可能重复:
How do I format the date and time That is received from database
我想只在“参与日期”列中显示日期,并且只在“在时间内”显示时间 我的代码是这可以任何人帮助我,我是一个新手
$bd = mysql_connect($mysql_hostname, $mysql_user, $mysql_password)
or die("Opps some thing went wrong");
mysql_select_db($mysql_database, $bd) or die("Opps some thing went wrong");
$sql="SELECT * FROM `hr_inout_time` WHERE user_id='$login_id'";
$result = mysql_query($sql) or die(mysql_error());
echo "<table align='center' border='1' cellspacing='5' width='50%' height='50%'>
<th>In Comming Date</th>
<th>In Comming Time</th>
";
while($row=mysql_fetch_array($result)) {
$in_time=$row["user_in_time"];
$time_date =$row["time_date"];
echo "<tr>
<td align='center'>$time_date</td>
<td align='center'>$in_time</td>
</tr>";
}
echo('</table>');
?>
答案 0 :(得分:1)
简短回答 - 替换:
<td align='center'>$time_date</td>
<td align='center'>$in_time</td>
使用
<td align='center'>".explode(" ",$time_date)[0]."</td>
<td align='center'>".explode(" ",$in_time)[1]."</td>
答案 1 :(得分:0)
更改
$in_time=$row["user_in_time"];
$time_date =$row["time_date"];
到
$in_time = date('H:i', strtotime($row["user_in_time"]));
$time_date = date('d F Y', strtotime($row["time_date"]));
将H:i
和d F Y
更改为您首选的日期/时间格式。
答案 2 :(得分:0)
您可以使用SQL完成所有操作,因此您无需在PHP脚本中进行转换。由于您的日期看起来存储为DATETIME值,您可以使用DATE()和TIME()函数返回值的那些部分
$sql = "SELECT hr_inout_time.*,
DATE(hr_inout_time.time_date) as time_date,
TIME(hr_inout_time.user_in_time) as user_in_time
FROM `hr_inout_time`
WHERE user_id='$login_id'";
答案 3 :(得分:0)
$in_time = $row["user_in_time"];
$time_date = $row["time_date"];
为:
$in_time = date('H:i:s', strtotime($row["user_in_time"]));
$time_date = date('Y-m-d', strtotime($row["time_date"]));