关于一些 json 的问题

转换成数组处理

最后面的
},
前面有？

@onlyhot 源代码是数组 但是不知道数组怎么搞

@imlonghao673 fupdz 前面那个{就是 只是后面忘记加了个]

唯一方式：遍历重构

循环，然后判断 fupdz 一致则放一起

@qiayue 循环采用的是数据库的 while

function convertFupdzData($source) {
$fupdz_arr = [];

foreach($source as $item) {
if(array_key_exists($item['fupdz'], $fupdz_arr)) {
$fupdz_arr[$item['fupdz']][] = $item;
} else {
$fupdz_arr[$item['fupdz']] = [$item];
}
}

$result = [];
foreach($fupdz_arr as $fupdz => $data) {
$result[] = [
"fupdz" => strval($fupdz),
"data" => $data
];
}

return $result;
}

@mhycy 我试试

@mhycy

$query = DB::query("SELECT * FROM " . DB::table('forum_forum') . " WHERE type='group' and status=1 ORDER BY fid ASC");
$ar=array();
while ($soft = DB::fetch($query)) {
$querys = DB::query("SELECT * FROM " . DB::table('forum_forum') . " WHERE fup=".$soft['fid']." and status=1 ORDER BY fid ASC");
while ($softs = DB::fetch($querys)) {
$queryss = DB::query("SELECT * FROM " . DB::table('forum_forumfield') . " WHERE fid=".$softs['fid']." ORDER BY fid ASC");
while ($softss = DB::fetch($queryss)) {
$ar[] = array("fupdz"=>$soft['fid'],"data"=>array(array("id"=>$softs['fid'],"fup"=>$softs['fup'],"name"=>iconv('gb2312','utf-8',$softs['name']),"px"=>$softs['displayorder'],"icon"=>$softss['icon'])));
}
}
}


如何添加你刚发的那个 这是源代码

@byxiaozhe #10 我只能说你这个代码写的真是 666 啊

@byxiaozhe

。。。。你这直接改数据库查询不就完了

```javascript
const sort = (arr) => {
let hash = {}
let newArr = []
let i = 0
for (let item of foo) {
if (hash[item.fupdz] == undefined) {
hash[item.fupdz] = i
newArr.push(item)
i++
} else {
const idx = hash[item.fupdz]
const data = item.data
data.forEach(item => newArr[idx].data.push(item))
}
}
return newArr
}
```
这样写应该没问题，楼主试试

v2 的回复功能太坑，不支持 markdown 还把代码缩进给吃了....