php - 变量数量与参数数量不匹配——是的，它们匹配

Question

因此，根据建议，我正在尝试包含准备好的语句。

但是 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";
    }
}

我明白了

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

而我在查询中有两个问号，在 bind_Param 中有两个变量。

如何解决？

编辑:将\'?\' 替换为 ?导致

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

Answer 1

不要引用你的 ? 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";
    }
}