算法题。该怎么解决
算法题。
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。 <br />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<br />求高效思路或方法<br />stdClass Object<br />(<br /> [INFO] => stdClass Object<br /> (<br /> [A] => 1<br /> [B] => 2<br /> [C] => 3<br /> [C] => 4<br /> [D] => stdClass Object<br /> (<br /> [AA] => 11<br /> [BB] => 22<br /> [CC] => 33<br /> [CC] => 44<br /> )<br /><br /> )<br /><br />)<br />
------解决方案--------------------
对象不可能出现同名属性,请调整一下
------解决方案--------------------
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
------解决方案--------------------
不是很明白为什么要有这样的需求,可以这样写
$s = '{"A":1,"B":2,"C":3,"D":{"AA":11,"BB":22,"CC":33}}';<br />$o = json_decode($s); //模拟一个对象<br /><br />$t = serialize($o);<br />$t = preg_replace_callback('/("[A-Z]+")(;.+?;)/', function($r) { return strtolower($r[1]) . $r[2]; }, $t);<br />$o = unserialize($t);<br /><br />print_r($o);
stdClass Object<br />(<br /> [a] => 1<br /> [b] => 2<br /> [c] => 3<br /> [d] => stdClass Object<br /> (<br /> [aa] => 11<br /> [bb] => 22<br /> [cc] => 33<br /> )<br />)<br />
如果真是 json 则可写作
$s = '{"A":1,"B":2,"C":3,"D":{"AA":11,"BB":22,"CC":33}}';<br />$s = preg_replace_callback('/"[A-Z]+":/', function($r) { return strtolower($r[0]); }, $s);<br />print_r(json_decode($s));<br />
stdClass Object<br />(<br /> [a] => 1<br /> [b] => 2<br /> [c] => 3<br /> [d] => stdClass Object<br /> (<br /> [aa] => 11<br /> [bb] => 22<br /> [cc] => 33<br /> )<br /><br />)<br />
------解决方案--------------------
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。
<br /><?php<br />$s = '{"A":1,"B":2,"C":3,"D":{"AA":11,"BB":22,"CC":33}}';<br />$obj = json_decode($s);<br /><br />print_r($obj);<br /><br />$str = json_encode($obj);<br />$str = preg_replace_callback('/"(.*?)":/', 'tolower', $str);<br />$obj = json_decode($str);<br /><br />print_r($obj);<br /><br />function tolower($matches){<br /> return '"'.strtolower($matches[1]).'":';<br />}<br />?><br />

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