我想查看两个小时之间的差异，但是当用户没有注销时，我以字符串形式给它一个手动小时，然后我尝试将其转换为时间并检查差异。然而，结果总是$d_time_in，即使这两个字符串不是空的，它们具有相同的格式。
$start_timestamp = mktime(0, 0, 0, $_POST[selectMonth], $_POST[selectDay], $_POST[selectYear]);
$end_timestamp = $start_timestamp + 86399;
$q1 = "SELECT c.user_id, c.date, c.time_in, c.time_out, u.user_firstname, u.user_lastname
FROM control_time c
LEFT JOIN user u
ON u.user_id = c.user_id
WHERE c.date >= $start_timestamp AND c.date <= $end_timestamp";
$r1 = mysql_query($q1) or die(mysql_error());
$num_rows = mysql_num_rows($r1);
if($num_rows > 0){
$display .= "<table border='"0'" cellspacing='"1'" cellpadding='"4'">
<tr>
<td bgcolor='"#CCCCCC'"><strong>Usuario</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Fecha</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Ingreso</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Salida</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Trabajadas</strong></td>
</tr>";
while($time_info = mysql_fetch_assoc($r1)){
if($time_info[time_in] > ($time_info[date]+34200)){
$in_bg = "#F3F781";
}else{
$in_bg = "#DDDDDD";
}
//specify time in
$d_time_in = date("G:i a", $time_info[time_in]);
if($time_info[time_out] == 0){
if(date('w', $time_info[date]) == 6){
$d_time_out = '15:00 pm';
}else{
$d_time_out = '19:00 pm';
}
$out_time = strtotime($d_time_out);
$out_bg = "#F5DA81";
}else{
$d_time_out = date("G:i a", $time_info[time_out]);
$out_time = strtotime($d_time_out);
$out_bg = "#DDDDDD";
}
$difference_hours = strtotime($d_time_in) - $out_time;
if($difference_hours != 0){
$d_time_worked = date("G:i",$difference_hours);
}else{
$d_time_worked= "00:00";
}
$display .= "
<tr>
<td bgcolor='"#DDDDDD'">$time_info[user_firstname] $time_info[user_lastname]</td>
<td bgcolor='"#DDDDDD'">".date("d-m-y", $time_info[date])."</td>
<td bgcolor='"$in_bg'">$d_time_in</td>
<td bgcolor='"$out_bg'">$d_time_out</td>
<td bgcolor='"#DDDDDD'">$d_time_worked</td>
</tr>";
}
$display .= "</table>";
}
很难说问题出在哪里。
但是试着测试最后一个块并注释整个上面的代码。
$difference_hours = strtotime($d_time_in) - strtotime($d_time_out);
将$d_time_in和$d_time_out设置为两个常量时间戳
如果这段代码运行良好(您得到了时差)，请转到上面的块并测试它。
$d_time_in = date("G:i a", $time_info[time_in]);
相同的逻辑。使用两个预设的确定时间戳进行测试。
好了，看了100遍之后，我找到了解决办法。
$start_timestamp = mktime(0, 0, 0, $_POST[selectMonth], $_POST[selectDay], $_POST[selectYear]);
$end_timestamp = mktime(0, 0, 0, $_POST[selectMonthE], $_POST[selectDayE], $_POST[selectYearE]);
$end_timestamp = $end_timestamp + 86399;
$q1 = "SELECT c.user_id, c.date, c.time_in, c.time_out, u.user_firstname, u.user_lastname
FROM control_time c
LEFT JOIN user u
ON u.user_id = c.user_id
WHERE c.date >= $start_timestamp AND c.date <= $end_timestamp";
$r1 = mysql_query($q1) or die(mysql_error());
$num_rows = mysql_num_rows($r1);
if($num_rows > 0){
$display .= "<table border='"0'" cellspacing='"1'" cellpadding='"4'">
<tr>
<td bgcolor='"#CCCCCC'"><strong>Usuario</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Fecha</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Ingreso</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Salida</strong></td>
<td bgcolor='"#CCCCCC'"><strong>Trabajadas</strong></td>
</tr>";
while($time_info = mysql_fetch_assoc($r1)){
if($time_info[time_in] > ($time_info[date]+34200)){
$in_bg = "#F3F781";
}else{
$in_bg = "#DDDDDD";
}
//specify time in
$d_time_in = date("G:i:s", $time_info[time_in]);
$in_time = strtotime($d_time_in);
//specify time out
if($time_info[time_out] == 0){
//if time out is equal to 0 lets set it manually
if(date('w', $time_info[date]) == 6){
//if its saturday
$d_time_out = '15:00:00';
}else{
//if its any other day
$d_time_out = '19:00:00';
}
//convert out hour to time
$out_time = strtotime($d_time_out);
$out_bg = "#F5DA81";
}else{
//if time out != 0
$d_time_out = date("G:i:s", $time_info[time_out]);
//convert out hour to time
$out_time = strtotime($d_time_out);
$out_bg = "#DDDDDD";
}
//see differente between the two
$difference_hours = $out_time - $in_time;
if($difference_hours > 0){
//show how many hours worked
$d_time_worked = gmdate("H:i:s", $difference_hours);;
}else{
$d_time_worked= "00:00";
}
$display .= "
<tr>
<td bgcolor='"#DDDDDD'">$time_info[user_firstname] $time_info[user_lastname]</td>
<td bgcolor='"#DDDDDD'">".date("d-m-y", $time_info[date])."</td>
<td bgcolor='"$in_bg'">$d_time_in</td>
<td bgcolor='"$out_bg'">$d_time_out</td>
<td bgcolor='"#DDDDDD'">$d_time_worked</td>
</tr>";
}
$display .= "</table>";
}
谢谢大家的帮助!