2009-11-06 159 views

回答

3

如何:

$query = "SELECT * FROM `TableA` WHERE `field1` = '$a' AND `field2` = '$b'"; 

請記住mysql_real_escape_string()$a$b

+0

歡呼,並且$ b周圍的引號解決了我將要問的問題(爲什麼電子郵件地址不起作用) – Daniel 2009-11-06 01:51:04

+0

cherios丹尼爾! – mauris 2009-11-06 01:54:44

1

SELECT * FROM表A,其中字段1 = $ a和場2 = $ B

1
SELECT * FROM TablaA WHERE `field1` = $a AND `field2` = $b 

$ a和$ b需要報價,如果他們可能不是數字。出於某種原因,我頭腦中有數字。

1

您的查詢是有點畸形,但你接近:

$a = mysql_real_escape_string($foo); 
$b = mysql_real_escape_string($bar); 

$sql = " 
SELECT 
    * 
FROM 
    `TablaA` 
WHERE 
    `field1` = '{$a}' 
    AND `field2` = '{$b}' 
"; 

使用準備好的發言將是逃避要好很多,但你可能沒有準備好扳手就會將你拋進計劃。請記住,只要您對這些東西有信心,請查看「準備好的語句」和「mysqli」擴展名。

0
<?php 
$dbhost = 'localhost'; 
$dbuser = 'root'; 
$dbpass = 'password'; 

$conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); 

$dbname = 'petstore'; 
mysql_select_db($dbname); 

$a = mysql_real_escape_string($input1); 
$b = mysql_real_escape_string($input2); 

$q = mysql_query("SELECT * FROM `TableA` WHERE `field1`='$a' AND `field2`='$b'"); 

?> 

不知道你是否也需要連接。