php mysql select statement; Not getting the output

Question

I want to out put from a database the total sms sent from an application.

In this manner: date 2016-06-09 | total sent 4 | delivered 3 | failed 2|

| date       | total_sent | delivered | failed |
+------------+------------+-----------+--------+
| 2016-06-09 |          4 |         3 |      1 |
+------------+------------+-----------+--------+

However my code in php does not work. Some one please lead me...

$sql = mysql_query("SELECT DATE(time_sent) AS date,COUNT(*) as total_sent,SUM(status = 'Success') as delivered,SUM(status = 'failed') AS failed FROM smsdb GROUP BY DATE(time_sent) ");

$result1 = mysqli_query($con, $sql);
$resultarr = mysqli_fetch_assoc($result1);
$time_sent = $resultarr['time_sent'];


echo $time_sent;   

Answer 1

Make $sql variable a string: $sql = "SELECT DATE(time_sent) AS date,COUNT(*) as total_sent,SUM(status = 'Success') as delivered,SUM(status = 'failed') AS failed FROM smsdb GROUP BY DATE(time_sent) ";

The second parameter for mysqli_query must be a string : http://php.net/manual/ro/mysqli.query.php

EDIT

You want the value of time_sent, but in your $resultarr you don't have that key as you have renamed it to date in your mysql query: DATE(time_sent) AS date. The keys of your array are: 'date', 'total_sent', 'delivered' and 'failed'. Try this: $time_sent = $resultarr['date'];