Question

我需要选择五年后没有跟进检查的患者名单，并将其保存到其他餐桌。当我加载患者列表时，这个过程会做，但是我的代码没有用。请帮帮我

//this is my code for getting the date of the patient last check up

<?php
include('connect.php');
$sql = mysql_query("SELECT date as date FROM tblnpatient"); 
while ($row = mysql_fetch_assoc($sql)) { 
$dt = $row['date']; 
}

?>

这是我在五年后将记录放入新表格的代码

<script>
function arch()
{ var c ='<?php echo $dt; ?>';
var cc= c.getFullYear
var dd= new date();
var cd = dd.getFullYear();
var bb = cd - cc;

if(bb > 5 || (bb == 5)) {
<?php
include('connect.php');
$sql = mysql_query("SELECT * FROM tblnpatient"); 
while ($row = mysql_fetch_assoc($sql)) { 
$pnum=$row['pnum'];
$pname=$row['pname'];
$addr=$row['addr'];
$bdate=$row['bdate'];
$age=$row['age'];
$sex=$row['sex'];
$civil=$row['civil'];
$cont=$row['cont'];
$fname=$row['fname'];
$mname=$row['mname'];
$sname=$row['sname'];
$complaints=$row['complaints'];
$bp=$row['bp'];
$bt=$row['bt'];
$weight=$row['weight'];
$date=$row['date'];

$update=mysql_query("INSERT INTO tblarch(pnum, pname, addr, bdate, cont, fname, mname,     sname, sex, civil, age, bp, bt, weight, complaints, date)
   VALUES('$pnum','$pname','$addr','$bdate','$cont','$fname','$mname','$sname','$sex','$civil','$age','$bp','$bt','$weight','$complaints','$date')");

        exit();}
 ?>

}

Answer 1

您可以更改查询以轻松检查5年时间范围。

SELECT date as date FROM tblnpatient WHERE date <= DATE_SUB(NOW(),INTERVAL 5 YEAR);

Mysql Date_Sub doc

这意味着您现在只需将数据插入新表中，而无需任何用于检查日期的Javascript代码。

此外，如果两个表足够相似，您也可以不需要任何PHP代码。

示例：

INSERT INTO tblarch (pnum, pname, addr, bdate, cont, fname, mname, sname, sex, civil, age, bp, bt, weight, complaints, date) 
SELECT (pnum, pname, addr, bdate, cont, fname, mname, sname, sex, civil, age, bp, bt, weight, complaints, date) 
FROM tblnpatient WHERE date <= DATE_SUB(NOW(),INTERVAL 5 YEAR);

五年后将数据放入其他表格中

1 个答案: