如何将不同的值合并为自己的键上的数组
如何将不同的值合并为自己的键上的数组
提问于 2019-08-31 21:46:39
我有包含相同键的数组，但值可以不同。我有包含相同键的数组，但值可以不同。
我的数组：
$arr1 = json_decode('{
"Chain": "Monaco Chain",
"Style": "Monaco Swarovski",
"Gauge": "9.50 mm",
"Length": "30.00",
"Color": "YELLOW",
"Karat": "21",
"Lock": "Long",
"2 Tone": "Yes",
"Alternate": "Yes"
}', true);
$arr2 = json_decode('{
"Chain": "Monaco Chain",
"Style": "Monaco Swarovski",
"Gauge": "9.50 mm",
"Length": "30.00",
"Color": "GREEN",
"Karat": "21",
"Lock": "Long",
"2 Tone": "Yes",
"Alternate": "Yes"
}', true);
$arr3 = json_decode('{
"Chain": "Monaco Chain",
"Style": "Monaco Swarovski",
"Gauge": "9.50 mm",
"Length": "30.00",
"Color": "YELLOW",
"Karat": "21",
"Lock": "Long",
"2 Tone": "No",
"Alternate": "Yes"
}', true);
$arr4 = json_decode('{
"Chain": "Monaco Chain",
"Style": "Monaco Swarovski",
"Gauge": "9.50 mm",
"Length": "300.00",
"Color": "YELLOW",
"Karat": "21",
"Lock": "Long",
"2 Tone": "Yes",
"Alternate": "Yes"
}', true)我必须将不同的值合并为一个数组，并将它们保存在各自的键中，因此我应该得到以下输出：
'{
"Chain": "Monaco Chain",
"Style": "Monaco Swarovski",
"Gauge": "9.50 mm",
"Length": ["30.00", "300.00"],
"Color": ["YELLOW", "GREEN"],
"Karat": "21",
"Lock": "Long",
"2 Tone": ["Yes", "No"],
"Alternate": "Yes"
}'注意:请不要关心json。
感谢您的帮助和时间。
回答 1
Stack Overflow用户
发布于 2019-08-31 22:59:53
function key_concat(...$arrays) {
if (count($arrays) == 1) return $arrays[0];
$temp = array_shift($arrays);
for ($i = 0; $i < count($arrays); $i++) {
foreach($arrays[$i] as $key => $value) {
if ((is_string($temp[$key]) && trim($temp[$key]) != trim($arrays[$i][$key])) ||
(is_array($temp[$key]) && array_search($arrays[$i][$key], $temp[$key]) === false)
) {
if (is_string($temp[$key])) {
$temp[$key] = Array($temp[$key], $arrays[$i][$key]);
} else if (is_array($temp[$key])) {
$temp[$key][] = $arrays[$i][$key];
}
}
}
}
return $temp;
}
print_r(key_concat($arr1, $arr2, $arr3, $arr4, $arr5));页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/57738551
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