按帳戶號篩選的SQL

我需要更新SQL以按照我想要查詢的帳號進行篩選：此代碼顯示對應於& memid = 3的值的帳戶。我認爲給$ memid a值會限制我的過濾。如何在輸入帳號的情況下創建sql，然後顯示相應的帳號。按帳戶號篩選的SQL

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

這裏是整個代碼：

所有的

<?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); 
?>

來源

2013-08-07 Mhinne

mysql_ *已棄用。看看mysqli或pdo ... – Aquillo

'WHERE Account_Number = your_acc_num'？ –

你可以粘貼成員和帳戶表的結構嗎？看起來你需要加入accountid專欄。 –

首先，如果你需要找回只有特定的領域，不要使用*。如果表成員和帳戶表具有相同名稱的字段，請使用AS表達式。也最好是使用JOIN連接表 SQL實例：

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結果你到$ row變量數組與您在設置AS語句唯一鍵：

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

PS：英語不是我的母語，感謝您的理解，並隨時在必要時糾正我。

來源

2013-08-07 11:32:52 Ivanshum

謝謝，但它沒有奏效。 – Mhinne

按帳戶號篩選的SQL

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