使用MYSQLI_REPORT_STRICT如何使mysqli抛出异常？ [英] How to make mysqli throw exceptions using MYSQLI_REPORT_STRICT?

问题描述

在mysqli中有一个名为 mysqli_report（）的函数，看起来像PDO的 setAttribute（）方法与其 ERRMODE _ * 常量。手册说：

There is a function in mysqli, called mysqli_report(), which looks like a counterpart for PDO's setAttribute() method with its ERRMODE_* constants. The manual says:

MYSQLI_REPORT_STRICT 抛出mysqli_sql_exception错误而不是警告

所以，考虑到 PDO :: ERRMODE_EXCEPTION ，我试过这个代码

So, having PDO::ERRMODE_EXCEPTION in mind, i tried this code

mysqli_report(MYSQLI_REPORT_STRICT);
$mysqli->query("foo");

但是，令我失望的是，它根本不产生任何例外或警告。

but, to my disappointment, it produced no exception nor warning at all.

所以，这里有一个问题：有没有办法告诉mysqli抛出异常而不使用 MYSQLI_REPORT_ALL ？

So, here goes the question: is there a way to tell mysqli to throw exceptions without using MYSQLI_REPORT_ALL?

推荐答案

经过一些研究，我终于知道函数的参数是一个位掩码，一个必须组合几个值来获得所需的结果。最终组合不是过于逻辑的，但它的工作原理，在忽略警告时抛出查询错误的异常。

After some research I've finally learned that the function's parameter is a bitmask, and one have to combine several values to get the desired result. The final combination is not overly logical, but it works as intended, throwing an exception on a query error while ignoring warnings.

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

将产生所需的结果：

致命错误：未捕获的异常'mysqli_sql_exception'与消息'您的SQL语法有错误;检查与您的MySQL服务器版本对应的手册，以便在第1行'foo'附近使用正确的语法

Fatal error: Uncaught exception 'mysqli_sql_exception' with message 'You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'foo' at line 1'

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