通过javascript将json转换为树数据

Posted 2023-01-11

我有这种数据。

[
    
        "dbname": "db1",
        "tblname": "tbl1",
        "colname": "ID_1",
        "RowCounts": "50"
    ,    
    
        "dbname": "db1",
        "tblname": "tbl1",
        "colname": "ID_2",
        "RowCounts": "50"
    ,  
    "dbname": "db1",
    "tblname": "tbl1",
    "colname": "ID_3",
    "RowCounts": "50"
  ,
    
    "dbname": "db1",
    "tblname": "tbl1",
    "colname": "ID_4",
    "RowCounts": "30"
  ,
    "dbname": "db1",
    "tblname": "tbl2",
    "colname": "ID_1",
    "RowCounts": "20"
  ,
    "dbname": "db1",
    "tblname": "tbl2",
    "colname": "ID_2",
    "RowCounts": "30"
  ,
    "dbname": "db1",
    "tblname": "tbl2",
    "colname": "ID_3",
    "RowCounts": "10"
  ,
    "dbname": "db1",
    "tblname": "tbl2",
    "colname": "ID_4",
    "RowCounts": "60"
  ,
    "dbname": "db1",
    "tblname": "tbl2",
    "colname": "ID_5",
    "RowCounts": "30"
  ,
    "dbname": "db1",
    "tblname": "tbl3",
    "colname": "ID_6",
    "RowCounts": "20"
  
]

而且我想转换成这种格式。

["name":"db1", 
"rowcount":500, //sum of child row counts
"children":[
"name":"tbl1",
"rowcount":200, //sum of children row counts
"children":[
"name":"ID_1",
"rowcount":50
,
"name":"ID_2",
"rowcount":50
],
"name":"tbl2",
"rowcount":200,
"children":[
]]]

我该怎么做？我尝试过此方法，但没有用。

我导入了csv文件，但其结构无法与树形结构配合使用，因此我需要对其进行转换。它具有3个字段，即dbname，tblname，colname和行数。只需将行计数视为价值或其他东西即可。只是需要合并。

function createTreeData(structure) 
        const node = (name, parent = null, row_cnt = 0) => (
            name,
            parent,
            row_cnt,
            children: []
        );
        const addNode = (parent, child) => (parent.children.push(child), child);
        const findNamed = (name, parent) => 
            for (const child of parent.children) 
                if (child.name === name) 
                    return child
                
                const found = findNamed(name, child);
                if (found) 
                    return found
                
            
        
        const TOP_NAME = "Top",
            top = node(TOP_NAME);
        for (const children of structure) 
            let par = top;
            for (const name of children) 
                const found = findNamed(name, par);
                par = found ? found : addNode(par, node(name, par.name));
            
                              
        return top;
    
它创建树结构，但不创建行数，并将其子级的行数相加。 

任何帮助将不胜感激。

我有这种数据。 [“” dbname“：” db1“，” tblname“：” tbl1“，” colname“：” ID_1“，” RowCounts“：” 50“，” dbname“：” db1“，” tblname“：.. 。

答案

您可以使用一组用于嵌套结构的键，减少此结构，并向最内部的数组添加一个新对象。在途中添加行数。