Php function that takes SQL query as a parameter and returns JSON string

Question

This is my function so far, how do I convert the item to json and print it?

$query = 'SELECT * FROM department';
function dataGenerator($query) {
$conn = oci_new_connect('u','p','h') or die ("dbdn");
$stid = oci_parse($conn, $query);
$r = oci_execute($stid);

while ($row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS)) {
    foreach ($row as $item) {
        print ($item); }
}
}
dataGenerator("SELECT * FROM department");

What have you tried? Nothing in your function suggests you're attempting to create a JSON string. For example, have you looked up json_encode? — Robert
– Robert, Commented Jan 24, 2018 at 15:05
Just from a coding point of view - make the database connection outside the function (better to make 1 connection per page) and pass in as a parameter. — Nigel Ren
– Nigel Ren, Commented Jan 24, 2018 at 15:20

Arun Kumaresh · Accepted Answer · 2018-01-24 15:15:30Z

1

use json_encode and your function should have a return

function dataGenerator($query) {
$conn = oci_new_connect('u','p','h') or die ("dbdn");
$stid = oci_parse($conn, $query);
$r = oci_execute($stid);

while ($row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS)) {

          $result[] = $row;
}
return json_encode($result);
}

answered Jan 24, 2018 at 15:15

Arun Kumaresh

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2 Comments

multimediadev Over a year ago

How do i log this json data onto console as a object

Arun Kumaresh Over a year ago

console.log('<?php echo $your_json ?>')

Ali Faris · Accepted Answer · 2018-01-24 15:08:10Z

1

make array of items and then use json_encode to generate the json as string

$items = [];
while ($row = oci_fetch_array($stid, OCI_ASSOC+OCI_RETURN_NULLS)) {
    foreach ($row as $item) {
        $items[] = $item;
    }
}

$json = json_encode(array("items" => $items))

edited Jan 24, 2018 at 15:08

answered Jan 24, 2018 at 15:06

Ali Faris

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