Question

我正在尝试将我的响应作为嵌套的json格式

[  
   {  
      "recipe_id":"33",
      "e_name":"crispy",
      "e_desc":"Crispy Gingersnaps",
      "calries":"500",
      "steps":[  
         {  
            "step_id":"22",
            "recipe_id":"33",
            "step_number":3,
            "instruction":'here is the instructions',            
         }
         {  
            "step_id":"23",
            "recipe_id":"34",
            "step_number":4,
            "instruction":'here is the instructions',            
         }
        {  
            "step_id":"23",
            "recipe_id":"35",
            "step_number":5,
            "instruction":'here is the instructions',            
         }
      ]
   }
]

这是我的代码

$query  = new \yii\db\Query();
            $recipes = $query->select(['recipe.recipe_id','recipe.image','recipe.'.$name,'recipe.'.$small_desc,'recipe.person_count',
            'recipe.calories','recipe.period','recipe.'.$desc,'recipe_step.instruction'])->from('recipe')
            ->where(['recipe.recipe_id' => $recipe_id])            
            ->innerJoin("recipe_step","recipe.recipe_id=recipe_step.recipe_id");

我得到结果，但是json格式不符合我的预期

Answer 1

我不知道是否存在仅通过查询实现此行为的选项。但是您可以实现与步骤关系的接受关系，并使用yii2方法添加步骤。

食谱模型

public function getStep()
{
  return $this->hasMany(Step::className(), ['recipe_id' => 'id']);
}

您希望在哪里获得

$model = Recipe::find()->where([Recipe::tableName() . '.id' => $recipe_id])->innerJoin('step')->asArray()->one();

因此，您获得了已描述的数据结构。

Answer 2

您需要使用all（）执行查询，以在查询结束时获得期望的结果。

 $recipes = $query->select(['recipe.recipe_id','recipe.image','recipe.'.$name,'recipe.'.$small_desc,'recipe.person_count',
        'recipe.calories','recipe.period','recipe.'.$desc,'recipe_step.instruction'])->from('recipe')
        ->where(['recipe.recipe_id' => $recipe_id])            
        ->innerJoin("recipe_step","recipe.recipe_id=recipe_step.recipe_id")->all();

返回嵌套Json格式的响应

2 个答案: