给定以下代码:
$objResellerList - DB::table('ResellerMaster as RM')
->leftJoin('StateMaster as SM','SM.id',RM.id_StateMaster')
->leftJoin('RegionMaster as REM','REM.id','=','RM.id_RegionMaster')
->leftJoin('DealerGroupMaster as
我们有一份用Laravel5.2写的旧申请书。我们目前正试图升级到5.4,而阻碍我们的最后一件事是whereHas错误。下面是原始sql查询:
# Laravel 5.4 Query:
select * from `jobs` where exists (select * from `channels` where `jobs`.`channel_id` = `channels`.`id` OR (`id` = 'RetroSnickers' or `channelid` like 'RetroSnickers' or `channelname` like
我在转换以下SQL查询以使用Laravel的查询构建器时遇到了很多问题。
SELECT * FROM gifts
JOIN giftcategory ON gifts.id = giftcategory.giftid
JOIN giftoccasions ON gifts.id = giftoccasions.giftid
JOIN giftrelationship ON gifts.id = giftrelationship.giftid
WHERE (gifts.gender = 'any' OR gifts.gender = 'male')
AND g
我尝试将两个SQL请求放入一个laravel数组中。 这些是我的SQL SELECT * FROM `transactions` WHERE `plan` LIKE '1050' ORDER BY `updated_at` DESC
SELECT * FROM `transactions` WHERE `user` LIKE '1050' ORDER BY `updated_at` DESC 我想让所有事务有“计划”== "1050“和所有事务有”用户“== "1050”在一个变量。 这对我不起作用: $ct=transactions::whe
如何在我的orWhere周围创建括号:
public function categories()
{
return $this->belongsToMany('App\Category', 'user_category')->orWhere('default', 1);
}
因此,它被转换为:
where (`user_category`.`user_id` = ? or `default` = 1)
目前,这里的括号丢失了,并使整个查询变得混乱。例如,我尝试过:
public function categories()
{
我有以下问题?当你在我们的平台上搜索foo时,你会得到一个空白页面的搜索结果。如果用户输入的搜索查询不在我们的平台上,并带有尖刻的小响应,如“对不起,我们的平台上没有foo's”,我想放一个搜索结果。
在控制器下的searchcontroller.php中,我有以下内容
<?php
class SearchController extends BaseController
{
public function getIndex()
{
$search = Request::get('q');
if (empty($search)) {
如何将此SQL转换为Laravel5.5雄辩格式
select * from 'arm_articles' where ('article_tag' like '%standard%' or 'article_topic' like '%standard%' or 'article_details' like '%standard%' or 'article_type' like '%standard%') and ( ('id' be
$user= Select from user where 'email' = $email AND 'specialty' =null OR 'address' = null OR 'country' =null OR 'state' = null;
这是我的代码,但它不能正常工作。如果任何指定列的值为null,则返回该行。
$doctor= Doctor::where('email',$email)->where(function ($query) {
我想用laravel实现正确的查询,如果有人能提供帮助的话
我有这个问题
SELECT * FROM `users` WHERE ((`is_verified` = 1) AND (`first_name` like '%%' or `middle_name` like '%%' or `last_name` like '%%' or `email` like '%%'))
在我的代码中有这样的代码
$user->where([
['is_verified', '=