[英]where clause in inner join query
請幫助我有關此查詢:
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
";
}
該查詢可以完美地運行而無需:
WHERE task.user_id = '$userpno'
我也嘗試過這種方式:
WHERE task.user_id = $userpno
但這是行不通的。
錯誤是:
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in C:\wamp\www\admin\classes\Task.php on line 22
請問我如何放置WHERE子句。
WHERE子句在查詢末尾
SELECT task.employee_id , task.user_id , task.service_id, service.name AS servicename ,service.description AS servicedescription, employee.name AS employeename, employee.pic_path AS employeepicture,employee.pic_path
FROM task
INNER JOIN employee ON employee.pno = task.employee_id
INNER JOIN user ON user.pno = task.user_id
INNER JOIN service ON service.service_id = task.service_id
where task.user_id = '$userpno'
查詢結構為:SELECT,FROM(在此處聯接),WHERE
你太早了
$this->query =
"SELECT task.employee_id , task.user_id , task.service_id, service.name AS servicename ,service.description AS servicedescription, employee.name AS employeename, employee.pic_path AS employeepicture,employee.pic_path
FROM task INNER JOIN employee ON employee.pno = task.employee_id INNER JOIN user ON user.pno = task.user_id INNER JOIN service ON service.service_id = task.service_id
WHERE task.user_id = '$userpno'";
正確運行的查詢將返回資源,失敗的查詢將返回false
您的WHERE子句后面有JOIN子句。 那是無效的,因此您的查詢由於失敗而返回false
。
作為參考,SELECT查詢的某些部分必須采用以下文檔中概述的順序/格式: http : //dev.mysql.com/doc/refman/5.0/en/select.html
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'";
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