试图获取非对象MySQLi结果的属性 [英] Trying to get property of non-object MySQLi result

问题描述

有一些我在苦苦挣扎的PHP代码-搜索了Google等，并尝试了提及的所有内容，但是由于某种原因，我无法解决它.

Got a bit of PHP code I'm struggling with - had a search around Google etc. and tried everything mentioned, but for some reason I'm having trouble solving it.

问题是:

我有一些代码正在查询数据库中是否存在特定用户.

I have some code that is querying a database for the presence of a particular user.

代码(这是类中的方法)

The code (it's a method inside a class)

<?php
global $mysqli;
// Query specified database for value
$q = 'SELECT id FROM ' . $database . ' WHERE username = \'' . $username . '\'';
$r = $mysqli->query($q);
var_dump($r);
if ($r->num_rows) {
    // If row found username exists so return false
    return false;
}
...
?>

我已经var转储了查询结果($ r)并得到了它:

I've var dumped the result of the query ($r) and got this:

object(mysqli_result)#4 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows"]=> int(1) ["type"]=> int(0) }

这是正确的，上面应该只有1行.

This is correct, there should only be 1 row as above.

我确实收到此错误，链接到显示if ($r->num_rows) {

I do get this error linking to the line saying if ($r->num_rows) {

注意:尝试在第LINE行的FILE中获取非对象的属性

Notice: Trying to get property of non-object in FILE on line LINE

但是我不知道为什么，因为对象是有效的(如上所述)，并且应该可以正常工作.据我所知，似乎一切顺利，我只是想知道为什么会出现错误.我敢肯定这很简单，但我会很感激.

but I don't know why since the object is valid (as above) and it should be working fine. From what I can tell it seems to be going through alright, I'm just wondering why there's an error. I'm sure it's something simple but I'd appreciate any help.

推荐答案

$sql = "SELECT * FROM table";
$result = $conn->query($sql);

if (!$result) {
    trigger_error('Invalid query: ' . $conn->error);
}

使用mysqli_error()函数检查错误

check the error with mysqli_error() function

您的查询可能有错误.

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