在对象上调用函数时，为什么在非对象上出现此函数调用错误?

本文介绍了在对象上调用函数时，为什么在非对象上出现此函数调用错误?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

代码:

public function countDaysWithoutEvents(){
    $sql = "SELECT 7 - COUNT(*) AS NumDaysWithoutEvents
            FROM
            (SELECT d.date
                FROM cali_events e
                LEFT JOIN cali_dates d
                ON e.event_id = d.event_id
                WHERE YEARWEEK(d.date) = YEARWEEK(CURRENT_DATE())
                AND c.category_id = ?
                GROUP BY DAY(d.date)
            ) AS UniqueDates";

    $stmt = $this->link->prepare($sql);
    $stmt->bind_param('i', $this->locationID);
    $stmt->execute();

    $stmt->bind_result($count);
    $stmt->close();

    return $count;
}

$this->link->prepare($sql)为MySQLi创建一个准备好的语句.

$this->link->prepare($sql) creates a prepared statement for MySQLi.

为什么会出现此错误?

推荐答案

AND c.category_id = ?-查询中没有表别名c.

AND c.category_id = ? - there is no table alias c in your query.

除了尝试

$stmt = $this->link->prepare($sql);
if (!$stmt) {
  throw new ErrorException($this->link->error, $this->link->errno);
}

if (!$stmt->bind_param('i', $this->locationID) || !$stmt->execute()) {
  throw new ErrorException($stmt->error, $stmt->errno);
}

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