在我的情况下,我在 mySQL 中有一个日期字段,我曾经通过下面的代码获取变量的值。当我查询诸如“08-28-1989”之类的完整日期时,这是正确的。
mysql:
SELECT COUNT(*)
FROM content
where dateReceived='{$approved_date}'
and description='{$description}'
PHP:
$approved_year = $_POST["approved_year"];
$approved_month = $_POST["approved_month"];
$approved_day = $_POST["approved_day"];
$approved_dt = strtotime($approved_year."-".$approved_month."-".$approved_day);
$approved_date = date("Y-m-d",$approved_dt);
如果我说用户只输入“08”(仅限月份),我怎样才能使它更加动态,然后它会显示该月的所有记录,而不管日期和年份。或者用户只需输入月份和年份,然后将其查询到我的数据库。
这也是我的 HTML:
<p>
Date Approved
<select name="approved_month">
<?php
$month = array(" ","January","February","March","April",
"May","June","July","August",
"Septembe","October","November","December");
for($x=0;$x<12;$x++){
echo "<option value=\"".$x."\">".$month[$x]."</option>";
}
?>
</select>
<input type="text" name="approved_day" value="" size="2" />
<input type="text" name="approved_year" value="" size="10" />
</p>