Laravel whereJsonContains 查询不起作用

Question

我正在尝试这个查询：

$order = Order::whereJsonContains('products->id',$serial_no)->first();

但它不起作用，请帮我解决这个问题。提前致谢。

[
  {
    "id": 24,
    "name": "test product",
    "qnt": 3,
    "sale_price": 18,
    "buy_price": 14,
    "stock": 3,
    "total": 54,
    "company": {
      "id": 1,
      "name": "test company",
      "contact": "01521482467,5555555",
      "address": "fdsa fdsa fdsa fdsa fdsa",
      "is_active": 1
    },
    "brand": {
      "id": 1,
      "name": "Test Brand",
      "is_active": 1
    },
    "category": {
      "id": 1,
      "name": "Test",
      "is_active": 1
    }
  },
  {
    "id": 28,
    "name": "test2",
    "qnt": 1,
    "sale_price": 18,
    "buy_price": 12,
    "stock": 1,
    "total": 18,
    "company": {
      "id": 1,
      "name": "test company",
      "contact": "01521482467,5555555",
      "address": "fdsa fdsa fdsa fdsa fdsa",
      "is_active": 1
    },
    "brand": {
      "id": 1,
      "name": "Test Brand",
      "is_active": 1
    },
    "category": {
      "id": 2,
      "name": "Test2",
      "is_active": 1
    }
  }
]

Answer 1

试试这些

$order = Order::whereJsonContains('products', ['id' => $serial_no])->first();

当你的 json 看起来像这样时

{
    "id": 24,
    "name": "test product",
    "qnt": 3,
    "sale_price": 18,
    "buy_price": 14,
    "stock": 3,
    "total": 54,
    "company": {
      "id": 1,
      "name": "test company",
      "contact": "01521482467,5555555",
      "address": "fdsa fdsa fdsa fdsa fdsa",
      "is_active": 1
 }

然后您的代码将正常工作。 但是你的JSON 在Array 中点赞

[
   {
        "id": 24,
        "name": "test product",
        "qnt": 3,
        "sale_price": 18,
        "buy_price": 14,
        "stock": 3,
        "total": 54,
        "company": {
          "id": 1,
          "name": "test company",
          "contact": "01521482467,5555555",
          "address": "fdsa fdsa fdsa fdsa fdsa",
          "is_active": 1
     }
]

所以你可以使用这些语法来解决你的问题

$order = Order::whereJsonContains('field_name', ['object_key' => 'object_value'])->first();

Answer 2

使用 get() 代替 first() 并重试。

Answer 3

@md abdul karim 请检查您的 mysql 版本，以及 mysql json_contains 与 mysql 5.7 或更高版本一起使用。