数据相似时不向数据库插入值答案

【问题标题】：Not inserting values to database when data is similar数据相似时不向数据库插入值
【发布时间】：2015-11-02 07:35:34
【问题描述】：

我正在开发一个工资系统。 The admin is able to successfully save these data but when an employee ID, which is already in the database, is selected, it is not added to the database.这是我的代码：

$sql = "INSERT INTO payroll VALUES('$id','$period','$paidDays','$hourlyRate','$hoursworked','$overtime','$overtimePay','$undertime','undertimePay','$sss','$philHealth','$pagibig','$OtherDeductions','$tax','$grossPay','$netPay')";

我需要解决这个问题才能记录员工的薪资历史。

请注意此代码有效。只是当输入了类似的员工时，并没有添加到数据库中。谢谢！

已编辑：这是table structure

【问题讨论】：

您希望将其添加到数据库中吗？还是您打算避免这种情况？
@Sina，我相信她想添加到数据库，因为他想拥有“工资历史”。您是否有机会将该列设为“唯一索引”？
你好新浪！是的，它应该被添加到数据库中。
嗨 Mustafa，我没有让它独一无二
你能把你的表结构贴出来

标签： php mysql sql

【解决方案1】：

看你的表结构，id一定是PRIMARY KEY，所以选择现有id时不能添加。
由于主键不能重复，所以如果你想拥有那个过程，就必须改变结构。

【讨论】：

【解决方案2】：

我假设 '..similiar employee...' 是由 ID 定义的，这可能会给您一些提示：

 Insert Into Payroll
 Select id,period,paidDays,hourlyRate,hoursworked, overtime,overtimePay,
 undertime,undertimePay,sss,philHealth,pagibig,OtherDeductions,
 tax,grossPay,netPay
 FROM (
 select '$id' AS ID,'$period' AS period,'$paidDays' AS  paidDays,
 '$hourlyRate' AS hourlyRate, '$hoursworked' AS hoursworked,
 '$overtime' AS overtime,'$overtimePay' AS overtimePay,
 '$undertime' AS undertime,'undertimePay' AS undertimePay,
 '$sss' AS sss,'$philHealth' AS philHealth,'$pagibig' AS pagibig,
 '$OtherDeductions' AS OtherDeductions,'$tax' AS tax,
 '$grossPay' AS grossPay,'$netPay' AS netPay, 0 AS JUM
  UNION ALL
  select id,period,paidDays,hourlyRate,hoursworked,  overtime,
  overtimePay, undertime,undertimePay,sss,philHealth,pagibig,
  OtherDeductions,tax,grossPay,netPay, 1 AS JUM
  from [YourTable]
  Where [ID] = [EmployeeID]) AS T
  GROUP BY id,period,paidDays,hourlyRate,hoursworked, overtime,
  overtimePay, undertime,undertimePay,sss,philHealth,pagibig,
  OtherDeductions,tax,grossPay,netPay
  HAVING SUM(JUM) <0

【讨论】：