$ar1=array(array('id'=>1,"name"=>"张1","sign"=>""),array('id'=>2,"name"=>"张2","sign"=>""),array('id'=>5,"name"=>"张5","sign"=>""));
$ar2=array(array('id'=>1,"name"=>"张1"),array('id'=>2,"name"=>"张2"));
如果ar2的id与ar1的id相同，那么ar1的sign则复制给1，否则不赋值，怎么解决？
------解决思路----------------------
$ar1=array(array('id'=>1,"name"=>"张1","sign"=>""),array('id'=>2,"name"=>"张2","sign"=>""),array('id'=>5,"name"=>"张5","sign"=>""));
$ar2=array(array('id'=>1,"name"=>"张1"),array('id'=>2,"name"=>"张2"));
//先将 $ar2 规格化一下
foreach($ar2 as $v) $r[$v['id']] = $v;
$ar2 = $r;
//接下来就简单了
foreach($ar1 as &$v) {
if(isset($ar2[$v['id']])) $v['sign'] = 1;
}
print_r($ar1);
Array
(
[0] => Array
(
[id] => 1
[name] => 张1
[sign] => 1
)
[1] => Array
(
[id] => 2
[name] => 张2
[sign] => 1
)
[2] => Array
(
[id] => 5
[name] => 张5
[sign] =>
)
)