How do I get the result from a MySQL DATEDIFF query as a variable to pass to a PHP function?

Question

I am trying to get the difference in seconds between the date/time in the column 'last_visited' in my table 'users' and the current date/time for a particular user. I think the following query should do it:

$query ="SELECT unix_timestamp(NOW()) - unix_timestamp(last_visit) from users WHERE username='$user'";
$result=mysql_query($query) or die(mysql_error());

However, I am not sure how to get the result from this as a variable so that I can pass it to a PHP function. Could someone help with this?

have you looked at the `$result`variable using `var_dump` ? Also, please don't use the `mysql_*` methods, there deprecated. — Florian, Aug 12 '12 at 21:32

score 1 · Answer 1 · answered Aug 12 '12 at 21:33

1

$query ="SELECT unix_timestamp(NOW()) - unix_timestamp(last_visit) AS time_difference from users WHERE username='$user'";
                                                                   ^

I guess it will be easier to use this variable if it has a proper name.

answered Aug 12 '12 at 21:33

Tim

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Thanks. How do I then pass 'time_difference' to my 'time_since' function as in `$visited = time_since (*SECONDS NEED TO GO HERE*);` – Nick Aug 12 '12 at 21:37
**c 2**'s comment above tells how to read data from the query result. – Tim Aug 12 '12 at 21:42
@Tim: Always link. What's above or below varies, because this site sometimes prefers your answer first, than the other. And welcome to SO! – hakre Aug 12 '12 at 21:52

score 1 · Accepted Answer · answered Aug 12 '12 at 21:35

Return the result with a name:

$query ="SELECT unix_timestamp(NOW()) - unix_timestamp(last_visit) AS time_diff from users WHERE username='$user'";
$result=mysql_query($query) or die(mysql_error());

while($row=mysql_fetch_assoc($result)) 
{ 
 $diff= $row['time_diff']
}

How do I get the result from a MySQL DATEDIFF query as a variable to pass to a PHP function?

2 Answers2