問題描述

1、目前有一個很任性的接口API，它提供的數(shù)據(jù)非常不合理2、數(shù)據(jù)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、我想將它們合并成一個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'}]}]

是否有好的辦法？

問題解答

額，就是不想寫循環(huán)

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重構(gòu)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));回答2：

php的，一步步來

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)));//整理數(shù)據(jù)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; }} }} } //生成樹 $arr = getTree($arr); //刪除id、pid deleteKey($arr, ’id’, ’pid’); return $arr; //刪除不需要的鍵id、pid}//獲取樹function getTree($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 $tree;}//刪除不需要的鍵function deleteKey(&$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]); }}

正常輸出

[ {'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' }] }] }]回答3：

let obj1 = {};let obj={}let obj2 = eval(’(’ + (JSON.stringify(obj) + JSON.stringify(obj1)).replace(/}{/, ’,’) + ’)’);

很好用的，不過瀏覽器可能會提示eval是有害的，這個想要去掉，可以百度有方法，具體我忘了