我試圖使一個MySQL選擇用於mysql_query功能的使用,但我不斷收到此錯誤:你的SQL語法有錯誤。 MySQL的Selectgiving誤差與LIKE和
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
我已經竭盡所能,併到處搜尋,但似乎沒有任何正在工作。所有的幫助表示讚賞。
$Qry = "SELECT
Jobs.ID,
'Jobs.Status',
Jobs.JobNum,
'Orgs.Nme',
'Persons.FirstNme',
'Persons.LastNme',
'JobTypes.JobType',
'Jobs.Dsc',
'Jobs.Notes'
FROM Jobs ";
if($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";
SOLUTION:
SELECT ...
FROM ...
LEFT JOIN ...
LEFT JOIN ...
WHERE ...
ORDER BY ...
我的WHERE是在錯誤的地方,它應是左側的兩個JOINS後。
WHERE應該在JOIN之後和ORDER BY之前。 – andrewsi