_Ninja · Accepted Answer · 2017-09-27 08:10:52Z

0

Your json is probably an array of objects, so you can loop through them and use something like:

echo $dateResult[$counter]->role_id;
//$counter will be your loop incrementer;

answered Sep 27, 2017 at 8:10

Node_Ninja

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1 Comment

Wai Hung Over a year ago

but it does not return 4

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