在我的情况,我有MySQL中的日期字段,我通过使用下面的代码来获取值变量。这是正确的,当我查询完整的日期,如“08-28-1989”。查询月,日和年中的日期
MySQL的:
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);
我怎样才能让更多的动态,如果我说,用户只需输入 “08”(月只),那么它会显示所有记录在那个月不分日日。或者,用户只需输入月份和年份,并将其查询到我的数据库。
这里是我的HTML也:
<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>
向我们显示您正在用于按日期获取结果的数据库查询。 –