Question

所以，我写了一些查询代码返回错误：

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given

我知道这是由查询的布尔响应引起的，我已经检查过了，返回的布尔值等于true。所以我不明白为什么没有数据数组的响应......这是我的代码：

$data = mysqli_multi_query($connection, 'UPDATE teams SET teams.teamViews = teams.TeamViews
 + 1 WHERE (teams.teamID, \''.$userToken.'\') NOT IN (SELECT teams_views.teamId,
teams_views.'.$viewType.' FROM teams_views) AND teams.teamUrl = \''.TEAM_URL.'\';
INSERT INTO teams_views (teamId, '.$viewType.') SELECT t.teamId, \''.$userToken.'\'
FROM teams t WHERE t.teamUrl = \''.TEAM_URL.'\' AND NOT EXISTS (SELECT \''.$userToken.'\'
FROM teams_views WHERE t.teamId = teamId);
SELECT * FROM teams WHERE teams.teamUrl = \''.TEAM_URL.'\';');
$dataRow = mysqli_fetch_array($data, MYSQLI_ASSOC);

SQL中有三个查询 - 更新，插入和选择。

如何更改我的查询或PHP以返回数据，而不是布尔值？感谢

Answer 1

根据spencer7593的评论中的建议，我的问题是使用mysqli_store_result，mysqli_free_result和mysqli_next_result的组合解决的。以下是用于执行此操作的函数：

function multi_queries($query, $numQueries) {
    $connection = new database_connection();
    $data = mysqli_multi_query($connection->connection, $query) or die(mysqli_error($connection->connection));
    $data = mysqli_store_result($connection->connection);
    if (sizeof($data) > 0) {
        $this->success = true;
        do {
            if ($result = mysqli_store_result($connection->connection)) {
                while ($row = mysqli_fetch_row($result)) {
                    $this->data[sizeof($this->data)] = $row;
                }
                mysqli_free_result($result);
            }
        } while (mysqli_next_result($connection->connection));
    }
    $connection->close_connection();
}

mysqli_multi_query（） - 返回true而不是返回数据

1 个答案: