Question

我正在开发一个系统，它应该输出一天内发送的短信总数。它还应该分开交付的和失败的。它应该只将那天的记录放在那里......就像那天的记录一样。我的数据库是smsdb。

$sql = "CREATE TABLE smsdb (
     id INT(6) AUTO_INCREMENT PRIMARY KEY,
     sms_text VARCHAR(300) NOT NULL,
     receiver_number VARCHAR(30) NOT NULL,
     time_sent DATETIME,
     status VARCHAR(20) )";


//I tried to use the command to display todays records but I get all records even for other days.

$sql = ("SELECT DATE(time_sent) AS date, COUNT(*) as total_sent, SUM(status = 'Success') as delivered,SUM(status = 'failed') AS failed FROM smsdb GROUP BY DATE(time_sent) ");

我只想要今天的总记录。亲切的是，有人帮助我

Answer 1

您没有加入日期过滤器。您可以将其作为变量传递，因此您可以选择其他日期：

$sql = ("SELECT DATE(time_sent) AS date, COUNT(*) as total_sent, SUM(status = 'Success') as delivered,SUM(status = 'failed') AS failed FROM smsdb WHERE DATE(time_sent) = '$date_selected'");

请注意，您确实应该为此绑定参数，但这会给您一个想法。

如果您总是希望它是当天的另一个选项，请使用CURDATE()功能：

$sql = ("SELECT DATE(time_sent) AS date, COUNT(*) as total_sent, SUM(status = 'Success') as delivered,SUM(status = 'failed') AS failed FROM smsdb WHERE DATE(time_sent) = CURDATE()");

Answer 2

只需过滤当前日期的where条件：

SELECT
    DATE(time_sent) AS date,
    COUNT(*) as total_sent,
    SUM(case when status = 'Success' then 1 else 0 end) as delivered,
    SUM(case when status = 'failed' then 1 else 0 end) AS failed
FROM smsdb
WHERE date(time_sent)=curdate()
GROUP BY DATE(time_sent)

PHP Mysql select语句只显示当天的记录

2 个答案: