Question

我的日期输入日期为：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));

结果：2016-10-20

到目前为止一直很好......

我只需要显示我选择的行日期。

+------+-------------+---------------------+
| 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 |
+------+-------------+---------------------+

那是我的结果代码，但......

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

为什么不工作？

Answer 1

应该是

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

Answer 2

您的字段名称在哪里？

试试这个

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

Answer 3

试试这个：

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

Answer 4

您必须使用as，

指定列名以执行条件

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

Answer 5

您也可以尝试以下查询以获得预期结果

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

这将比较确切的日期。

SELECT * FROM dblist WHERE日期？

5 个答案: