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PHP显示基于昨天和今天的数据

[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

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";

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