变量数量不匹配参数数量 - 是的，他们做

Question

所以根据一个建议，我试图包括准备好的语句。

但PHP不接受我的代码。

$stmt = $link->prepare('SELECT COUNT(*) FROM `table` WHERE `company` = CONVERT(_utf8 \'?\' USING latin1) COLLATE latin1_german1_ci AND `password` = CONVERT(_utf8 \'?\' USING latin1) COLLATE latin1_german1_ci'); 
$stmt->bind_Param('ss', $firmaP, $kennwP); 
$firmaP = utf8_encode($_POST['company']); 
$kennwP = utf8_encode($_POST['password']); 
if ($stmt->execute()) { 
    if($row = $stmt->fetch()) { 
     echo "OK"; 
    } 
    else 
    { 
     echo "NO"; 
    } 
} 

我得到

Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement

而我在查询的两个questionmarks和bind_Param两个变量。

如何解决这个问题？

编辑：替换\'？\'？导致

Fatal error: Call to a member function bind_param() on a non-object

Answer 1

不要引用您的? S作为如下：

$stmt = $link->prepare('SELECT COUNT(*) FROM `table` WHERE `company` = CONVERT(_utf8 ? USING latin1) COLLATE latin1_german1_ci AND `password` = CONVERT(_utf8 ? USING latin1) COLLATE latin1_german1_ci'); 
$firmaP = utf8_encode($_POST['company']); 
$kennwP = utf8_encode($_POST['password']); 
$stmt->bind_Param('ss', $firmaP, $kennwP); 
if ($stmt->execute()) { 
    if($row = $stmt->fetch()) { 
     echo "OK"; 
    } 
    else 
    { 
     echo "NO"; 
    } 
}