我有一个需要在PHP中以json格式编码的数据库查询,我需要以特定格式编码数据。
我想要的格式是
{
"Car tyre Showroom":{"ABCD":"1234567","CDEF":"90000000","PPPP":"1000000"},
"Bike tyre Showroom":{"AFGH":"32124235","AAAAA":"9111111111"},
"Car Driving School":{"AMNB ":"45565778"},
"Car battery shop":{"PLQA":"4235346456"}
}
但我的格式为
{
"Car tyre Showroom":{"PPPP":"1000000"}
},{
"Bike tyre Showroom":{"AAAAA":"9111111111"}
},{
"Car Driving School":{"AMNB ":"45565778"}
},{
"Car battery shop":{"PLQA":"4235346456"}
}
我的数据库查询是:
$query = "select S.SpecificCategoryName,A.* from specificcategories S,areaspecificdealers A where A.SpecificCategoryId=S.SpecificCategoryId and A.LocationCode=(Select LocationCode from arealist where LocationName='".$location."')";
我用这种方式编码json,
for($col = 0; $col < count($result); $col++)
{
$values[$result[$col]['SpecificCategoryName']]= array($result[$col]['ClientName']=>$result[$col]['PhoneNumber']);
}
echo json_encode($ values);
请以上述方式帮助如何编码数据。
答案 0 :(得分:1)
这应该可以解决问题:
$result=array();
for($row = 0; $row < count($result); $row++)
{
$result[$result[$row]['SpecificCategoryName']]= array($result[$row]['DealerName1']=>$result[$row]['PhoneNumber1']);
}
echo json_encode($result);