php 如何调试为什么最简单的 MySQL 查询返回 false？
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How can I debug why simplest MySQL query returns false?
提问by tatiana_c
I work with xampp. I performed MySQL connection:
我和 xampp 一起工作。我执行了 MySQL 连接：
$connection = mysql_connect($host , $user , $passw);
mysql_select_db($db, $connection);
I received output with echo command (by check the boolean returned values) that connection is established and the database $db is found.
我收到了带有 echo 命令的输出（通过检查布尔返回值），该连接已建立并且数据库 $db 已找到。
But the simplest query like:
但最简单的查询如：
$query = mysql_query("SELECT * FROM 'users'");
returns false. How can I debug why?Thanks.
返回假。我怎样才能调试为什么？谢谢。
回答by Your Common Sense
An obligatory update: as mysql ext is no more, here are answers for two remaining MySQL APIs which I written on my site based on the experience from answering 1000s questions on Stack Overflow:
强制性更新：由于 mysql ext 已不复存在，以下是我根据在 Stack Overflow 上回答 1000 多个问题的经验在我的网站上编写的其余两个 MySQL API 的答案：
- How to report errors in mysqli
- How to connect to MySQL using PDO(with the aim of the proper error reporting).
- 如何在mysqli中报告错误
- 如何使用 PDO 连接到 MySQL（目的是正确报告错误）。
In short, for mysqi the following line have to be added before mysqli_connect()call:
简而言之，对于 mysqi，必须在mysqli_connect()调用前添加以下行：
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
while for PDO the proper error mode have to be set, for example
而对于 PDO，必须设置正确的错误模式，例如
$pdo->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );
As of the old mysql ext,
从旧的 mysql ext 开始，
To get an error from mysql_query()you have to use mysql_error()function.
So always run all your queries this way, at least until you develop a more advanced query handler:
要从mysql_query()你那里得到错误，你必须使用mysql_error()函数。
因此，始终以这种方式运行所有查询，至少在您开发更高级的查询处理程序之前：
$query = "SELECT * FROM 'users'";
$result = mysql_query($query) or trigger_error(mysql_error()." ".$query);
the problem with your current query is 'users'part. Single quotes have to be used to delimit strings while for the identifiers you have to use backticks:
您当前查询的问题是'users'一部分。必须使用单引号来分隔字符串，而对于标识符，您必须使用反引号：
SELECT * FROM `users`
In order to see these errors during development, add these lines at the top of your code to be sure you can see every error occurred
为了在开发过程中看到这些错误，请在代码顶部添加这些行，以确保您可以看到发生的每个错误
ini_set('display_errors',1);
error_reporting(E_ALL);
on the production server, however, the value on the first line should be changed from 1 to 0
但是，在生产服务器上，第一行的值应该从 1 更改为 0
回答by Ynhockey
Use the mysql_error() function:
使用 mysql_error() 函数：
$query = mysql_query("SELECT * FROM 'users'") or die(mysql_error());
EDIT: Per Col. Shrapnel's comment: you should never use die() outside of a test environment. In general it's bad practice when writing code that's even intended for production.
编辑：Per Col. Shrapnel 的评论：你不应该在测试环境之外使用 die()。通常，在编写甚至用于生产的代码时，这是不好的做法。
Here is some more information: http://www.phpfreaks.com/blog/or-die-must-die
回答by mrroot5
Based on Your Common Senseanswer this is an object oriented style:
根据您的常识答案，这是一种面向对象的风格：
$query = "SELECT * FROM 'users'";
$result = $mysqli -> query($query) or trigger_error($mysqli -> error." ".$query);
