我创建两个函数我想从第一个函数获取location_4值作为$ id并将其values数组传递给第二个函数我是如何做到的?这是我的代码:
function getbussinessdetail($profit_id)
{
$select="select a.account_id,a.title,a.budget,a.location_4,a.renewal_date,l.name from listing a,listinglevel l where l.value=a.level and profit_instructor_id='".$profit_id."' group by a.account_id";
$result = $GLOBALS ['mysqli']->query ($select) or die ($GLOBALS ['mysqli']->error . __LINE__);
while($rs=$result->fetch_assoc ())
{
$res[]=$rs;
}
return $res;
}
function getCityName($id)
{
if($id>0)
{
$select="select name from Location_4 where id=$id";
$result = $GLOBALS ['mysqli1']->query ($select) or die ($GLOBALS ['mysqli1']->error . __LINE__);
$rs=$result->fetch_assoc ();
return $rs['name'];
}
}
答案 0 :(得分:0)
试试这个:
$profit_id = 4;
$res = array();
$res = getbussinessdetail($profit_id);
if(count($res) > 0)
{
$cityNames = array();
foreach($res as $id)
{
$cityNames[] = getCityNames($id['location_4']);
}
}
var_dump($cityNames);
或者,只需更新第一个查询并删除第二个函数:
$select="select a.account_id,a.title,a.budget,a.location_4,a.renewal_date,
l.name from listing a,listinglevel l where l.value=a.level
and profit_instructor_id='".$profit_id."' group by a.account_id";
有了这个:
$select= "select a.account_id,a.title,a.budget,
(select name from Location_4 where id=a.location_4) as cityName,
a.renewal_date, l.name
from listing a,listinglevel l
where l.value=a.level
and profit_instructor_id='".$profit_id."' group by a.account_id";