PHP - MySQL results to JSON

Question

I'm trying to understand how to convert MySQL results to a JSON format so that I can then use this JSON later on with Javascript to build a HTML table. However my code just produces lot's of null values and I don't yet understand why.

$result = mysqli_query($con, "SELECT * FROM Customers");

$test = json_encode($result);

print $test;

Output:

{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}

I have, for example, fields such as "CustomerID" and "Name", and even they don't show up in the JSON result.

What am I doing wrong? Thanks

Possible duplicate - stackoverflow.com/questions/13945071/… — Jay Blanchard
– Jay Blanchard, Commented Apr 21, 2014 at 16:18
not sure if you are aware, when using 'json_decode', using a second parameter of 'true' will force the conversion to arrays rather than objects. i.e. $fooArray = json_decode($json, true);. it may save some hassle. — Ryan Vincent
– Ryan Vincent, Commented Apr 21, 2014 at 18:27

Alexey Palamar · Accepted Answer · 2014-04-21 16:32:48Z

19

$result = mysqli_query($con, "SELECT * FROM Customers");   
while($row = mysqli_fetch_assoc($result))
    $test[] = $row; 
print json_encode($test);

edited Apr 21, 2014 at 16:32

answered Apr 21, 2014 at 16:17

Alexey Palamar

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3 Comments

user9993 Over a year ago

Thanks, that works. It shows the first customer in the database, is there some way I can loop through it to show all the customers?

Alexey Palamar Over a year ago

Done. Try to use now.

CrandellWS Over a year ago

for my setup I was doing it different this may help someone down the road

$link = mysqli_connect("localhost", "root", "password", "db");     $results = $link->query($sql); while ($row = $results->fetch_assoc())      $resultArray[] = $row;  echo "<pre>" . json_encode($resultArray, JSON_PRETTY_PRINT) . "</pre>";

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