Heim > Artikel > Backend-Entwicklung > php变换数组格式
如何将此数组
<code> [0] => Array ( [三月外] => Array ( [id] => 1 [age] => 22 [birthday] => 10月04日 [username] => 胡迪 [distance_birthday_time] => 214 ) ) [1] => Array ( [三月外] => Array ( [id] => 6 [age] => 24 [birthday] => 12月04日 [username] => 发给 [distance_birthday_time] => 275 ) ) </code>
变成
<code>[0] => Array ( [三月外] => Array ( [0]=>Array ( [id] => 1 [age] => 22 [birthday] => 10月04日 [username] => 胡迪 [distance_birthday_time] => 214 ) [1]=>Array ( [id] => 6 [age] => 24 [birthday] => 12月04日 [username] => 发给 [distance_birthday_time] => 275 ) ) ) </code>
如何将此数组
<code> [0] => Array ( [三月外] => Array ( [id] => 1 [age] => 22 [birthday] => 10月04日 [username] => 胡迪 [distance_birthday_time] => 214 ) ) [1] => Array ( [三月外] => Array ( [id] => 6 [age] => 24 [birthday] => 12月04日 [username] => 发给 [distance_birthday_time] => 275 ) ) </code>
变成
<code>[0] => Array ( [三月外] => Array ( [0]=>Array ( [id] => 1 [age] => 22 [birthday] => 10月04日 [username] => 胡迪 [distance_birthday_time] => 214 ) [1]=>Array ( [id] => 6 [age] => 24 [birthday] => 12月04日 [username] => 发给 [distance_birthday_time] => 275 ) ) ) </code>
笨点的办法吧,还是去foreach 一一比较去吧,可用key来判断当前的key是否存在,存在的话,直接追加到数据后面,否则就当成新元素增加
你这个就只是这种情况 吗?数组里有没有可能是多个元素的呢?
你是想把相同key的合并吧,
<code>foreach($a as $k=>$v){ $newArray[$k][] = $v; } $myAarray[0] = $newArray; </code>