$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"));登入後複製
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？
回复讨论(解决方案)
$ar1=array(array('id'=>1,"name"=>"张1","sign"=>""),array('id'=>2,"name"=>"张2","sign"=>""),array('id'=>5,"name"=>"张5","sign"=>""));$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);登入後複製
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] => ))
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