SELECT * FROM dblist的日期？

Question

我有輸入日期的日期：20-10-2016

echo "<form method='post' action='rapor.php'>";
echo "<input type='date' name='pickdate' value=".date("Y-m-d")."> <input type='submit' value='Git'>";
echo "</form>";

$pickeddate = strtr($_POST['pickdate'], '/', '-');
echo date('Y-m-d', strtotime($pickeddate)); 

結果如下：2016-10-20

到現在為止還挺好..

我只需要顯示我選擇的行日期。

+------+-------------+---------------------+
| name |  mail       | dateandtime         |
+------+-------------+---------------------+
| AAA  |   a@a.com   | 2016-04-20 06:44:19 |
| BDC  |   b@c.com   | 2016-10-21 06:44:19 |
| CDD  |   c@d.com   | 2016-04-10 06:44:19 |
| EED  |   e@d.com   | 2016-10-20 06:44:19 |
| SAS  |   a@s.com   | 2016-04-10 06:44:19 |
+------+-------------+---------------------+

多數民眾贊成在我的結果代碼，但是....

$result = mysql_query('SELECT * FROM dblist WHERE ('$pickeddate%')  ');

為什么不工作？

Answer 1

它應該是

$newdate = date('Y-m-d', strtotime($pickeddate)); 
$result = mysql_query('SELECT * FROM dblist WHERE dateandtime LIKE  "'.$newdate.'%"  ');

Answer 2

您的字段名稱在哪里？

嘗試這個

$result = mysql_query('SELECT * FROM dblist WHERE dateandtime LIKE ('$pickeddate%')  ');

Answer 3

嘗試這個：

$pickeddate = date('Y-m-d', strtotime($_POST['pickdate']));
SELECT * FROM dblist WHERE date(dateandtime) LIKE '$pickeddate%';

Answer 4

您必須指定列名稱才能使用as做條件，

$result = mysql_query('SELECT * FROM dblist WHERE dateandtime LIKE  "'.$pickeddate.'%"  ');

Answer 5

您也可以在下面的查詢中查詢您的預期結果

$newdate = date('Y-m-d', strtotime($pickeddate)); 
$result = mysql_query('SELECT * FROM dblist WHERE DATE(dateandtime) = "'.$newdate.'"');

這將是比較確切的日期。