[英]php merge arrays based on matching column name with multiple entries or duplicates
I have an array of arrays like this : 我有一个这样的数组数组:
$data = array (
'data1' => array (
0 =>
array (
0 => 'ID',
1 => 'PinCode',
2 => 'Date',
),
1 =>
array (
0 => '101',
1 => '454075',
2 => '2012-03-03',
),
2 =>
array (
0 => '103',
1 => '786075',
2 => '2012-09-05',
),
),
'data2' => array (
0 =>
array (
0 => 'Balance',
1 => 'ID',
),
1 =>
array (
0 => '4533',
1 => '101',
)
),
'data3' => array (
0 =>
array (
0 => 'Active',
1 => 'ID',
),
1 =>
array (
0 => 'Yes',
1 => '101',
),
2 =>
array (
0 => 'No',
1 => '103',
)
),
);
Here is the current working answer by user @Nigel Ren I have to this question here. 这是用户@Nigel Ren的当前工作答案,我在这里对此有疑问 。
$store = [];
$headers = [];
foreach ( $data as $set ) {
$headerRow = array_shift($set);
// Collect all header columns
$headers = array_merge($headers, $headerRow);
foreach ( $set as $index => $list ){
// Create associative list of data so they can be combined (i.e. ID fields)
$list = array_combine($headerRow, $list);
// Use ID value as key and create if needed
if ( !isset($store[$list["ID"]]) ) {
$store[$list["ID"]] = $list;
}
else {
$store[$list["ID"]] = array_merge($store[$list["ID"]], $list);
}
}
}
$headers = array_unique($headers);
$output = [ 'output' => [$headers]];
// Create template array, so that missing fields will be set to null
$blank = array_fill_keys($headers, null);
foreach ( $store as $dataRow ) {
// Fill in the fields for this ID and then change to numeric keys
$output['output'][] = array_values(array_merge($blank, $dataRow));
}
print_r($output);
Above code works perfectly for the above given array. 上面的代码非常适合上面给定的数组。
Here are the new cases I need to handle: 这是我需要处理的新情况:
CASE 1 : When the data contains only one array where the ID are same : 情况1:当数据仅包含ID相同的一个数组时:
$data = array (
'data1' => array (
0 =>
array (
0 => 'ID',
1 => 'PinCode',
2 => 'Date',
),
1 =>
array (
0 => '101',
1 => '454075',
2 => '2012-03-03',
),
2 =>
array (
0 => '101',
1 => '786075',
2 => '2012-09-05',
),
),
'data2' => array (
0 =>
array (
0 => 'Balance',
1 => 'ID',
),
),
'data3' => array (
0 =>
array (
0 => 'Active',
1 => 'ID',
),
),
);
In this case Nigel's answer doesn't output both the both records for the same ID. 在这种情况下,Nigel的答案不会同时输出相同ID的两条记录。 It outputs just a single record. 它仅输出一条记录。 Expected output when there's a single array with duplicate ID's present : 当存在具有重复ID的单个数组时的预期输出:
Array
(
[output] => Array
(
[0] => Array
(
[0] => ID
[1] => PinCode
[2] => Date
)
[1] => Array
(
[0] => 101
[1] => 454075
[2] => 2012-03-03
)
[2] => Array
(
[0] => 101
[1] => 786075
[2] => 2012-09-05
)
)
)
CASE 2 : 案例2:
When any of the array's other than the first array contain's entries for multiple same ID. 当除第一个数组以外的任何其他数组包含多个相同ID的条目时。
$data = array (
'data1' => array (
0 =>
array (
0 => 'ID',
1 => 'PinCode',
2 => 'Date',
),
1 =>
array (
0 => '101',
1 => '454075',
2 => '2012-03-03',
),
2 =>
array (
0 => '103',
1 => '786075',
2 => '2012-09-05',
),
),
'data2' => array (
0 =>
array (
0 => 'Order',
1 => 'ID',
),
1 =>
array (
0 => 'Colgate',
1 => '101',
),
2 =>
array (
0 => 'Waffles',
1 => '101',
)
),
);
Expected output in this case : 在这种情况下的预期输出:
Array
(
[output] => Array
(
[0] => Array
(
[0] => ID
[1] => PinCode
[2] => Date
[3] => Order
)
[1] => Array
(
[0] => 101
[1] => 454075
[2] => 2012-03-03
[3] => Colgate
)
[2] => Array
(
[0] => 101
[1] => 454075
[2] => 2012-03-03
[3] => Waffles
)
[3] => Array
(
[0] => 103
[1] => 786075
[2] => 2012-09-05
[3] =>
)
)
)
How do I do it? 我该怎么做?
you can get the desired output by using a loop: 您可以使用循环获得所需的输出:
$desired_array = array();
$n_data1 = count($data['data1']);
$n_data2 = count($data['data2']);
$n_data3 = count($data['data3']);
for ($i=0; $i < $n_data1; $i++) {
if ($n_data2 > $i) {
foreach ($data['data2'][$i] as $key => $value) {
if(!in_array($value,$data['data1'][$i])){
$data['data1'][$i][]=$value;
}
}
} else {
$data['data1'][$i][]= null;
}
}
for ($i=0; $i < $n_data1; $i++) {
if ($n_data3 > $i) {
foreach ($data['data3'][$i] as $key => $value) {
if(!in_array($value,$data['data1'][$i])){
$data['data1'][$i][]=$value;
}
}
} else {
$data['data1'][$i][]= null;
}
}
$desired_array = $data['data1'];
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