Question

此编码要求用户输入他们的员工ID，然后搜索数据库并将他们的购买信息显示在表格中（orderID，shippingDate等）。我希望能够创建和IF语句（非常确定它和IF语句），确保输入的人员ID存在于数据库中，如果它继续下面的代码，否则它只显示没有表写工作人员id不存在。不包含html文件，因为它不需要。

<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml" xml:lang="en" lang="en">
<head>
<title>Prac 2 Task 3</title>
</head>
<body>

<?php

$staffid = $_GET["staffID"];

if($staffid != $_GET["staffID"]){
echo "Id doesnt Exist";
}
else{
true;
}
?>




<?php
$conn = mysql_connect("localhost", "twa291", "twa291up");
mysql_select_db("factory291", $conn)
or die ('Database not found ' . mysql_error() );

$sql = "SELECT orderID, orderDate, orderDate, shippingDate, staffName FROM purchase, 
staff 
WHERE 
purchase.staffID = staff.staffID AND staff.staffID= '$staffid' ORDER BY purchase.orderDate ASC";

$rs = mysql_query($sql, $conn)
or die ('Problem with query' . mysql_error());



?>


<table border="1" summary="Staff Orders">
<tr>
<th>Order ID</th>
<th>Order Date</th>
<th>Shipping Date</th>
<th>Staff Name</th>
</tr>

<?php
while ($row = mysql_fetch_array($rs)) { ?>

<tr>

<td><?php echo $row["orderID"]?></td>
<td><?php echo $row["orderDate"]?></td>
<td><?php echo $row["shippingDate"]?></td>
<td><?php echo $row["staffName"]?></td>

</tr>


<?php   }
mysql_close($conn); ?>
</table>
</body>
</html>

感谢您的帮助。

Answer 1

在显示html之前先查询

<?php

$conn = mysql_connect("localhost", "twa291", "twa291up");
mysql_select_db("factory291", $conn)
or die ('Database not found ' . mysql_error() );

$sql = "SELECT orderID FROM purchase WHERE 
purchase.staffID = " . $_GET["staffID"];;

$rs = mysql_query($sql, $conn)
or die ('Problem with query' . mysql_error()); 




if($rs != null ){
 //show table
}
else{
  echo "Id doesnt Exist";
}
?>

只需修复代码，逻辑就像这样

Php和sql试图弄清楚用户名是否存在

1 个答案: