獲取最新的行插入到SQL一定ID

Question

我創建一個薪水報告

這是我的查詢：

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

，輸出是： output並通過滾動向下，this is the continuation of the output。

總結一下，查詢顯示某個ID的所有記錄，但我希望它顯示數據庫中該特定ID的最後一個插入數據。這怎麼會發生？

Answer 1

要獲得最新的行，您必須訂購id desc並獲得一個記錄集limit 1。

因此，代碼將

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

Answer 2

試試這個

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

如果最新的ID是最大的價值。

Answer 3

使用這個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";