如何在不同环境中获取MySQLi错误信息

In my local/development environment, the MySQLi query is performing OK. However, when I upload it on my web host environment, I get this error:

在我的本地/开发环境中，MySQLi查询执行正常。但是，当我在我的Web主机环境中上传它时，我收到此错误：

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

致命错误：在...中的非对象上调用成员函数bind_param（）

Here is the code:

这是代码：

global $mysqli;
$stmt = $mysqli->prepare("SELECT id, description FROM tbl_page_answer_category WHERE cur_own_id = ?");
$stmt->bind_param('i', $cur_id);
$stmt->execute();
$stmt->bind_result($uid, $desc);

To check my query, I tried to execute the query via control panel phpMyAdmin and the result is OK.

为了检查我的查询，我尝试通过控制面板phpMyAdmin执行查询，结果是正常的。

1 个解决方案

#1

First of all, always have this line before MySQLi connect in all your environments:

首先，在MySQLi连接所有环境之前，始终拥有此行：

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

After that all MySQL errors will be transferred into PHP exceptions. Uncaught exception, in turn, makes a PHP fatal error. Thus, in case of a MySQL error, you'll get a conventional PHP error. That will instantly make you aware of the error cause. A stack trace will lead you to the exact spot where error occurred.

之后，所有MySQL错误都将转移到PHP异常中。反过来，未捕获的异常会导致PHP致命错误。因此，如果出现MySQL错误，您将收到传统的PHP错误。这将立即让您意识到错误原因。堆栈跟踪将引导您到达发生错误的确切位置。

Note that you have to be able to see PHP errors in general. And here indeed goes the matter of different environments:

请注意，您必须能够查看PHP错误。这确实是不同环境的问题：

On a live site you have to peek into error logs, so, settings have to be

在实际站点上，您必须查看错误日志，因此必须进行设置
```
error_reporting(E_ALL);
ini_set('display_errors', 0);
ini_set('log_errors', 1);
```
While on a local development server it's OK to make errors on the screen:

在本地开发服务器上，可以在屏幕上显示错误：
```
error_reporting(E_ALL);
ini_set('display_errors', 1);
```

And a little list of what you should not:

还有一些你不应该做的清单：

Use error suppression operator (@)
使用错误抑制运算符（@）
Use die() or echo or any other function to print the error message on screen unconditionally. PHP can echo it all right already, no assistance is required.
使用die（）或echo或任何其他函数无条件地在屏幕上打印错误消息。 PHP可以很好地回应它，不需要任何帮助。
Testing the query result manually (like if($result)) just makes no sense. Either your query failed and you will already get the error exception, or it was all right and there is nothing to test.
手动测试查询结果（如if（$ result））只是没有意义。您的查询失败并且您已经获得错误异常，或者它没有问题，并且没有任何可测试的内容。
Use try..catch for echoing the error message. Again PHP can do it better, way better.
使用try..catch回显错误消息。 PHP可以做得更好，更好。

#1