将数组转换为更具可读性和逻辑性的结构答案

【问题标题】：convert an array to more readable and logical structure将数组转换为更具可读性和逻辑性的结构
【发布时间】：2012-06-26 21:03:04
【问题描述】：

我有一个包含大约 200 条记录的数组，如下所示：

 [0] => Array
        (
            [sid] => 54
            [sname] => vISL User book
            [sdate] => 2012-03-01 00:00:00
            [pid] => 262
            [cid] => 95
            [at] => 
            [amount] => 0
            [psdate] => 2012-03-14 00:00:00
            [first_name] => Alan
            [last_name] => Darshan
            [pcid] => 95
            [cname] => Invisal
        )
 [1] => Array
        (
            [sid] => 54
            [sname] => vISL User book
            [sdate] => 2012-03-01 00:00:00
            [pid] => 263
            [cid] => 94
            [at] => 
            [amount] => 0
            [psdate] => 2012-03-14 00:00:00
            [first_name] => umed
            [last_name] => sarkar
            [pcid] => 92
            [cname] => Tisal
        )

我想将此数组转换为以下格式。原因是 sid、sname 和 sdate 组合可能有多个 pid、cid ... cname 值：

获得所需数组结构的最佳方法是什么？

【问题讨论】：

你是如何读取数据的？
these questions 的答案通常是您必须自己创建这个自定义数组。

标签： php arrays

【解决方案1】：

我会说遍历数组，然后使用var_export 获取输出：

$tmp = Array();
$out = Array();
foreach($input as $row) {
    $tmp[$row['sid']][] = $row;
}
foreach($tmp as $key=>$rows) {
    $tmp[$key] = Array(
        "sid"  =>$rows[0]['sid'],
        "sname"=>$rows[0]['sname'],
        "sdate"=>$rows[0]['sdate'],
        "parr" =>Array()
    );
    foreach($rows as $row) {
        $tmp[$key]['parr'][] = Array(
            "pid"       =>$row['pid'],
            "cid"       =>$row['cid'],
            "at"        =>$row['at'],
            "amount"    =>$row['amount'],
            "psdate"    =>$row['psdate'],
            "first_name"=>$row['first_name'],
            "last_name" =>$row['last_name'],
            "pcid"      =>$row['pcid'],
            "cname"     =>$row['cname']
        );
    }
}
$out = array_values($tmp);
file_put_contents("tmp.php",'<?php $data = '.var_export($out,true).'; ?>');

【讨论】：

【解决方案2】：

$old_data = /* same as in exemple */

$sids = array();

foreach($old_data as $row)
{
    if(!isset($sids[$row['sid']]))
    {
        $sids[$row['sid']] = array();
        foreach(array('sid', 'sname', 'sdate') as $keys)
        {
            $sids[$row['sid']][$keys] = $row[$keys];
        }
        $sids[$row['sid']]['parr'] = array();
    }

    $current_parr = array();
    foreach(array('pid', 'cid', 'at', 'amount', 'psdate', 'first_name', 'last_name', 'pcid', 'cname') as $keys)
    {
        $current_parr[$keys] = $row[$keys];
    }

    $sids[$row['sid']]['parr'][] = $current_parr;
}

【讨论】：

【解决方案3】：

我假设您从连接查询中获得了这些数组结果。如果您将这些数据与数据库分开，我认为这会更容易。所以你的问题类似于这个帖子：enter link description here

不管怎样，你可以试试这个：

$pk = 'sid';                // record pk
$pkKeys = array(
    'pid' => array(         // subRecord pk
        'parr' => array(    // subRecords array key => array(subRecord names)
            'pid',
            'cid',
            'amount',
            'psdate',
            'pcid',
        ),
    ),
);

$resultArr = array();
array_walk($records, 
    function ($val, $key) use (&$resultArr, $pkKeys, $pk) {
        $mapped = array();
        foreach ($val as $itemKey => $item) {
            if (in_array($itemKey, array_keys($pkKeys))) {
                $populateData = $pkKeys[$itemKey];
                $subRecordPk = key($populateData);
                foreach($populateData[$subRecordPk] as $populateKey) {
                    if ($subRecordPk !== $populateKey) {
                        $resultArr[$val[$pk]][$subRecordPk][$val[$itemKey]][$populateKey] 
                            = $val[$populateKey];
                        $mapped[$populateKey] = $populateKey;
                    }
                }
            } else if (!in_array($itemKey, $mapped)){
                $resultArr[$val[$pk]][$itemKey] = $val[$itemKey];
            }
        }
    }
);

【讨论】：