INSERT

Question

我得到：

Warning: mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement in E:\XAMPP\htdocs\account\lib\register.php on line 73

当我使用这段代码时：

if($stmt = $conn -> prepare("INSERT INTO login(user, pass) VALUES(?, ?)")) {

  /* Bind parameters s - string, b - blob, i - int, etc */
  $stmt -> bind_param("ss", $user, $pw);

  /* Execute it */
  $stmt -> execute();

  /* Bind results */
  $stmt -> bind_result($user, $pw);

  /* Close statement */
  $stmt -> close();
  $userId = $conn->insert_id;
}

我不明白，为什么每次都这样，我的代码片段有什么问题吗？

Answer 1

您正在尝试 bind_result 一个未返回任何结果的语句。

删除此行。 $stmt -> bind_result($user, $pw);

INSERT - 绑定变量的数量与准备好的语句中的字段数量不匹配

INSERT - Number of bind variables doesn't match number of fields in prepared statement

php

mysqli