php变换数组格式
- WBOY原创
- 2016-06-06 20:37:361005浏览
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>
回复内容：
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>
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