我正在尝试在 PHP 发布脚本上提取查询,该脚本将提取信息
SELECT `id`
FROM `leads`
WHERE `status` = 'Passed'
AND `campaign_id` = '22d0cf4f-0f81-28b1-d8dc-4f046bd1d8ee'
原来的行是
$id_query = mysql_query("SELECT `id` FROM `leads` WHERE `status` = 'Passed' AND 'campaign_id' = '22d0cf4f-0f81-28b1-d8dc-4f046bd1d8ee'");
$id = mysql_fetch_row($id_query);
echo $id[0];
因此,我想添加一个日期变量,该变量将选择大于 65 年的时间:
BETWEEN (current_date - "DateField") > 365.0 * 65.0
但显然这是行不通的。我正在考虑使用 JOIN 并添加
AND 'datesubmitted_c' BETWEEN Date_Add(curDate(), INTERVAL -65 YEARS) AND curDate()'
$id_query = mysql_query("SELECT `id` FROM `leads` WHERE `status` = 'Passed' AND 'campaign_id' = '22d0cf4f-0f81-28b1-d8dc-4f046bd1d8ee' AND 'datesubmitted_c' BETWEEN Date_Add(curDate(), INTERVAL -65 YEARS) AND curDate()'");
不确定使用联接是否合适?
I think你正在寻找DATE_ADD方法,以便您可以在范围之间进行检查。就像是:
SELECT *
FROM mytable
WHERE _condition_1_
AND date_column BETWEEN CURDATE() AND DATE_ADD(CURDATE(), INTERVAL 65 YEAR)
这将检查日期列是否date_column
日期介于今天和 65 年后。
本文内容由网友自发贡献,版权归原作者所有,本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容,请联系:hwhale#tublm.com(使用前将#替换为@)