2013-04-04 82 views
1

此代碼giveing我這個錯誤的號碼:非法參數編號:綁定變量的數量不匹配的令牌數量PHP-SQL的非法參數編號:綁定變量的數量不符令牌

$fields[':email']=$_GET["email"]; 
$fields[':password']=$_GET["password"]; 
$fields[':telefono']=$_GET["telefono"]; 
$fields[':old_email']=$_GET["old_email"]; 
$fields[':isAdmin']=0; 
$qry.="UPDATE utente SET Email:=email, Password:=password ,Telefono:=telefono, isAdmin=:isAdmin WHERE Email=:old_email"; 
$sth=$CONNESSIONE->prepare($qry); 
$sth->execute($fields); 

(5){[「:email」] => string(5)「test2」[「:password」] => string(1)「b」[「:telefono」 ] => string(7)「0415600」[「:old_email」] => string(4)「test」[「:isAdmin」] => int(0)}

and my table is this

CREATE TABLE IF NOT EXISTS `utente` (
    `Email` varchar(150) COLLATE utf8_unicode_ci NOT NULL, 
    `Password` varchar(150) COLLATE utf8_unicode_ci NOT NULL, 
    `Telefono` varchar(150) COLLATE utf8_unicode_ci NOT NULL, 
    `isAdmin` tinyint(1) NOT NULL, 
    UNIQUE KEY `Email` (`Email`) 
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci; 

爲什麼這不起作用?

回答

6

改變這種

$qry.="Email:=email, Password:=password ,Telefono:=telefono, isAdmin=:isAdmin WHERE Email=:old_email"; 

$qry.="Email=:email, Password=:password ,Telefono=:telefono, isAdmin=:isAdmin WHERE Email=:old_email"; 

=

+1

之前輸錯:非常感謝你 – 2013-04-04 12:54:15

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