blocks|key|324319|text|这是为我工作的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|324320|$user_info+=+DB::table('usermetas')
+++++++++++++++++->select('browser',+DB::raw('count(*)+as+total'))
+++++++++++++++++->groupBy('browser')
+++++++++++++++++->get();|code-block|syntax|javascript|324321|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This is working for me:

<pre><code>$user_info = DB::table('usermetas')
 -&gt;select('browser', DB::raw('count(*) as total'))
 -&gt;groupBy('browser')
 -&gt;get();
</code></pre>

blocks|key|323637|text|这适用于我(Laravel+5.1)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323638|$user_info+=+Usermeta::groupBy('browser')->select('browser',+DB::raw('count(*)+as+total'))->get();|code-block|syntax|javascript|323639|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This works for me (Laravel 5.1):

<pre><code>$user_info = Usermeta::groupBy('browser')-&gt;select('browser', DB::raw('count(*) as total'))-&gt;get();
</code></pre>

blocks|key|2224186|text|谢谢安东尼奥|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2224187|我刚在末尾添加了lists命令，所以它只返回一个带键和计数的数组：|offset|length|style|CODE|2224188|Laravel+4
|blockquote|2224189|2224190|$user_info+=+DB::table('usermetas')
++++->select('browser',+DB::raw('count(*)+as+total'))
++++->groupBy('browser')
++++->lists('total','browser');|code-block|syntax|javascript|2224191|Laravel+5.1
|2224192|2224193|$user_info+=+DB::table('usermetas')
++++->select('browser',+DB::raw('count(*)+as+total'))
++++->groupBy('browser')
++++->lists('total','browser')->all();|2224194|Laravel+5.2%2B
|2224195|2224196|$user_info+=+DB::table('usermetas')
++++->select('browser',+DB::raw('count(*)+as+total'))
++++->groupBy('browser')
++++->pluck('total','browser');|2224197|entityMap^0|0|8|5|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|13|8|@$D|14|E|15|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|16|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|17|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|18|8|@]|9|@]|A|$O|P]]|$1|Q|3|R|5|J|7|19|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|T|3|U|5|N|7|1B|8|@]|9|@]|A|$O|P]]|$1|V|3|W|5|J|7|1C|8|@]|9|@]|A|$]]|$1|X|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|Y|3|Z|5|N|7|1E|8|@]|9|@]|A|$O|P]]|$1|10|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]]|11|$]]

Thanks Antonio,
I've just added the <code>lists</code> command at the end so it will only return one array with key and count:
<blockquote>
Laravel 4
</blockquote>
<pre><code>$user_info = DB::table('usermetas')
 -&gt;select('browser', DB::raw('count(*) as total'))
 -&gt;groupBy('browser')
 -&gt;lists('total','browser');
</code></pre>
<blockquote>
Laravel 5.1
</blockquote>
<pre><code>$user_info = DB::table('usermetas')
 -&gt;select('browser', DB::raw('count(*) as total'))
 -&gt;groupBy('browser')
 -&gt;lists('total','browser')-&gt;all();
</code></pre>
<blockquote>
Laravel 5.2+
</blockquote>
<pre><code>$user_info = DB::table('usermetas')
 -&gt;select('browser', DB::raw('count(*) as total'))
 -&gt;groupBy('browser')
 -&gt;pluck('total','browser');
</code></pre>

blocks|key|324374|text|如果您想获得集合，请使用groupBy并计数：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|324375|$collection+=+ModelName::groupBy('group_id')
->selectRaw('count(*)+as+total,+group_id')
->get();|code-block|syntax|javascript|324376|干杯!|324377|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you want to get collection, groupBy and count:

<pre><code>$collection = ModelName::groupBy('group_id')
-&gt;selectRaw('count(*) as total, group_id')
-&gt;get();
</code></pre>

Cheers!

blocks|key|1081827|text|开放config/database.php|type|ordered-list-item|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1081828|在strict连接设置中查找mysql键|BOLD|1081829|将值设置为false|1081830|unstyled|1081831|1081832|entityMap^0|2|J|0|1|6|E|5|1|6|E|5|0|5|5|5|5|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|H]|$9|V|A|W|B|H]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|I|3|J|5|6|7|11|8|@$9|12|A|13|B|H]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|K|3|-4|5|L|7|16|8|@]|D|@]|E|$]]|$1|M|3|-4|5|L|7|17|8|@]|D|@]|E|$]]|$1|N|3|-4|5|L|7|18|8|@]|D|@]|E|$]]]|O|$]]

