Create json from array in PHP

Question

I have this little sql code:

try {
        $stmt = $conn->prepare("SELECT APPID FROM COMMENTROOM WHERE BADGEID=? GROUP BY APPID");
        $stmt->execute(array($badgeID));
            while ($row = $stmt->fetch(PDO::FETCH_ASSOC)) {
                $row2[$i][0] = $row['APPID'];
                $i++;
            }
        } catch(PDOException $e) {
            echo 'ERROR: ' . $e->getMessage();
            $server_dir = $_SERVER['HTTP_HOST'] . rtrim(dirname($_SERVER['PHP_SELF']), '/\\');
            header('Location: http://' . $server_dir);
            exit();
    }

When I now print out the result with print(json_encode($row2)); I get the following:

[["0000000021"],["0000000037"],["0000000038"],["0000000039"],["0000000128"],["0000000130"]]

Since I already have this data I don't need to query the database. But I still need to forward it in the same format to my Java app. When I use

$output[] = json_encode($row2);
print(json_encode($output));

I get a different output:

["[[\"0000000021\"],[\"0000000037\"],[\"0000000038\"],[\"0000000039\"],[\"0000000128\"],[\"0000000130\"]]"]

I checked a few stackoverflow questions but I didn't find any that addresses the same problem.

Steve · Accepted Answer · 2014-03-18 13:13:32Z

2

You are json encoding it twice

$output[] = json_encode($row2);
print(json_encode($output));

Just do it once:

$output[] = $row2;
print(json_encode($output));

answered Mar 18, 2014 at 13:13

Steve

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6 Comments

erdomester Over a year ago

This adds another bracket to the result. So it begins with [[["0000000021" instead of [["0000000021"

Steve Over a year ago

I dont understand what you are trying to do here at all - what was the purpose of creating the output array in the 1st place?

erdomester Over a year ago

Actually I messed up the question. I have an array with 6 fields. These fields are populated from 3 SELECT FROM queries. Is there a way to transform this array into an output that normally result from an sql query? The output I need is like [{"APPNAME":"Enhanced Email","LINK":"http... I did this by creating a populating a temporary table then querying it but I think it's not the best way to do.

Steve Over a year ago

@erdomester Please edit your question with your exact expected output so i can help you

erdomester Over a year ago

I accepted this solution since you answered it. However the real answer is print(json_encode($row2;));.

|

Zsolt Tolvaly · Accepted Answer · 2014-03-18 14:25:12Z

0

EDIT Now I understand what you want...

$retval = array();
//as this while you can add as many items to the array as you want
while( $row = $stmt->fetch(PDO::FETCH_OBJ) ){
    $retval[] = $row;
}
echo json_encode( $retval );

Regards, hotzu

edited Mar 18, 2014 at 14:25

answered Mar 18, 2014 at 13:16

Zsolt Tolvaly

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2 Comments

erdomester Over a year ago

This changes the result. It now begins with {"data":[["0000000021", instead of [["0000000021",

Zsolt Tolvaly Over a year ago

where do you want to use it? In a Javascript or a Java application?

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