php - 您的 SQL 语法有误。 MySQL Selectgiving 错误与 LIKE 和

Question

我正在尝试使 MySQL 选择用于 mysql_query 函数，但我不断收到此错误:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'LEFT JOIN Orgs ON Jobs.CustOrgID = Orgs.ID 
            LEFT JOIN Persons ON Jobs.CustPer' at line 11

我已经尝试了所有方法并到处搜索，但似乎没有任何效果。感谢所有帮助。

 $Qry = "SELECT 

     Jobs.ID, 
     'Jobs.Status',
     Jobs.JobNum, 
     'Orgs.Nme', 
     'Persons.FirstNme',
     'Persons.LastNme',
     'JobTypes.JobType',
     'Jobs.Dsc',
     'Jobs.Notes'
     FROM Jobs ";

     if($column !== null && $text !== null) {
        $Qry .= "WHERE " . $column . " LIKE '%" . $text . "%' ";
     }

     $Qry .= "LEFT JOIN Orgs ON Jobs.CustOrgID = Orgs.ID 
         LEFT JOIN Persons ON Jobs.CustPersonID = Persons.ID
         LEFT JOIN JobTypes ON Jobs.JobTypeID = JobTypes.ID 
         ORDER BY JobNum";

解决方案:

     SELECT ...
     FROM ...
     LEFT JOIN ...
     LEFT JOIN ...
     WHERE ...
     ORDER BY ...

我的WHERE放错了，应该在两个LEFT JOINS之后。

Answer 1

您将 WHERE 插入错误的位置。它必须在连接之后出现:

SELECT ...
FROM ...
LEFT JOIN ...
LEFT JOIN ...
WHERE ...
ORDER BY ...