我有一个“ conversation_message”表,并按“ sender_type”列(管理员/用户)分开了发送者角色。管理员和用户都在不同的表中。但是当我调用模型时,显示错误在null上调用成员函数addEagerConstraints()
表格列和数据
| id | id_group | id_reply | id_sender | sender_type | message
| 1 | 1 | null | 3 | admin | Hi, I'm admin
| 2 | 1 | 1 | 3 | admin | I wanna give u promo
| 3 | 1 | 2 | 18 | user | What's promo ?
我已经尝试过使用列值作为条件,但是它不起作用。
Conversation_message.php
public function sender(){
switch($this->sender_type){
case "user":
return $this->belongsTo(User::class, 'id_sender', 'id');
case "admin":
return $this->belongsTo(Admin::class, 'id_sender', 'id');
default:
return;
}
}
InboxController.php
public function message_detail_chat($groupId){
$data = [];
$data['messages'] = Conversation_message::with('replies')
->with('sender')
->with('recipients')
->where(['id_group' => $groupId])->get();
}
我希望按列值'sender_type'使用条件模型,但实际输出是错误的。
答案 0 :(得分:0)
laravel提供查询范围,您可以从此处https://laravel.com/docs/5.8/eloquent#local-scopes
进行读取#include <stdio.h>
int sum (int m, char n){
return m+n;
}
int main(){
char ch;
int c;
printf("Enter an integer and a character separated by a blank> ");
scanf("%d %c",&c, &ch);
if((c >= '0' && c <= '9')||(ch >= '0' && ch <= '9')){
int cs = sum(c, ch - 0);
printf("Character '%d' represents a digit. Sum of '%d' and '%c' is %d" , c, c, ch - 0, cs);
}
return 0;
}
现在您可以像这样访问发件人
function userSender(){
$this->belongsTo(User::class, 'id_sender', 'id');
}
function adminSender(){
return $this->belongsTo(Admin::class, 'id_sender', 'id');
}
public function scopeSender($query)
{
return $query
->when($this->sender_type === 'user',function($q){
return $q->with('userSender');
})
->when($this->sender_type === 'admin',function($q){
return $q->with('adminSender');
});
}
这应该给您正确的发件人。希望能帮助到你。