嵌套数组递归-NodeJS

时间:2018-09-04 13:34:17

标签: javascript arrays node.js recursion

我有一个嵌套的数组,其中包含动态数量的子级,我想基于此数组生成条件树。

数组的示例:

[
 {
  condition: 'conditionA',
  children: [
    {
      condition: 'conditionA_1',
      children: [
        ...
      ]
    },
  ]
 },
 {
  condition: 'conditionB',
  children: [
    ...
  ]
 }
]

我想生成一个包含以下条件语句的字符串

if (conditionA) {
  if (conditionA_1) {
    ...
  }
} else if (conditionB) {
  ...
}

有人对如何正确处理有任何想法吗?

谢谢。

2 个答案:

答案 0 :(得分:0)

无缩进:

只需将数组中的每个节点map递归到if(condition) { ... },然后将结果块与" else "结合起来:

function makeSource(arr) {
  return arr.map(function(node) {
    return "if (" + node.condition + ") { " + (node.children? makeSource(node.children): "") + " }";
  }).join(" else ");
}

演示:

function makeSource(arr) {
  return arr.map(function(node) {
    return "if (" + node.condition + ") { " + (node.children? makeSource(node.children): "") + " }";
  }).join(" else ");
}

var array = [ { condition: 'conditionA', children: [ { condition: 'conditionA_1'} ] }, { condition: 'conditionB' } ];

var source = makeSource(array);

console.log(source);

带有缩进:

要实现缩进,我们需要一个变量来保存当前块的深度。根据变量depth,只需在结果字符串的每一行之前重复空格字符即可。在每个递归调用中创建depth

function makeSource(arr, depth = 0) {
  return arr.map(function(node) {
    var str = " ".repeat(depth * 2) + "if (" + node.condition + ") {\n";
    if(node.children) {
      str += makeSource(node.children, depth + 1);
    } else {
      str += " ".repeat((depth + 1) * 2);                      // unecessary, it just indents the empty line to where the code should be
    }
    return str + "\n" + " ".repeat(depth * 2) + "}";
  }).join(" else ");
}

* 2部分代表缩进编号。如果您想缩进4个空格,则将其替换为* 4

演示:

function makeSource(arr, depth = 0) {
  return arr.map(function(node) {
    var str = " ".repeat(depth * 2) + "if (" + node.condition + ") {\n";
    if(node.children) {
      str += makeSource(node.children, depth + 1);
    } else {
      str += " ".repeat((depth + 1) * 2);
    }
    return str + "\n" + " ".repeat(depth * 2) + "}";
  }).join(" else ");
}

var array = [ { condition: 'conditionA', children: [ { condition: 'conditionA_1'} ] }, { condition: 'conditionB' } ];

var source = makeSource(array);

console.log(source);

答案 1 :(得分:0)

const input = [
 {
  condition: 'conditionA',
  children: [
    {
      condition: 'conditionA_1',
      children: [
        
      ]
    },
  ]
 },
 {
  condition: 'conditionB',
  children: [
    
  ]
 }
]

const createSource = arr =>{
  let source = "";
  
  if(arr.length === 0){
    return source;
  }
  
  arr.forEach((value,index) =>{
  
    source+="if( "+value.condition+" ){";
    if(value.children){
       source+=createSource(value.children)
    }
    source+="}" + (index+1  < arr.length ? " else " : "");
  })
  
  return source;
}
console.log(createSource(input))