获取最新的行插入到SQL一定ID

Question

我创建一个薪水报告

这是我的查询：

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

，输出是： output并通过滚动向下，this is the continuation of the output。

总结一下，查询显示某个ID的所有记录，但我希望它显示数据库中该特定ID的最后一个插入数据。这怎么会发生？

Answer 1

要获得最新的行，您必须订购id desc并获得一个记录集limit 1。

因此，代码将

$sql = "SELECT * 
FROM payroll, employees 
WHERE employees.username='$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是最大的价值。

Answer 3

使用这个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";