其中內子句聯接查詢

請幫我對此查詢：其中內子句聯接查詢

function viewServices($userpno) 
{ 
    echo $userpno; 

    $this->query = " 
SELECT task.employee_id , task.user_id , task.service_id, service.name AS servicename , 
     service.description AS servicedescription, employee.name AS employeename, employee.pic_path AS employeepicture, 
     employee.pic_path 
FROM task where task.user_id = '$userpno' 
INNER JOIN employee ON employee.pno = task.employee_id 
INNER JOIN user ON user.pno = task.user_id 
INNER JOIN service ON service.service_id = task.service_id 
"; 
}

查詢工作完全沒有：

WHERE task.user_id = $userpno

：

WHERE task.user_id = '$userpno'

我已經以這種方式也嘗試

但它不起作用。

的錯誤是：

Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\wamp\www\admin\classes\Task.php on line 22

請quide我，我怎樣才能把WHERE子句。

來源

2011-03-26 umar

你知道你在做什麼這裏是一個潛在的SQL注入攻擊？您*必須*通過mysql_real_escape_string發送userpno值。請參閱http://php.net/manual/fr/function.mysql-real-escape-string.php和http://en.wikipedia.org/wiki/SQL_injection。 – 2011-03-27 01:56:44

@FrançoisBeausoleil我們不確定'$ userpno'是從哪裏來的，所以我們不能肯定地說。也許這個函數直接從另一個查詢結果中傳遞userID。但是，如果它的價值來源於用戶的輸入，那麼你是絕對正確的。 – Wiseguy 2011-03-27 04:32:24

嘗試：

$this->query = "SELECT task.employee_id , task.user_id , task.service_id, 
service.name AS servicename ,service.description AS servicedescription, 
employee.name AS employeename, employee.pic_path AS employeepicture,employee.pic_path 
FROM task 
INNER JOIN employee ON employee.pno = task.employee_id 
INNER JOIN user ON user.pno = task.user_id 
INNER JOIN service ON service.service_id = task.service_id 
where task.user_id = '$userpno'";

來源

2011-03-26 22:45:02 waqasahmed

np anytime bro :) – waqasahmed 2011-03-27 01:07:29

在您的WHERE子句後面有JOIN子句。這是無效的，所以你的查詢返回false，因爲它失敗了。

作爲參考，SELECT查詢的部分必須在訂單/格式如這裏的文檔概述： http://dev.mysql.com/doc/refman/5.0/en/select.html

來源

2011-03-26 22:41:29 Wiseguy

WHERE子句推移在查詢

SELECT task.employee_id , task.user_id , task.service_id, service.name AS servicename ,service.description AS servicedescription, employee.name AS employeename, employee.pic_path AS employeepicture,employee.pic_path 
FROM task 
INNER JOIN employee ON employee.pno = task.employee_id 
INNER JOIN user ON user.pno = task.user_id 
INNER JOIN service ON service.service_id = task.service_id 
where task.user_id = '$userpno'

來源

2011-03-26 22:41:32 Dalen

你很好， – Dalen 2011-03-26 22:56:01

查詢結構體的端是：SELECT，FROM（連接在這裏），WHERE

你有WHERE太快

$this->query = 
    "SELECT task.employee_id , task.user_id , task.service_id, service.name AS servicename ,service.description AS servicedescription, employee.name AS employeename, employee.pic_path AS employeepicture,employee.pic_path 
    FROM task INNER JOIN employee ON employee.pno = task.employee_id INNER JOIN user ON user.pno = task.user_id INNER JOIN service ON service.service_id = task.service_id 
    WHERE task.user_id = '$userpno'";

個

查詢運行正確返回的資源，那些無法返回false

來源

2011-03-26 22:44:37

沒問題....... – 2011-03-26 22:53:25

其中內子句聯接查詢

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