算法题。
程序员文章站
2022-04-02 09:37:13
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对象原有属性值全部是大写,需求是将对象的属性值全部转化成小写么。 通常思路是 对象转化成数组(遍历转化成小写)再转化成对象 这样的思路很繁琐递归数据源且效率低求高效思路或方法stdClass Object( [INFO] => stdClass Object ( [A] => 1 [B] => 2 [C] => 3 [C] => 4 [D] => stdClass Object ( [AA] => 11 [BB] => 22 [CC] => 33 [CC] => 44 ) ))
回复讨论(解决方案)
对象不可能出现同名属性,请调整一下
对象不可能出现同名属性,请调整一下
stdClass Object( [INFO] => stdClass Object ( [A] => 1 [B] => 2 [C] => 3 [D] => stdClass Object ( [AA] => 11 [BB] => 22 [CC] => 33 [CC] => 44 ) ) )
版主有好的思路么
转化为数组了过后可以用spl中的数组迭代器来遍历,不知道速度会不会快一点
不是很明白为什么要有这样的需求,可以这样写
$s = '{"A":1,"B":2,"C":3,"D":{"AA":11,"BB":22,"CC":33}}';$o = json_decode($s); //模拟一个对象$t = serialize($o);$t = preg_replace_callback('/("[A-Z]+")(;.+?;)/', function($r) { return strtolower($r[1]) . $r[2]; }, $t);$o = unserialize($t);print_r($o);
stdClass Object( [a] => 1 [b] => 2 [c] => 3 [d] => stdClass Object ( [aa] => 11 [bb] => 22 [cc] => 33 ))
如果真是 json 则可写作
$s = '{"A":1,"B":2,"C":3,"D":{"AA":11,"BB":22,"CC":33}}';$s = preg_replace_callback('/"[A-Z]+":/', function($r) { return strtolower($r[0]); }, $s);print_r(json_decode($s));
stdClass Object( [a] => 1 [b] => 2 [c] => 3 [d] => stdClass Object ( [aa] => 11 [bb] => 22 [cc] => 33 ))
1.先json_encode把对象转字符串
2.使用preg_replace_callback 调用 strtolower把 "*": 转为小写,相当于把遍历改成正则替换。
3.json_decode 把字符串转为对象。