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Laravel Eloquent查询模型与关系(Laravel Eloquent query model with relationship)

 我有2个模型与公司和DamageReport的关系。  
 DamageReport始终通过关键company_id链接到公司。  
 因此，DamageReport中的company_id等于公司中的id。  
 很简单吧？ 现在我的目标是在知道DamageReport的id时查询公司。  
 例如  
 我有一行DamageReport表：  
id company_id

6  1
 
 ID公司的记录是：  
id name

1  Company 1
 
 所以在我的控制器中我有DamageReport id（6）并且需要查询id为1的公司。  
 我在我的模特中建立了这样的关系  
 公司型号：  
/**
 * The Damage Reprots that belong to the Company.
 */
public function damageReports()
{
    return $this->belongsToMany('App\DamageReport');
}
 
 DamageReport模型：  
/**
 * The company of a damagereport
 *
 */
public function company()
{
    return $this->belongsTo('App\Company');
}
 
 现在在我的控制器中我试过这样的事情，但我老实说没有任何线索  
$company = new Company;

$company = $company->company($damageReportId);

dd($company);

I have 2 models with a relationship Company and DamageReport. 
A DamageReport is always linked to a Company by the key company_id. 
So company_id in DamageReport equals id in Company. 
Very simple, right? Now my goal is to query the Company when I know the id of the DamageReport. 
For example 
I have a row of the DamageReport table: 
id company_id

6  1
 
And the record of Company with id is: 
id name

1  Company 1
 
So in my controller I have the DamageReport id (6) and need to query company with id 1. 
I've set up a relationship like this in my models 
Company model: 
/**
 * The Damage Reprots that belong to the Company.
 */
public function damageReports()
{
    return $this->belongsToMany('App\DamageReport');
}
 
DamageReport model: 
/**
 * The company of a damagereport
 *
 */
public function company()
{
    return $this->belongsTo('App\Company');
}
 
Now in my controller I tried something like this but I honestly have no clue  
$company = new Company;

$company = $company->company($damageReportId);

dd($company);

原文：https://stackoverflow.com/questions/46315758

更新时间：2023-01-09 13:01

最满意答案

 假设条件是A >=1 & B ==0我们可以使用data.table  
library(data.table)
i1 <- setDT(df1)[, grp := rleid(A >= 1 & B==0)][, .I[A >= 1 & B==0 & seq_len(.N)>2], grp]$V1
df1[i1, IncrementalCounter := seq_len(.N)][is.na(IncrementalCounter), 
            IncrementalCounter := 0][, grp := NULL][]   
#    A B IncrementalCounter
# 1: 1 2                  0
# 2: 4 3                  0
# 3: 3 2                  0
# 4: 2 0                  0
# 5: 1 0                  0
# 6: 2 0                  1
# 7: 3 2                  0
# 8: 1 2                  0
# 9: 3 2                  0
#10: 2 0                  0
#11: 2 0                  0
#12: 3 0                  2
#13: 4 0                  3
#14: 2 0                  4
#15: 9 1                  0
 
 
 我们也可以使用rle来做base R  
rl <- with(df1, rle(A >=1 & B ==0))
r2 <- inverse.rle(within.list(rl, {i1 <- which(values)
             lengths[i1-1] <- lengths[i1-1] + 2
             lengths[i1] <- lengths[i1] - 2

         }))

cumsum(r2)*r2
#[1] 0 0 0 0 0 1 0 0 0 0 0 2 3 4 0
 
 数据  
df1 <- structure(list(A = c(1L, 4L, 3L, 2L, 1L, 2L, 3L, 1L, 3L, 2L, 
2L, 3L, 4L, 2L, 9L), B = c(2L, 3L, 2L, 0L, 0L, 0L, 2L, 2L, 2L, 
0L, 0L, 0L, 0L, 0L, 1L)), .Names = c("A", "B"), class = "data.frame", 
row.names = c(NA, -15L))

Assuming that the condition is A >=1 & B ==0 we can use data.table  
library(data.table)
i1 <- setDT(df1)[, grp := rleid(A >= 1 & B==0)][, .I[A >= 1 & B==0 & seq_len(.N)>2], grp]$V1
df1[i1, IncrementalCounter := seq_len(.N)][is.na(IncrementalCounter), 
            IncrementalCounter := 0][, grp := NULL][]   
#    A B IncrementalCounter
# 1: 1 2                  0
# 2: 4 3                  0
# 3: 3 2                  0
# 4: 2 0                  0
# 5: 1 0                  0
# 6: 2 0                  1
# 7: 3 2                  0
# 8: 1 2                  0
# 9: 3 2                  0
#10: 2 0                  0
#11: 2 0                  0
#12: 3 0                  2
#13: 4 0                  3
#14: 2 0                  4
#15: 9 1                  0
 
 
We can also do with base R using rle 
rl <- with(df1, rle(A >=1 & B ==0))
r2 <- inverse.rle(within.list(rl, {i1 <- which(values)
             lengths[i1-1] <- lengths[i1-1] + 2
             lengths[i1] <- lengths[i1] - 2

         }))

cumsum(r2)*r2
#[1] 0 0 0 0 0 1 0 0 0 0 0 2 3 4 0
 
data 
df1 <- structure(list(A = c(1L, 4L, 3L, 2L, 1L, 2L, 3L, 1L, 3L, 2L, 
2L, 3L, 4L, 2L, 9L), B = c(2L, 3L, 2L, 0L, 0L, 0L, 2L, 2L, 2L, 
0L, 0L, 0L, 0L, 0L, 1L)), .Names = c("A", "B"), class = "data.frame", 
row.names = c(NA, -15L))

Laravel Eloquent查询模型与关系(Laravel Eloquent query model with relationship)

最满意答案

数据

data

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