<ol>
<li>Open <code>config/database.php</code></li>
<li>Find <code>strict</code> key inside <code>mysql</code> connection settings</li>
<li>Set the value to <code>false</code></li>
</ol>

blocks|key|2224213|text|也是这样，更整洁一点。getQuery()只返回底层构建器，它已经包含表引用。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2224214|$browser_total_raw+=+DB::raw('count(*)+as+total');
$user_info+=+Usermeta::getQuery()
++++->select('browser',+$browser_total_raw)
++++->groupBy('browser')
++++->pluck('total','browser');|code-block|syntax|javascript|2224215|entityMap^0|B|A|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Works that way as well, a bit more tidy.
<code>getQuery()</code> just returns the underlying builder, which already contains the table reference.
<pre><code>$browser_total_raw = DB::raw('count(*) as total');
$user_info = Usermeta::getQuery()
 -&gt;select('browser', $browser_total_raw)
 -&gt;groupBy('browser')
 -&gt;pluck('total','browser');
</code></pre>

blocks|key|323641|text|试试这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323642|->groupBy('state_id','locality')
++->havingRaw('count+>+1+')
++->having('items.name','LIKE',"%25$keyword%25")
++->orHavingRaw('brand+LIKE+?',array("%25$keyword%25"))|code-block|syntax|javascript|323643|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

<h2>Try with this</h2>

<pre><code>-&gt;groupBy('state_id','locality')
 -&gt;havingRaw('count &gt; 1 ')
 -&gt;having('items.name','LIKE',"%$keyword%")
 -&gt;orHavingRaw('brand LIKE ?',array("%$keyword%"))
</code></pre>

blocks|key|323651|text|$post+=+Post::select(DB::raw('count(*)+as+user_count,+category_id'))
++++++++++++++->groupBy('category_id')
++++++++++++++->get();|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|323652|这是一个按类别计算员额的结果的例子。|unstyled|323653|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>$post = Post::select(DB::raw('count(*) as user_count, category_id'))
 -&gt;groupBy('category_id')
 -&gt;get();
</code></pre>
This is an example which results count of post by category.

blocks|key|323659|text|Laravel版本8|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323660|删除DB的依赖项|323661|+++++$counts+=+Model::whereIn('agent_id',+$agents)
++++++++->orderBy('total',+'asc')
++++++++->selectRaw('agent_id,+count(*)+as+total')
++++++++->groupBy('agent_id')
++++++++->pluck('total','agent_id')->all();|code-block|syntax|javascript|323662|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Laravel Version 8
Removed the dependency of DB
<pre><code> $counts = Model::whereIn('agent_id', $agents)
 -&gt;orderBy('total', 'asc')
 -&gt;selectRaw('agent_id, count(*) as total')
 -&gt;groupBy('agent_id')
 -&gt;pluck('total','agent_id')-&gt;all();
</code></pre>

blocks|key|323668|text|另一种方式是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323669|$data+=+Usermeta::orderBy('browser')->selectRaw('browser,+count(*)+as+total')->get()|code-block|syntax|javascript|323670|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another way would be this:
<pre class="lang-php prettyprint-override"><code>
$data = Usermeta::orderBy('browser')-&gt;selectRaw('browser, count(*) as total')-&gt;get()

</code></pre>

blocks|key|1081837|text|如果您想得到排序的数据，也可以使用这个|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1081838|$category_id+=+Post::orderBy('count',+'desc')
++++->select(DB::raw('category_id,count(*)+as+count'))
++++->groupBy('category_id')
++++->get();|code-block|syntax|javascript|1081839|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to get sorted data use this also
<pre><code>$category_id = Post::orderBy('count', 'desc')
 -&gt;select(DB::raw('category_id,count(*) as count'))
 -&gt;groupBy('category_id')
 -&gt;get();
</code></pre>

blocks|key|324387|text|在Laravel+8中，您可以使用countBy()获得一个组的总数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|324388|检查相同的文档。https://laravel.com/docs/8.x/collections#method-countBy|324389|entityMap|0|LINK|mutability|MUTABLE|url|https://laravel.com/docs/8.x/collections#method-countBy^0|H|9|0|8|1J|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@$9|T|A|U|1|V]]|E|$]]|$1|H|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

In Laravel 8 you can use <code>countBy()</code> to get the total count of a group.
Check the documentation on the same.
<a href="https://laravel.com/docs/8.x/collections#method-countBy" rel="nofollow noreferrer">https://laravel.com/docs/8.x/collections#method-countBy</a>

