Question

我需要一些帮助。我需要这个查询

SELECT *, STR_TO_DATE( end_date,  "%d/%m/%Y" ) AS DATE
FROM  `resume_experience` 
WHERE  `resume_id` =1
ORDER BY DATE DESC

以codeigniter活动记录格式编写。这就是我写的。

$result = $this->db->select('*,STR_TO_DATE(end_date,%d/%m/%Y) AS DATE')
            ->from('resume_experience')
            ->order_by('DATE', "desc")
            ->where($where)
            ->get()
            ->result_array();
    return $result;

它给了我以下错误。

您的SQL语法有错误;检查与您的MySQL服务器版本相对应的手册，以便在＃{; FROM（resume_experience）附近使用正确的语法WHERE resume_id =＆＃39; 1＆＃39; ORDER BY DATE desc＆＃39;在第2行

SELECT *, STR_TO_DATE(end_date, `%d/%m/%Y)` AS DATE FROM (`resume_experience`) WHERE `resume_id` = '1' ORDER BY `DATE` desc

非常感谢任何帮助。

谢谢和等待，

艾哈迈德

Answer 1

STR_TO_DATE()需要围绕格式字符串的引号：

$this->db->select('*,STR_TO_DATE(end_date,"%d/%m/%Y") AS DATE')

Answer 2

您需要在FALSE中将第二个参数作为select()传递，因此它不会引用STR_TO_DATE(end_date,%d/%m/%Y) AS DATE

的bacticks

$this->db->select("*,STR_TO_DATE(end_date,'%d/%m/%Y') AS DATE",FALSE)

See active record

在Codeigniter中写入此查询

2 个答案: