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array:1 [
1649 => array:2 [
"FirstName" => "jack"
"lastName" => "straw"
"mergedWITh" array:3 [
"id" =>'1650'
"id" =>'1651'
"id" =>'1652'
]
]
]
而不是……
array:4 [
1649 => array:2 [
"firstName" => "jack"
"lastName" => "straw"
]
1650 => array:2 [
"firstName" => "jack"
"lastName" => "straw"
]
1651 => array:2 [
"firstName" => "jack"
"lastName" => "straw"
]
1652 => array:2 [
"firstName" => "jack"
"lastName" => "straw"
]
]
我有一个循环运行,可以拉出重复项并找到组中的最低ID,但不确定将它们合并为一个的正确方法.
我展示的代码是搜索结果,该搜索已在这些特定字段中标识了具有重复条目的ID.我只是想进一步细化它不删除,但在id 1649的末尾添加一个字段,表示mergedWith(1650,1651,1652)
krsort($input); //group foreach ($input as $id => $PErson) { // overwrite the id each time,but since the input is sorted by id in descending order,// the last one will be the lowest id $names[$person['lastName']][$person['firstName']] = $id; } // ungroup to get the result foreach ($names as $lastName => $firstNames) { foreach ($firstNames as $firstName => $id) { $result[$id] = ['firstName' => $firstName,'lastName' => $lastName]; } }
编辑：根据您更新的问题,没有太多不同.只保留所有ID而不是单个ID.
krsort($input);
foreach ($input as $id => $person) {
// append instead of overwrite ↓
$names[$person['lastName']][$person['firstName']][] = $id;
}
foreach ($names as $lastName => $firstNames) {
foreach ($firstNames as $firstName => $ids) {
// $ids is already in descending order based on the initial krsort
$id = array_pop($ids); // removes the last (lowest) id and returns it
$result[$id] = [
'firstName' => $firstName,'lastName' => $lastName,'merged_with' => implode(',',$ids)
];
}
}
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