blocks|key|2224305|text|简单的解决方案(用Laravel+9和Spatie/权限进行测试)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2224306|主计长：|2224307|//Get+permissions+group+by+guard+name(3+in+my+case:+web,+admin+and+api)
$permissions+=+Permission::get()->groupBy('guard_name');|code-block|syntax|javascript|2224308|查看：|2224309|@foreach($permissions+as+$guard+=>+$perm)
++<div+class="form-group">
++++<label+for="permission">Permissions+({{+ucfirst($guard)+}})+{{+count($perm)+}}</label>
++++<select+name="permission[]"+id="permission"+class="form-control+@error('permission')+is-invalid+@enderror"+multiple>
++++++@foreach($perm+as+$value)
++++++++<option+value="{{+$value->id+}}">{{+$value->name+}}</option>
++++++@endforeach
++++</select>
++++@error('permission')
++++++<div+class="invalid-feedback">
++++++++{{+$message+}}
++++++</div>
++++@enderror
++</div>
@endforeach|2224310|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Simple solution(tested with Laravel 9 and Spatie/Permissions).
Controller:
<pre><code>//Get permissions group by guard name(3 in my case: web, admin and api)
$permissions = Permission::get()-&gt;groupBy('guard_name');
</code></pre>
View:
<pre><code>@foreach($permissions as $guard =&gt; $perm)
 &lt;div class=&quot;form-group&quot;&gt;
 &lt;label for=&quot;permission&quot;&gt;Permissions ({{ ucfirst($guard) }}) {{ count($perm) }}&lt;/label&gt;
 &lt;select name=&quot;permission[]&quot; id=&quot;permission&quot; class=&quot;form-control @error('permission') is-invalid @enderror&quot; multiple&gt;
 @foreach($perm as $value)
 &lt;option value=&quot;{{ $value-&gt;id }}&quot;&gt;{{ $value-&gt;name }}&lt;/option&gt;
 @endforeach
 &lt;/select&gt;
 @error('permission')
 &lt;div class=&quot;invalid-feedback&quot;&gt;
 {{ $message }}
 &lt;/div&gt;
 @enderror
 &lt;/div&gt;
@endforeach
</code></pre>

blocks|key|323691|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|323692|在作曲家中使用"illuminate/database":+"%5E9.38"。所以“应该”是最迟在这篇文章发表的时候。|offset|length|style|CODE|323693|我真的不知道为什么他们认为返回第一张唱片是正确的选择。即|323694|当前的实现对包括groupBy语句在内的查询没有意义。为什么N_1应该是“正确的”结果？由于@taylorotwell正确地指出了计数子查询结果的一些性能问题，我们为什么不在php端修复它，检查是否有组语句，如果有，则执行N_1+%2B+N_2+%2B.%2B+N_M？
|blockquote|323695|323696|https://github.com/laravel/ideas/issues/1693#issuecomment-621167890|323697|323698|包装查询并进行计数似乎对我有效。|323699|$records+=+...

$record_count+=+DB::table(+"fake"+);
$record_count->fromSub($records->select(DB::raw(1)),"query");
$record_count->count();|code-block|syntax|javascript|323700|创建一个“假”查询生成器|unordered-list-item|323701|从(.)处添加一个分词"+from“查询将“子查询”设置为选择1，而不是返回巨大的列数据。不确定是否需要这样做，但在我看来，这似乎是个好主意|323702|323703|323704|323705|323706|select+count(*)+as+aggregate+from+(select+1+from+...+group+by+...)+as+`query`|323707|子是$records将正常执行的“查询”。|323708|323709|323710|https://github.com/laravel/framework/issues/44081#issuecomment-1301816710|323711|entityMap|0|LINK|mutability|MUTABLE|url|1^0|0|7|U|0|0|0|0|0|1V|0|0|0|0|0|0|0|0|0|0|0|0|2|8|0|0|0|0|21|1|0^^$0|@$1|2|3|-4|4|5|6|1L|7|@]|8|@]|9|$]]|$1|A|3|B|4|5|6|1M|7|@$C|1N|D|1O|E|F]]|8|@]|9|$]]|$1|G|3|H|4|5|6|1P|7|@]|8|@]|9|$]]|$1|I|3|J|4|K|6|1Q|7|@]|8|@]|9|$]]|$1|L|3|-4|4|5|6|1R|7|@]|8|@]|9|$]]|$1|M|3|N|4|5|6|1S|7|@]|8|@$C|1T|D|1U|1|1V]]|9|$]]|$1|O|3|-4|4|5|6|1W|7|@]|8|@]|9|$]]|$1|P|3|Q|4|5|6|1X|7|@]|8|@]|9|$]]|$1|R|3|S|4|T|6|1Y|7|@]|8|@]|9|$U|V]]|$1|W|3|X|4|Y|6|1Z|7|@]|8|@]|9|$]]|$1|Z|3|10|4|Y|6|20|7|@]|8|@]|9|$]]|$1|11|3|-4|4|Y|6|21|7|@]|8|@]|9|$]]|$1|12|3|-4|4|5|6|22|7|@]|8|@]|9|$]]|$1|13|3|-4|4|5|6|23|7|@]|8|@]|9|$]]|$1|14|3|-4|4|5|6|24|7|@]|8|@]|9|$]]|$1|15|3|16|4|T|6|25|7|@]|8|@]|9|$U|V]]|$1|17|3|18|4|5|6|26|7|@$C|27|D|28|E|F]]|8|@]|9|$]]|$1|19|3|-4|4|5|6|29|7|@]|8|@]|9|$]]|$1|1A|3|-4|4|5|6|2A|7|@]|8|@]|9|$]]|$1|1B|3|1C|4|5|6|2B|7|@]|8|@$C|2C|D|2D|1|2E]]|9|$]]|$1|1D|3|-4|4|5|6|2F|7|@]|8|@]|9|$]]]|1E|$1F|$4|1G|1H|1I|9|$1J|N]]|1K|$4|1G|1H|1I|9|$1J|1C]]]]

