如何仅在 PHP 中显示日期？
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How to display Date only in PHP?
提问by srahifah
How to display date only(without time) with format (Y:m:d) for table Return Date? When I choose the date in sql, it'll view as picture1(eg: 2012-04-10), after I save and display back it'll view as picture2(eg: Apr 10 2012 12:00:00). I would like to how to display as(eg: 2012-04-10) not display as (eg: Apr 10 2012 12:00:00) in table Return Date. Thanks
如何为表返回日期仅显示日期（无时间）和格式（Y:m:d）？当我在sql中选择日期时，它会显示为picture1（例如：2012-04-10），保存并显示回来后它会显示为picture2（例如：Apr 10 2012 12:00:00）。我想如何在表返回日期中显示为（例如：2012-04-10）而不是显示为（例如：Apr 10 2012 12:00:00）。谢谢
Picture 1
图片1
Picture 2
图2
under Result.php
在 Result.php 下
$query = "SELECT ....BE.[batch_id],BED.[ReturnDate]
FROM .... as BE
INNER JOIN .... as BED
ON BE.[batch_exception_id] = BED.[batch_exception_id]
WHERE BE.[process_date_time] between '$date1' and '$date2'";
while ($row = mssql_fetch_assoc($result)) {
$rDate = $row['ReturnDate'];
$beID = $row['batch_exception_id'];
$proc_dt = $row['process_date_time'];
$batchid = $row['batch_id'];
echo "<tr>";
echo "<td>" . $beID . "<input type='hidden' name='beID[$i]' value='$beID'/></td>";
echo "<td>" . $batchid . "<input type='hidden' name='batchid[$i]' value='$batchid'/></td>";
echo "<td>" . $proc_dt . "<input type='hidden' name='procDT[$i]' value='$proc_dt'/></td>";
echo "<td>";
$date="date[".$i."]";
echo "<input type='text' name='date[$i]' id='$date' value='$rDate' onclick=\"fPopCalendar('".$date."')\">";
$i++;
echo "</td>";
echo "</tr>";
}
Under Save.php
在 Save.php 下
//Select database
$selected = mssql_select_db($myDB, $link)or die("Couldn't open database $myDB");
for ($i=0; $i<$tot_rec;$i++) {
$ret_date=trim($arrretDate[$i]);
ECHO "$ret_date<br>";
if (strlen($ret_date)>0) {
$query = "UPDATE .....
SET [ReturnDate] = '$ret_date'
WHERE [batch_exception_id]= '$arrbeID[$i]'";
$result = mssql_query($query);}
}
//execute the SQL query
if ($result){
header("Location:....");}
else{
echo "Error Save";
}
回答by gmaliar
You can try this
date('Y:m:d', strtotime($date));
你可以试试这个
date('Y:m:d', strtotime($date));
回答by Mohamed Hassan
You can use: strftime("%Y-%m-%d",$date);. Also check thislink.
您可以使用：strftime("%Y-%m-%d",$date);。另请检查此链接。
Content from documentation:
文档中的内容：
strftime— Format a local time/date according to locale settings
Description:string strftime ( string $format [, int $timestamp = time() ] )
strftime— 根据区域设置格式化本地时间/日期
描述：string strftime ( string $format [, int $timestamp = time() ] )
%Y Four digit representation for the year Example: 2038
%m Two digit representation of the month 01 (for January) through 12 (for December)
%d Two-digit day of the month (with leading zeros) 01 to 31
回答by Vaishu
Try this..
尝试这个..
SELECT DATE( ReturnDate ) as dat FROM Batch
Return Date : sep 12 2012 12:24:12
返回日期：2012 年 9 月 12 日 12:24:12
dat : sep 12 2012
日期：2012 年 9 月 12 日
