Calculate DateDiff in PHP in minutes [closed] - php

Question

Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Questions concerning problems with code you've written must describe the specific problem — and include valid code to reproduce it — in the question itself. See SSCCE.org for guidance.
Closed 9 years ago.
Improve this question
<?php
$sql_apr = " SELECT SUM ( meter * minute ) FROM table";
$rs_apr = #mysql_query($sql_apr);
$total_apr = #mysql_fetch_array($rs_apr);
$try4 = $total_apr['SUM(meter * minute'];
while ($rs_t = #mysql_fetch_array($rs_t)) {
$minute = '';
$sql_t = "SELECT DATEDIFF(MINUTE,'e_date e_time','s_date s_time') AS minute";
$rs_t = #mysql_query($sql_t);
$minute = $rs_t['minute'];
}
?>

Answer 1

You are looking at the wrong result:
$rs_t=#mysql_query($sql_t);
$minute = $total_t['minute'];
should be
$rs_t=#mysql_query($sql_t);
$minute = $rs_t['minute'];
// ^ use the rs_t result, not the result from the first query
You are also result the variable that you are looping on. I highly doubt that this while loop will ever end. You are looping on the result from $rs_t and then you reassign $rs_t inside the loop.

Answer 2

why don't you do this in correct order
$query = "SELECT DATEDIFF(MINUTE,'e_date e_time','s_date s_time') AS minute";
$results = #mysql_query($query);
$row = #mysql_fetch_array($results);
$minute = $row['minute'];
print_r($minute);
$sql_apr = " SELECT SUM ( meter * " . $minute . " ) AS my_sum FROM table";
$rs_apr = #mysql_query($sql_apr);
$total_apr = #mysql_fetch_array($rs_apr);
$try4 = $total_apr['my_sum'];
You try to get result (in while) before you run query.
edit:
$query = "SELECT TIMESTAMPDIFF(MINUTE,'s_date s_time','e_date e_time') AS minute";