Since this is the top result when i search for <code>eloquent count with groupby returns only first</code>
using <code>&quot;illuminate/database&quot;: &quot;^9.38&quot;</code> in composer. so &quot;should&quot; be the latest at the time of this post
I honestly have no idea why they think that returning the first record is the right option.. ie
<blockquote>
IMHO the current implementation doesn't make sense for queries including groupBy statements.
Why should be N_1 be the &quot;right&quot; result?
</blockquote>
<blockquote>
Since @taylorotwell rightly pointed out some performance issues with counting the subquery results, why don't we fix that on the php side, by checking if there are any group statements, and if so, performing a N_1 + N_2 + .... + N_M ?
</blockquote>
<a href="https://github.com/laravel/ideas/issues/1693#issuecomment-621167890" rel="nofollow noreferrer">https://github.com/laravel/ideas/issues/1693#issuecomment-621167890</a>
<hr />
Wrapping the query and doing a count seems to work for me
<pre><code>$records = ...

$record_count = DB::table( &quot;fake&quot; );
$record_count-&gt;fromSub($records-&gt;select(DB::raw(1)),&quot;query&quot;);
$record_count-&gt;count();
</code></pre>
<ul>
<li>create a &quot;fake&quot; query builder</li>
<li>add a sub &quot;from&quot; from (...) query set the &quot;sub query&quot; to select 1 instead of returning huge column data. not sure if this is needed but in my mind it seems like a good idea</li>
<li>do the normal -&gt;count()</li>
</ul>
returns the &quot;expected&quot; result since it executes:
<pre><code>select count(*) as aggregate from (select 1 from ... group by ...) as `query`
</code></pre>
The sub is the &quot;query&quot; that <code>$records</code> would normaly execute
<hr />
My use-case is for pagination (not using laravel). so i get the record count then pass it to the paginator then call <code>-&gt;forPage()-&gt;get()</code>
<a href="https://github.com/laravel/framework/issues/44081#issuecomment-1301816710" rel="nofollow noreferrer">https://github.com/laravel/framework/issues/44081#issuecomment-1301816710</a>

blocks|key|2224229|text|这里有一种更多的Laravel处理组的方法，而不需要使用raw语句。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2224230|$sources+=+$sources->where('age','>',+31)->groupBy('age');

$output+=+null;
foreach($sources+as+$key+=>+$source)+{
++++foreach($source+as+$item)+{
++++++++//get+each+item+in+the+group
++++}
++++$output[$key]+=+$source->count();
}|code-block|syntax|javascript|2224231|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a more Laravel way to handle group by without the need to use raw statements. 

<pre><code>$sources = $sources-&gt;where('age','&gt;', 31)-&gt;groupBy('age');

$output = null;
foreach($sources as $key =&gt; $source) {
 foreach($source as $item) {
 //get each item in the group
 }
 $output[$key] = $source-&gt;count();
}
</code></pre>

I have a table that contains, amongst other columns, a column of browser versions. And I simply want to know from the record-set, how many of each type of browser there are. So, I need to end up with something like this: Total Records: 10; Internet Explorer 8: 2; Chrome 25: 4; Firefox 20: 4. (All adding up to 10)

Here's my two pence:

<pre><code>$user_info = Usermeta::groupBy('browser')-&gt;get();
</code></pre>

Of course that just contains the 3 browsers and not the number of each. How can I do this?

Laravel Eloquent groupBy() AND also return count of each group

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个表，其中包含一列浏览器版本。我只想从记录中知道每种浏览器有多少种。因此，我需要这样的结果:总记录: 10；I 8: 2；Chrome 25: 4；Firefox 20: 4。(加起来10)这是我的两便士：$user_info = Usermeta::groupBy('browser')->get();当然，这只包含了3个浏览器，而不是每个浏览器的数量。我该怎么做？

问Laravel雄辩的groupBy()和每组的返回计数
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回答 15

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