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求好的步骤~

程序员文章站 2024-02-07 14:26:40
...
求好的方法~?
数据如:
INSERT INTO `vm_nav` (`id`, `ctype`, `cid`, `name`, `ifshow`, `vieworder`, `opennew`, `url`, `type`) VALUES
(1, NULL, 0, 'index', 0, 1, 0, '', 'nav'),
(2, NULL, 0, 'product', 1, 2, 0, '', 'nav'),
(3, NULL, 0, 'member', 1, 3, 0, '', 'nav'),
(4, NULL, 0, 'guide', 1, 4, 0, '', 'nav'),
(5, NULL, 0, 'kehu', 1, 5, 0, '', 'nav'),
(6, NULL, 0, 'about', 1, 6, 0, '', 'nav'),
(7, NULL, 0, 'tehui', 1, 7, 0, '', 'nav'),
(8, NULL, 3, 'member_index', 1, 1, 0, '', 'nav'),
(9, NULL, 3, 'member_info', 1, 2, 0, '', 'nav');


求这种数据
array(8) {
[0]=>
array(6) {
["id"]=>
string(1) "2"
["name"]=>
string(12) "product"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
[1]=>
array(6) {
["id"]=>
string(1) "3"
["name"]=>
string(18) "member"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
[2]=>
array(6) {
["id"]=>
string(1) "8"
["name"]=>
string(12) "member_index"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "3"
}
[3]=>
array(6) {
["id"]=>
string(1) "9"
["name"]=>
string(12) "member_info"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "3"
}
[4]=>
array(6) {
["id"]=>
string(1) "4"
["name"]=>
string(12) "guide"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
[5]=>
array(6) {
["id"]=>
string(1) "5"
["name"]=>
string(12) "kehu"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
[6]=>
array(6) {
["id"]=>
string(1) "6"
["name"]=>
string(12) "about"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
[7]=>
array(6) {
["id"]=>
string(1) "7"
["name"]=>
string(12) "tehui"
["opennew"]=>
string(1) "0"
["url"]=>
string(0) ""
["ctype"]=>
NULL
["cid"]=>
string(1) "0"
}
}


我用了一个笨办法:先foreach吧cid 插入到一个son成员里 然后再一个foreach 放到其后面,描述的不清楚,代码如下:

function nav_list(){
$navlist = get_navigator();
sort($navlist['nav']);
//$navlist['nav'] 就是取出所有数据 ,上面用sort排了下序
foreach($navlist['nav'] as $k=>$v)
{
if($v['cid'] != 0)
{
//插入到一个son成员里
$l[$v['cid']]['son'][] = $v;

}else{
$l[$v['id']] = $v;
}
}
foreach($l as $k=>$v){
if($v['son']){
//消除son成员
$tmp[] = array_diff_key($v,array('son'=>''));
foreach($v['son'] as $key=>$val){
//再一个foreach 放到其后面
$tmp[] = $val;
}
}else{
$tmp[] = $v;
}
}
return $tmp;
}


~ 怎么样会快捷一点 ~??? 求解
------解决思路----------------------
select id, ctype, cid, name, ifshow, vieworder, opennew, url, type from vm_nav where ifshow and cid=0
union
select cid as id, ctype, cid, name, ifshow, vieworder, opennew, url, type from vm_nav where ifshow and cid>0
order by id
求好的步骤~

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