merging two array into single inside json_encode

Question

I have to combine two array into single inside Json encode. my code is,

    $email  =   $_GET["email"];
    $password   =   $_GET["password"];

    $query      =   "SELECT * FROM tbl_user_login WHERE email='$email' AND password='$password' AND verification='1'";
    $result     =   mysqli_query($c, $query) or die(mysqli_error($c));
    $length     =   mysqli_num_rows($result);

    if($length == 1)
    {
        $var[]  =   array('status'=>"success");

        while($obj = mysqli_fetch_object($result)) 
        {
            $var[] = $obj;
        }
        echo '{"login":'.json_encode($var).'}';
    }
    else
    {
        $arr    =   array('status'=>"notfound");
        echo '{"login":['.json_encode($arr).']}';
    }

Now the result is,

{"login":[{"status":"success"},{"login_id":"1","name":"Jithin Varghese","password":"some","phone":"","email":"[email protected]","addr":"","city":"","state":"","pincode":"0","type":"STD","verification":"1"}]}

And the require output is,

{"login":[{"status":"success","login_id":"1","name":"Jithin Varghese","password":"some","phone":"","email":"[email protected]","addr":"","city":"","state":"","pincode":"0","type":"STD","verification":"1"}]}

How to combine array. I have tried a lot.

You must not use User Input in your SQL Query without validation, you're vunerable to an SQL Injection Attack. Please have a look at prepared statements — KhorneHoly
– KhorneHoly, Commented Dec 16, 2015 at 11:42

Mihai Matei · Accepted Answer · 2015-12-16 11:47:11Z

4

Change

$var[] = array('status'=>"success");
while($obj = mysqli_fetch_object($result)) 
{
    $var[] = $obj;
}

to

$var['status'] = "success";
// use the assoc fetch here.. to avoid casting to array
while($arr = mysqli_fetch_assoc($result)) 
{
    $var = array_merge($var, $arr);
}

edited Dec 16, 2015 at 11:47

answered Dec 16, 2015 at 11:43

Mihai Matei

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1 Comment

Jithin Varghese Over a year ago

now the result is

{"login":{"status":"success","0":{"login_id":"1","name":"Jithin Varghese","password":"some","phone":"","email":"[email protected]","addr":"","city":"","state":"","pincode":"0","type":"STD","verification":"1"}}}

Steve · Accepted Answer · 2015-12-16 11:47:22Z

You can use array_merge to get the exact output you request:

$email  =   $_GET["email"];
$password   =   $_GET["password"];

$query      =   "SELECT * FROM tbl_user_login WHERE email='$email' AND password='$password' AND verification='1'";
$result     =   mysqli_query($c, $query) or die(mysqli_error($c));
$length     =   mysqli_num_rows($result);

$response = [];
if($length == 1)
{
    $response['login'] = arry(array_merge(array('status'=>"success"), mysqli_fetch_assoc($result)));
}
else
{
    $response['login'] = array(array('status'=>"notfound"));
}

header('Content-Type: application/json');
echo json_encode($response);

Note that it seems unnecessary to have login property be an array when there is only one result, so it would make sense to remove the outer array wrap:

if($length == 1)
{
    $response['login'] = array_merge(array('status'=>"success"), mysqli_fetch_assoc($result));
}
else
{
    $response['login'] = array('status'=>"notfound");
}

memo · Accepted Answer · 2015-12-16 11:47:09Z

0

It's a wild guess, but try this.

$result = array(
    'login' => $var
);

echo json_encode($result);

answered Dec 16, 2015 at 11:47

memo

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