[英]PHP show data based on Yesterday and Today
I have an question about show data table based on date. 我有一个关于基于日期的显示数据表的问题。 In this case I want to show data based on today and yesterday.
在这种情况下,我想基于今天和昨天显示数据。
Design table : 设计表:
form_no | form_no | model_name |
model_name | prod_status |
prod_status | date_added
添加日期
Example data : 示例数据:
1 | 1 | acer |
宏碁| OK |
好的 12-APR-2013
12-APR-2013
2 | 2 | acer |
宏碁| NG |
NG | 11-APR-2013
11-APR-2013
So if in table I just want the data will show once, because the model name is same. 因此,如果在表格中我只想要显示一次数据,因为模型名称是相同的。 Table view Like this
表视图像这样
No | 没有| Model |
型号| Yesterday Status |
昨天状态| Today Status
今天状态
1 | 1 | acer |
宏碁| NG |
NG | OK
好
and here is my code : 这是我的代码:
$today = date("j-F-Y");
$yesterday = date("j-F-Y, time() - 60 * 60 * 24;");
$query = "SELECT model_name, prod_status, date_added FROM t_production_status WHERE date_added like '$today%'";
$result = mysql_query($query);
Now I'm done show data by today in table, now I want to display yesterday status to table view. 现在我已经在今天的表格中完成了显示数据,现在我想显示昨天状态到表格视图。
You can do it as below. 你可以这样做。
SELECT *
FROM tableName
WHERE date BETWEEN (CURDATE() - INTERVAL -1 DAY) AND CURDATE()
Use BETWEEN
in your query 在您的查询中使用
BETWEEN
$start_date = /* Yesterdays date */ ;
$end_date = /* Yesterdays date */ ;
$query = "SELECT * FROM table_name WHERE date_field BETWEEN $start_date AND $end_date";
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.