php - 警告 : mysqli_stmt_bind_result(): Number of bind variables doesn't match number of fields in prepared statement in

Question

这个问题已经有答案了:

mysqli_stmt::bind_result(): Number of bind variables doesn't match number of fields in prepared statement (2 个回答)

已关闭 1 年前。

我收到以下错误:

 Warning: mysqli_stmt_bind_result(): Number of bind variables doesn't match number of fields in prepared statement in  line 48

下面是第 48 行

mysqli_stmt_bind_result($stmt, $user_email, $user_pass);

下面是大部分代码

   // Using mysql_real_escape_string to ensure that the statements are legal SQL statements that can be stored
      $email = mysqli_real_escape_string($con, $_POST['email']);
      $password = mysqli_real_escape_string($con, $_POST['password']);


      /*

        Using Prepared statement to prevent SQL Injection

      */


      //  Create a perepared statement 
     if($stmt = mysqli_prepare($con, "SELECT * from admins WHERE user_email=? AND user_pass=?")) {


        // Bind the paramaters
        mysqli_stmt_bind_param($stmt, "ss", $email, $password);


        // Execute the query
        mysqli_stmt_execute($stmt);


        /*

         Usually we would bind the result to be able to retrieve their given value
         However, in this case we are not interested in retrieving values from the database

        */
           /* fetch value */
        mysqli_stmt_fetch($stmt);


        /* bind result variables */
         mysqli_stmt_bind_result($stmt, $user_email, $user_pass);

         echo $user_email;

连接已正确建立，即使没有在 myqsli_prepare 查询中设置条件，我仍然检索到相同的错误。

任何帮助将不胜感激。

Answer 1

问题是在 select 时您选择了 2 个以上的列，并且在 mysqli_stmt_bind_result 中只绑定(bind)了 2 个列，换句话说，select 中的列数必须与 mysqli_stmt_bind_result 中的变量数相同

要修复，请更改

SELECT * from admins WHERE user_email=? AND user_pass=?

致:

SELECT user_mail, user_pass from admins WHERE user_email=? AND user_pass=?"