$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？
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);
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] => ))