mysqli->查询未返回任何内容 [英] mysqli->query not returning anything

问题描述

我在使用MySQLi时遇到麻烦.我已经习惯了旧的mysql_query，所以我确定这是一个非常愚蠢的问题. 我无法根据他们输入的电子邮件从数据库返回ID.

I'm having trouble with MySQLi. I am used to the old mysql_query, so I'm sure this is an incredibly stupid question. I am having trouble returning the ID from a database based on the email they entered.

这是查询代码:

$userid = $mysqli->query("SELECT id FROM client WHERE email = $email");

回显$ userid不会返回任何内容.

Echoing $userid does not return anything.

我是MySQLi的新手，所以这可能很简单.

I am new to MySQLi, so this is probably something very simple.

谢谢.

推荐答案

mysqli: :query :

返回值

失败时返回FALSE.对于成功的SELECT，SHOW，DESCRIBE或EXPLAIN查询，mysqli_query()将返回mysqli_result对象.对于其他成功的查询，mysqli_query()将返回TRUE.

Return Values

Returns FALSE on failure. For successful SELECT, SHOW, DESCRIBE or EXPLAIN queries mysqli_query() will return a mysqli_result object. For other successful queries mysqli_query() will return TRUE.

因此结果应为TRUE，FALSE或 mysqli_result对象，而不是您期望的查询的实际结果.

So the result should be TRUE, FALSE or a mysqli_result object, not the actual result of the query like you're expecting.

您需要执行以下操作:

$result = $mysqli->query("SELECT id FROM client WHERE email = $email");
if($result === FALSE) {
    die("Uh oh something went wrong");
}

然后使用类似 mysqli_result::fetch_array() 之类的东西.

Then use something like mysqli_result::fetch_array().

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