Encode data to json with php

Question

I am doing a web service in php to encode data to json. My php code is:

    function getProgramDay($day)
{

    global  $DDBB_SERVER, $DDBB_USER, $DDBB_PASSWORD, $DDBB;



    $sql = "SELECT program.id, programa.name FROM `mybd` WHERE program.day = '" . $day . "'";


    $con = mysqli_connect($DDBB_SERVER, $DDBB_USER, $DDBB_PASSWORD, $DDBB);

    if (!$con) {
        die('Error de Conexión (' . mysqli_connect_errno() . '): ' . mysqli_connect_error());
    }

    if (!mysqli_set_charset($con, "utf8")) {
        die("Error loading character set utf8:" . mysqli_error($con));
    }

    $result = mysqli_query($con, $sql);

    $res = array();

    // Prepare data
    while ($row = mysqli_fetch_assoc($result)) {
        $res[] = $row;
    }

    // Free resultset
    mysqli_free_result($result);

    // Close connection
    mysqli_close($con);

    // Return data
    return $res;

}

And the result of this function in json is:

[{"id":"1","nombre":"Hello"}]

I would like that the result will be:

{"results":[{"id":"1","nombre":"Hello"}]}

How can I do this? I have tried but not works:

return "{'results':" .$res . "}";

Thank you so much :-)

warning your code might be vulnerable to sql injection attacks! — Daniel A. White
– Daniel A. White, Commented Jun 3, 2014 at 13:06

Veda · Accepted Answer · 2014-06-03 13:06:03Z

6

return json_encode(array("results"=>$res));

As said in the comment, if you want to return an array and create the json in the calling code, do it like this:

return array("results"=>$res);

answered Jun 3, 2014 at 13:06

Veda

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3 Comments

Steve Over a year ago

I expect that json_encode is called in the calling code (as he says he gets json), so he should not do it twice, just return the php array

Igor Over a year ago

Thanks, it puts "{\"results\": and using json validator it says that is not valid json, how can I quit this \? Thank you so much

Steve Over a year ago

@Igor thats because you are using json_encode twice - do as per the amended code above (or see my comment to your question)

Community · Accepted Answer · 2023-11-21 15:10:29Z

0

Your output

 [{"id":"1","nombre":"Hello"}]

is an array (witch you build with) // Prepare data while ($row = mysqli_fetch_assoc($result)) { $res[] = $row; }

What you want to have is an an object

object(stdClass)#6 (1) { ["results"] => array(1) { [0] => object(stdClass)#7 (2) { ["id"] => string(1) "1" ["nombre"] => string(5) "Hello" } } }

so change your Prepare data with mysqli_fetch_object and use json_encode

edited Nov 21, 2023 at 15:10

CommunityBot

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answered Jun 3, 2014 at 13:21

donald123

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