调用非对象上的成员函数 bind_param()(尽管经过研究仍无法解决)

时间:2023-03-06
本文介绍了调用非对象上的成员函数 bind_param()(尽管经过研究仍无法解决)的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着跟版网的小编来一起学习吧!

问题描述

$stmt = $mysqli->prepare('select Un from member where Lock = ? and Activated = ?');
$stmt -> bind_param("ss", 'N', 'Y');//This line gave the error
$stmt -> execute();
$stmt->store_result();//apply to prepare statement
$numRows = $stmt->num_rows;

if ($numRows > 0)//if have result
 while ($row = $stmt->fetch())

我上面的代码给了我一个在非对象上调用成员函数 bind_param()" 错误.我真的不明白为什么我会收到这个错误.我有正确的列名.

my above code gave me a "Call to a member function bind_param() on a non-object" error. I really don't get it why i getting this error. I have correct cols name.

我是 mysqli 的新手,想学习如何调试此类错误.

I am new to mysqli and would like to learn how to debug such error.

  1. 我的准备语句或 bind_param() 有什么问题?
  2. 请教我如何调试此类错误

推荐答案

在非对象上调用成员函数 bind_param() 意味着 $stmt,即你试图调用 bind_param on,不是一个对象.为什么它不是一个对象?因为 $mysqli->prepare 没有返回一个对象.为什么它没有返回一个对象?

Call to a member function bind_param() on a non-object means that $stmt, which you're trying to call bind_param on, is not an object. Why is it not an object? Because $mysqli->prepare did not return an object. Why did it not return an object?

mysqli_prepare() 返回一个语句对象或 FALSE.
http://www.php.net/manual/en/mysqli.prepare.php

mysqli_prepare() returns a statement object or FALSE if an error occurred.
http://www.php.net/manual/en/mysqli.prepare.php

所以这意味着一定发生了错误.你应该打开error_reporting,它可能会告诉你,或者检查$mysqli->error(),这也可能告诉你.

So that means an error must have occurred. You should turn on error_reporting, which will probably tell you, or examine $mysqli->error(), which may tell you as well.

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