laravel 关联查询问题
1.两个表，customers和contacts表
customers 客户表 有 id name address 字段
contacts 联系人表 有 id name tel qq customer_id字段
一个客户有多个联系人
2.模糊搜索天降联系人tel，列出customers表，如果用join方法会有重复的结果,laravel关联查询orm不能像tp的关联模型一样在customers结果中多一个contact数组存放符合的contacts吗？还要考虑到分页。。
更新
抱歉回复迟了，谢谢几个朋友的更新，晚上花时间使用了下面朋友的方法，首先试了下@daniel_wu魏 的方法，laravel手册上的预加载方法。
我需要搜索联系人电话号码中含有号码8的客户，客户4中联系人3和客户5联系人1、联系人2是符合条件的，目标就是查询出这个两个客户
代码如下
$list = Customer::with(['Contacts' => function ($query) use ($request) {
if ($request->has('tel')) {
$query->where('tel', 'like', '%' . $request->input('tel') . '%');
}
}])->get();然后我打印出$list
打印代码如下
foreach ($list as $data) {
echo $data->name;
foreach ($data->contacts as $item) {
echo ' ';
echo $item->name;
}
echo '<br>';
}打印结果如下
sql代码如下
select * from `customers` where `customers`.`deleted_at` is null select * from `contacts` where `contacts`.`deleted_at` is null and `contacts`.`customer_id` in ('37', '38', '39', '40', '41') and `tel` like '%8%'
可以看出不需要的客户1，2，3被查询了出来只是过滤了下符合条件的联系人。
然后来看@dawniii 的方法
$list = Customer::whereHas('Contacts', function ($query) use ($request) {
if ($request->has('tel')) {
$query->where('tel', 'like', '%' . $request->input('tel') . '%');
}
})->get();得到的结果
sql代码
select * from `customers` where `customers`.`deleted_at` is null and (select count(*) from `contacts` where `contacts`.`customer_id` = `customers`.`id` and `tel` like '%8%' and `contacts`.`deleted_at` is null) >= 1 select * from `contacts` where `contacts`.`deleted_at` is null and `contacts`.`customer_id` = '40' and `contacts`.`customer_id` is not null select * from `contacts` where `contacts`.`deleted_at` is null and `contacts`.`customer_id` = '41' and `contacts`.`customer_id` is not null
如果我的描述有问题，让大家回答有偏移，说声抱歉，
这次问题让我对laravel的orm有了很多理解，如果有什么错误，请大家指正，抛砖引玉，laravel相关的orm实战内容太少，谢谢大家热情回答。
你如果想要Customers的不重复的列表
$list = Customers::whereHas('Contacts', function($q)
{
$q->where('tel', 'like', '%foo%');
})->get();多看看手册吧
$list =Contacts::with(['Customers', function ($q) {
// sub query
}])->get()没记错的话，应该是这样
暂时没想到比较好的办法 用一个笨办法解决了
先查出符合条件的contacts表的customer_id,distinct()去重，然后customers表whereIn()得出的customer_id
ORM並不是要取代SQL，一些較複雜的query可能還是需要直接下SQL，只是ORM讓我們有更多的武器，如單純的新增 修改 刪除就非常適合使用ORM的方式。
ORM不适合复杂的对应关系，用回db类吧
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