0

我需要更新 SQL 以按我要查询的帐号进行过滤：此代码显示与 &memid = 3 的值对应的帐户。我认为给 $memid 一个值会限制我的过滤。我怎样才能做一个sql 我将在其中输入帐号，然后将显示对应的帐户。

$memid = 3; // example
$sql = "SELECT member.*, account.* FROM member, account WHERE member.mem_id = 
account.mem_id AND member.mem_id = '".$memid."' ";

这是整个代码：

<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'password';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$memid = 3; // example
$sql = "SELECT member.*, account.* FROM member, account WHERE member.mem_id =  
account.mem_id AND member.mem_id = '".$memid."' ";

mysql_select_db('databaseName');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
 die('Could not get data: ' . mysql_error());
}
 while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "Account Number:{$row['Account_Number']}  <br> ".
     "First Name: {$row['fname']} <br> ".
     "Last Name: {$row['lname']} <br> ".
     "Address: {$row['address']} <br> ".
     "Contact: {$row['contact']} <br> ".
     "Share Capital: {$row['Share_Capital']} <br> ".
     "Regular Savings: {$row['Regular_Savings']} <br> ".
     "Power Savings: {$row['Power_Savings']} <br> ".
     "--------------------------------<br>";
} 
echo "Fetched data successfully\n";
mysql_close($conn);
?>
4

1 回答 1

0

首先，如果您只需要检索某些字段，请不要使用 *. 如果表成员和帐户表有同名字段，则使用 AS 表达式。此外，最好使用 JOIN SQL 示例连接表：

SELECT a.mem_id AS memberid,a.Account_Number AS accnum, %etc.% FROM member AS a
INNER JOIN account AS b ON a.mem_id = b.mem_id 
WHERE a.mem_id = '%id%';

在 mysql_fetch_array 的结果中，您将使用您在 AS 语句中设置的唯一键获得 $row 变量数组：

Array ("memberid"=>"%id%","accnum"=>"%Account_Number from DB%", %etc.%)

PS：英语不是我的母语，感谢理解，如有需要请随时纠正我。

于 2013-08-07T11:32:52.043 回答