调用mysqli_error()时发出警告 [英] Warning when calling mysqli_error()

问题描述

可能重复:
PHP警告帮助?

Possible Duplicate:
PHP warning help?

我试图从数据库中联接三个表以显示用户选择的类别，但是出现以下错误.

I'm trying to join three tables from a database in-order to display a users selected categories but I get the following error.

Warning: mysqli_error() expects exactly 1 parameter, 0 given in

查询数据库时，我认为我做错了事.

I think I'm doing something wrong when I query my database.

这是下面的代码.

// Query member data from the database and ready it for display
$mysqli = new mysqli("localhost", "root", "", "sitename");
$dbc = mysqli_query($mysqli,"SELECT users.*, categories.*, users_categories.* FROM users_categories INNER JOIN users_categories ON users_categories.user_id = users.user_id JOIN categories ON users_categories.user_id = users.user_id WHERE users_categories.user_id=3");

if (!$dbc) {
    // There was an error...do something about it here...
    print mysqli_error();
}   

//Users entered category loop
while ($row = mysqli_fetch_assoc($dbc)) {
        if (! empty($row['category'])) {
                echo '<div class="skill-info">';
                echo '<p>' , htmlspecialchars($row['category']) , '</p>';
            }

    }

这是我的MySQL表结构.

Here is my MySQL tables structure.

CREATE TABLE users (
user_id INT UNSIGNED NOT NULL AUTO_INCREMENT,
user_name VARCHAR(255) NOT NULL,
PRIMARY KEY (user_id)
);

CREATE TABLE categories ( 
id INT UNSIGNED NOT NULL AUTO_INCREMENT, 
parent_id INT UNSIGNED NOT NULL DEFAULT 0, 
category VARCHAR(255) NOT NULL, 
url VARCHAR(255) NOT NULL, 
PRIMARY KEY (id), 
INDEX parent (parent_id)
);

CREATE TABLE users_categories (
id INT UNSIGNED NOT NULL AUTO_INCREMENT,
user_id INT UNSIGNED NOT NULL, 
category_id INT UNSIGNED NOT NULL, 
PRIMARY KEY (id)
);

现在我遇到以下错误

Not unique table/alias: 'users_categories'

我该如何解决?

感谢大家的帮助.我有什么办法可以奖励每个人，因为他们在stackoverflow上提供的帮助而不是一个人吗?

Thanks everyone for the help. Is there a way i can reward every one for there help here on stackoverflow instead of one person?

推荐答案

您忘记将必需的参数传递给 mysqli_error().更改

You forgot to pass a required argument to mysqli_error(). Change

print mysqli_error();

到

print mysqli_error($mysqli);

进行更正后，您将能够看到错误消息并进行进一步调试.

Once you make the correction, you'll be able to see the error message and debug further.

FYI:看来您奇怪地混合了mysqli的OO和过程样式用法.您应该坚持一个或另一个.

FYI: It looks like you're oddly mixing the OO and procedural style usage of mysqli. You should stick to either one or the other.

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