Question

我有Maxoffer模特：

protected $dates = [
        'start'
    ];

    public function setStartAttribute($date){
        $this->attributes['start']= Carbon::createFromFormat('m/d/Y h:i a', $date);
    }
    public function getStartAttribute($date){
        return (new Carbon($date))->toRfc2822String();
    }

在我的数据库中start的格式为：2016-02-02 00:00:00

现在，当我尝试运行WHERE查询时：

$maxoffer =  Maxoffer::where('article_id', $request->input('article_id'))
                    ->where('start', $request->input('start'))
                    ->first();
dd($maxoffer);

Input(start)格式：2016-02-02 12:00 am我也尝试以这种格式制作：2016-02-02 00:00:00但是再次不工作所以dd给我结果null 我遇到了一个问题因为我不知道如何比较开始和输入（开始）。如何以相同的格式制作它们......

我也无法改变：

public function getStartAttribute($date){
            return (new Carbon($date))->toRfc2822String();
        }

因为我需要这种格式用于其他事情。

那么如何才能实现这一目标？

Answer 1

在下面尝试将字符串转换为碳时间和Eloquent然后可以比较DB中的时间

#include <stdio.h>
#include <conio.h>
void plus_func();
void menus_func();
void mul_func();
void div_func();
void modulus_func();
void shiftleft_func();
void shiftright_func();
void and_func();
void or_func();
void not_func();
void doubleand_func();
void doubleor_func();
void notequal_func();
void xor_func();
void main(void){

int x,y, res;
int m;
int op1;
int op2;
int op3;
int i=0;
int k;
while(i==0){
 printf("Enter any key to start the calculator function program or escape to exit it \n");
 k=getch();
 if(k!=0x1b)
{
 printf("enter the first num or press escape to exit \n"); 
 x=getch();
 if(x==0x1b)
 {
    break;
 }
 else
 {
    op1=x-48;
 }
  printf("enter the second num or press escape to exit \n"); 
  y=getch();

  if(y==0x1b)
  {
    break;
  }
  else
  {
    op2=y-48;
  }
  printf("enter the type of the operation \n",m);
  m=getch();
  if(m==0x1b)
  {
    break;
  }
  else
  {
    op3=m-48;
  }
  switch(op3)
  {

     case 1: plus_func(op1,op2);break;

     case 2: menus_func(op1,op2);break;

     case 3: mul_func(op1,op2);break;

     case 4: div_func(op1,op2);break;

     case 5: modulus_func(op1,op2);break;

     case 6: shiftleft_func(op1,op2);break;

     case 7: shiftright_func(op1,op2);break;

     case 8: and_func(op1,op2);break;

     case 9: or_func(op1,op2);break;

     case 10: not_func(op1,op2);break;

     case 11: doubleand_func(op1,op2);break;

     case 12: doubleor_func(op1,op2);break;

     case 13: notequal_func(op1,op2);break;

     case 14: xor_func(op1,op2);break;

     default: printf("error");
     break;
  }
}
  else
  {
    break;
  }
}
}

  void plus_func(int a, int b)
{
    int c;
    c=a+b;
    printf("the result is %d \n",c);
}

  void menus_func(int a, int b)
{
    int c;
    c=a-b;
    printf("the result is %d \n",c);
}

  void mul_func(int a, int b)
{
    int c;
    c=a*b;
    printf("the result is %d \n",c);
}
  void div_func(int a, int b)
{
    int c;
    c=a/b;
    printf("the result is %d \n",c);
}
  void modulus_func(int a, int b)
{
    int c;
    c=a%b;
    printf("the result is %d \n",c);
}
  void shiftleft_func(int a, int b){
    int c;
    c=a<<b;
    printf("the result is %d \n",c);
}
  void shiftright_func(int a, int b)
{
    int c;
    c=a>>b;
    printf("the result is %d \n",c);
}
  void and_func(int a, int b)
{
    int c;
    c=a&b;
    printf("the result is %d \n",c);
}
  void or_func(int a, int b)
{
    int c;
    c=a|b;
    printf("the result is %d \n",c);
}
  void not_func(int a, int b)
{
    int c;
    c=a+b;
    c=~c;
    printf("the result is %d \n",c);
}
  void doubleand_func(int a, int b)
{
    int c;
    c=a&&b;
    printf("the result is %d \n",c);
}
  void doubleor_func(int a, int b)
{
    int c;
    c=a||b;
    printf("the result is %d \n",c);
}
  void notequal_func(int a, int b)
{
    int c;
    c=a+b;
    c=!c;
    printf("the result is %d \n",c);
}
  void xor_func(int a, int b)
{
    int c;
    c=a^b;
    printf("the result is %d \n",c);
}

比较时间 - 日期格式时的Laravel问题

1 个答案: