Question

雇员：

+------------+----------------+
|  col_name  |   data_type    |
+------------+----------------+
| cid        | int            |
| name       | string         |
| model      | string         |
| cylinders  | int            |
| price      | int            |
+------------+----------------+

车：

table

我想将此架构导出到JSON对象，以便我可以根据表格填充HTML下拉菜单 - 例如，employee菜单将car和{{1} }。选择employee将使用与该表对应的列名和类型填充另一个下拉列表。

鉴于此用例，数据库的最佳json表示是否为此？

{
    "employee": {
        "salary": "int", 
        "destination": "string", 
        "eid": "int", 
        "name": "string"
    }, 
    "car": {
        "price": "int", 
        "model": "string", 
        "cylinders": "int", 
        "name": "string", 
        "cid": "int"
    }
}

编辑：或者这会更合适吗？

{
    "employee": [
        {
            "type": "int", 
            "colname": "eid"
        }, 
        {
            "type": "string", 
            "colname": "name"
        }, 
        {
            "type": "int", 
            "colname": "salary"
        }, 
        {
            "type": "string", 
            "colname": "destination"
        }
    ], 
    "car": [
        {
            "type": "int", 
            "colname": "cid"
        }, 
        {
            "type": "string", 
            "colname": "name"
        }, 
        {
            "type": "string", 
            "colname": "model"
        }, 
        {
            "type": "int", 
            "colname": "cylinders"
        }, 
        {
            "type": "int", 
            "colname": "price"
        }
    ]
}

Answer 1

在第一个示例中，所有数据都存储在对象中。假设结构存储在var mytables中，您可以使用Object.keys(mytables)获取名称，返回['employee', 'car']。内部列的等效项：Object.keys(mytables['employee'].cols)返回['salary','destination','eid','name']。

在第二个例子中，我建议将表格作为列存储在数组中，如

[name: 'employee', 
 cols: [ {
           "type": "int", 
           "colname": "cid"
         }, ...]

然后，您可以通过访问mytables[i].name

轻松迭代数组并获取名称

for (t in tables){
  console.log(tables[t].name);
  for (c in tables[t].cols)
    console.log(" - ",tables[t].cols[c].colname, ": ", tables[t].cols[c].type);
}

在JSON中表示数据库模式

1 个答案: