es6 - javascript 将多个JSON对象合并成一个(带子父关系)
PHP中文网
PHP中文网 2017-06-30 09:58:35
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1、目前有一个很任性的接口API,它提供的数据非常不合理
2、数据如下

var json = [{"mainId": 1, "title": "abc", "createdate": "2017-06-28"},{"mainId": 2, "title": "ddddd", "createdate": "2017-06-25"}]; var json1 = [{"childId":1, "mainId": 1, "childname": "cname"},{"childId":2, "mainId": 1, "childname": "cname2"},{"childId":3, "mainId": 2, "childname": "cname3"},{"childId":4, "mainId": 2, "childname": "cname4"}]; var json2 = [{"childchildId":1, "childId": 1, "childname": "cname"},{"childchildId":2, "childId": 1, "childname": "cname2"},{"childchildId":3, "childId": 2, "childname": "cname3"},{"childchildId":4, "childId": 2, "childname": "cname4"},{"childchildId":5, "childId": 3, "childname": "cname"},{"childchildId":6, "childId": 3, "childname": "cname2"},{"childchildId":7, "childId": 4, "childname": "cname3"},{"childchildId":8, "childId": 4, "childname": "cname4"}];

3、我想将它们合并成一个json如下:

var result = [{"mainId":1, "title": "abc", "createdate": "2017-06-28", child:[{"childId":1, "mainId": 1, "childname": "cname", childchild:[{"childchildId":1, "childId": 1, "childname": "cname"},{"childchildId":2, "childId": 1, "childname": "cname2"}]},{"childId":2, "mainId": 1, "childname": "cname2", childchild:[{"childchildId":3, "childId": 2, "childname": "cname3"},{"childchildId":4, "childId": 2, "childname": "cname4"}]}]

是否有好的办法?

PHP中文网
PHP中文网

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全部回复 (3)
洪涛

额,就是不想写循环

let json2Obj = json2.reduce((acc, cur) => { let childId = cur.childId; if (!acc[childId]) { acc[childId] = []; } acc[childId].push(cur); return acc; }, {}) json1.forEach(item => { item.childchild = json2Obj[item.childId] }) let json1Obj = json1.reduce((acc, cur) => { let mainId = cur.mainId; if (!acc[mainId]) { acc[mainId] = []; } acc[mainId].push(cur); return acc; }, {}) json.forEach(item => { item.child = json1Obj[item.mainId]; }) console.log(JSON.stringify(json));

reduce重构下:

const toObj = (json, idStr) => json.reduce((acc, cur) => { let id = cur[idStr]; if (!acc[id]) { acc[id] = []; } acc[id].push(cur); return acc; }, {}) const json2Obj = toObj(json2, 'childId'); json1.forEach(item => item.childchild = json2Obj[item.childId]) const json1Obj = toObj(json1, 'mainId'); json.forEach(item => item.child = json1Obj[item.mainId]) console.log(JSON.stringify(json));
    滿天的星座

    php的,一步步来

    header('content-type:application/json;charset=utf8'); $json = '[{"mainId": 1, "title": "abc", "createdate": "2017-06-28"},{"mainId": 2, "title": "ddddd", "createdate": "2017-06-25"}]'; $json1 = '[{"childId":1, "mainId": 1, "childname": "cname"},{"childId":2, "mainId": 1, "childname": "cname2"},{"childId":3, "mainId": 2, "childname": "cname3"},{"childId":4, "mainId": 2, "childname": "cname4"}]'; $json2 = '[{"childchildId":1, "childId": 1, "childname": "cname"},{"childchildId":2, "childId": 1, "childname": "cname2"},{"childchildId":3, "childId": 2, "childname": "cname3"},{"childchildId":4, "childId": 2, "childname": "cname4"},{"childchildId":5, "childId": 3, "childname": "cname"},{"childchildId":6, "childId": 3, "childname": "cname2"},{"childchildId":7, "childId": 4, "childname": "cname3"},{"childchildId":8, "childId": 4, "childname": "cname4"}]'; $arr = array_merge(json_decode($json, true), json_decode($json1, true), json_decode($json2, true)); print_r(json_encode(dataFormat($arr))); //整理数据格式 function dataFormat($arr){ foreach ($arr as $key => &$value) { $value['id'] = $key; } foreach ($arr as $key => &$value) { if (!empty($value['mainId']) && empty($value['childId'])) { $value['pid'] = '100'; } if (!empty($value['mainId']) && !empty($value['childId'])) { foreach ($arr as $k => $v) { if (!empty($v['mainId']) && empty($v['childId'])) { if ($value['mainId'] == $v['mainId']) { $value['pid'] = $k; } } } } if (!empty($value['childchildId'])) { foreach ($arr as $k => $v) { if (!empty($v['mainId']) && !empty($v['childId'])) { if ($value['childId'] == $v['childId']) { $value['pid'] = $k; } } } } } //生成树 $arr = getTree($arr); //删除id、pid deleteKey($arr, 'id', 'pid'); return $arr; //删除不需要的键id、pid } //获取树 function getTree($items){ $tree = array(); foreach($items as $item){ if(isset($items[$item['pid']])){ $items[$item['pid']]['child'][] = &$items[$item['id']]; }else{ $tree[] = &$items[$item['id']]; } } return $tree; } //删除不需要的键 function deleteKey(&$arr, $id, $pid){ foreach ($arr as $key => &$value) { if (!empty($value['child']) && is_array($value['child'])) { deleteKey($value['child'], $id, $pid); } unset($value[$id], $value[$pid]); } }

    正常输出

    [ { "mainId": 1, "title": "abc", "createdate": "2017-06-28", "child": [ { "childId": 1, "mainId": 1, "childname": "cname", "child": [ { "childchildId": 1, "childId": 1, "childname": "cname" }, { "childchildId": 2, "childId": 1, "childname": "cname2" } ] }, { "childId": 2, "mainId": 1, "childname": "cname2", "child": [ { "childchildId": 3, "childId": 2, "childname": "cname3" }, { "childchildId": 4, "childId": 2, "childname": "cname4" } ] } ] }, { "mainId": 2, "title": "ddddd", "createdate": "2017-06-25", "child": [ { "childId": 3, "mainId": 2, "childname": "cname3", "child": [ { "childchildId": 5, "childId": 3, "childname": "cname" }, { "childchildId": 6, "childId": 3, "childname": "cname2" } ] }, { "childId": 4, "mainId": 2, "childname": "cname4", "child": [ { "childchildId": 7, "childId": 4, "childname": "cname3" }, { "childchildId": 8, "childId": 4, "childname": "cname4" } ] } ] } ]
      伊谢尔伦
      let obj1 = {}; let obj={} let obj2 = eval('(' + (JSON.stringify(obj) + JSON.stringify(obj1)).replace(/}{/, ',') + ')');

      很好用的,不过浏览器可能会提示eval是有害的,这个想要去掉,可以百度有方法,具体我忘了

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