Question

错误：

致命错误：调用成员函数 bind_param（）在非对象中 /var/www/web55/web/pdftest/events.php 在第76行

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。

为什么我收到此错误？

Answer 1

AND c.category_id = ? - 您的查询中没有表别名c。

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);
}

Answer 2

我认为问题显然在于准备功能..

该函数可能失败，在这种情况下$ stmt将为FALSE，因此没有bind_param方法作为成员。

来自php mysqli manual：
mysqli_prepare（）返回一个语句对象，如果发生错误则返回FALSE。

检查您的查询！也许你的SELECT语句有问题。并且在尝试对您认为是由prepare函数返回的对象执行任何成员函数之前，还要检查FALSE。

if($stmt === FALSE)
    die("Prepare failed... ");// Handle Error Here

// Normal flow resumes here
$stmt->bind_param("i","");

的修改

我怀疑由于子查询，该语句可能会出错：

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)

相反，为什么不写这样的查询：

public function countDaysWithoutEvents()
{
    $count = FALSE;

    $sql  = "SELECT COUNT(d.date) ";
    $sql .= " FROM cali_events e ";
    $sql .= "      LEFT JOIN cali_dates d ON e.event_id = d.event_id ";
    $sql .= " WHERE YEARWEEK(d.date) = YEARWEEK(CURRENT_DATE()) ";
    $sql .= "       AND c.category_id = ? ";
    $sql .= " GROUP BY DAY(d.date) ";

    $stmt = $this->link->prepare($sql);
    if($stmt !== FALSE)
    {                
        $stmt->bind_param('i', $this->locationID);
        $stmt->execute();
        $stmt->bind_result($count);
        $stmt->fetch();                    // I think you need to do a fetch
                                           // here to get the result data..
        $stmt->close();
    }else                                  // Or, provide your own error
        die("Error preparing Statement");  // handling here

    return (7 - $count);
}

P.S。我想你也错过了对 fetch 的调用.. （见上面的例子）

Answer 3

$ this-＆gt; link-＆gt;准备此语句不返回对象所以它给你错误

当我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？

错误：

代码：

3 个答案: