php array empty when returned without referencing an index

Question

I have the following PHP which gathers data from the DB and places each row into an array which it then places in a final array which is returned to the JavaScript:

$active = /*some SQL statement*/
$result = mysqli_query($GLOBALS['con'], $active);

$data = array();

while ($row = mysqli_fetch_array($result, MYSQL_ASSOC)){

    $value = array( 
        "id"=>$row["quest"],
        "received"=>$row["received"]
    );

    array_push($data, $value);
}

return json_encode($data);

I know that the SQL statement is working and that $value has values in it, but the array returned to the JavaScript page is always an empty string instead of a JSON object.

If I place my return statement inside my while loop, then the data is returned:

while ($row = mysqli_fetch_array($result, MYSQL_ASSOC)){

    $value = array( 
        "id"=>$row["quest"],
        "received"=>$row["received"]
    );

    array_push($data, $value);
    return json_encode($data);

}

But obviously this does not return the data as I want it, after the loop has completed.

And if I change my return statement to:

return json_encode($data[1]);

Then the data is also returned.

Please help me figure this one out.

return json_encode($data); this line should be outside the while loop — Surace, Aug 04 '16 at 11:35
Also use `MYSQLI_ASSOC` and not `MYSQL_ASSOC` I am pretty sure this should be producing some sort of error message are you looking at php error log. Add [error reporting](http://stackoverflow.com/questions/845021/how-to-get-useful-error-messages-in-php/845025#845025) to the top of your file(s) _while testing_ right after your opening PHP tag for example ` — RiggsFolly, Aug 04 '16 at 11:37

Haresh Vidja · Answer 1 · 2016-08-04T11:39:40.583

0

Try with below changed code

$active = /*some SQL statement*/
$result = mysqli_query($GLOBALS['con'], $active);

$data = array();

while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){
    $value = array( 
        "id"=>$row["quest"],
        "received"=>$row["received"]
    );
   $data[]=$value; // this is trick 
}

return json_encode($data);

edited Aug 04 '16 at 11:39

answered Aug 04 '16 at 11:36

Haresh Vidja

8,340
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MYSQLI_ASSOC and not MYSQL_ASSOC – RiggsFolly Aug 04 '16 at 11:38
This just returns an empty array like so: `[ ]` – ALR Aug 04 '16 at 11:39
@ALR Can you try with remove MYSQLI_ASSOC as a paramer? – Haresh Vidja Aug 04 '16 at 11:44
It returns a blank string still – ALR Aug 04 '16 at 11:47

score 0 · Answer 2 · edited Aug 04 '16 at 11:37

0

try this:

while ($row = mysqli_fetch_array($result, MYSQL_ASSOC)){

$value = array( 
    "id"=>$row["quest"],
    "received"=>$row["received"]
);

array_push($data, $value);

}

return json_encode($data);

edited Aug 04 '16 at 11:37

Tharif

13,794
9
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answered Aug 04 '16 at 11:36

Faridul Khan

1,741
1
16
27

MYSQLI_ASSOC and not MYSQL_ASSOC – RiggsFolly Aug 04 '16 at 11:37
No, still returns an empty string – ALR Aug 04 '16 at 11:41
print_r($value); array_push($data, $value); check $value is empty or not – Faridul Khan Aug 04 '16 at 11:46
`print_r($data)` displays the data perfectly. So why does it not work when I try to send the array? – ALR Aug 04 '16 at 11:50

php array empty when returned without referencing an index

2 Answers2