获取插入sql的最新行以获取特定ID

Question

我正在创建当前薪资报告

这是我的疑问：

$sql = "SELECT *
FROM payroll, employees
WHERE employees.username='$username'
AND payroll.id=employees.id";

输出是： output并向下滚动this is the continuation of the output。

总结一下，查询显示某个ID的所有记录，但我希望它在数据库中显示该特定ID的最后插入数据。怎么会发生这种情况？

Answer 1

要获取最新的行，您必须按id desc订购，并且只能获得一个记录集limit 1。

所以代码将是

$sql = "SELECT *
FROM payroll, employees
WHERE employees.username='$username'
AND payroll.id=employees.id ORDER BY employees.id DESC LIMIT 1 ";

Answer 2

试试这个

$sql = "SELECT *
FROM payroll, employees
WHERE employees.username='$username'
AND payroll.id=employees.id ORDER BY employees.id DESC  LIMIT 1 " 

如果最新的ID是最大值。

Answer 3

使用此MySQL ORDER BY 或 LIMIT

$sql = "SELECT *
FROM payroll, employees
WHERE employees.username='$username'
AND payroll.id=employees.id ORDER BY employees.id DESC LIMIT 1";

Answer 4

$sql = "SELECT t1.*,t1.*
FROM payroll t1 join employees t2 on t1.id = t2.id
WHERE t1.username='$username' order by t2.id desc limit 1";