简介：
我正在尝试合并2个表recipes和posts，然后将->paginate(5)添加到查询中。
但由于某种原因，我收到了这个错误：
基数违规：1222使用的SELECT语句有一个 不同的列数（SQL :(选择count（*）作为聚合来自
posts
代码：
$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
->where("user_id", "=", $id)
->union($recipes)
->paginate(5)->get();
我做错了吗？
如果没有->paginate(5)，则查询工作正常。
答案 0 :(得分：8)
你是对的，分页导致问题。现在，您可以创建视图并查询视图而不是实际的表格，或手动创建Paginator：
$page = Input::get('page', 1);
$paginate = 5;
$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
->where("user_id", "=", $id)
->union($recipes)
->get();
$slice = array_slice($items->toArray(), $paginate * ($page - 1), $paginate);
$result = Paginator::make($slice, count($items), $paginate);
return View::make('yourView',compact('result'));
答案 1 :(得分：5)
我已经遇到过这种问题。我发现一个帖子也不是关于pagination而是关于unions。
请参阅此链接：Sorting UNION queries with Laravel 4.1
@Mohamed Azher分享了一个很好的技巧，它适用于我的问题。
$query = $query1->union($query2);
$querySql = $query->toSql();
$query = DB::table(DB::raw("($querySql order by foo desc) as a"))->mergeBindings($query);
这会创建一个如下所示的SQL：
select * from (
(select a as foo from foo)
union
(select b as foo from bar)
) as a order by foo desc;
你可以像往常一样使用Laravel paginate $query->paginate(5)。var http = require('http');
var fs = require('fs');
var mysql = require('mysql');
var url = require('url');
var mime = require('mime');
var config = JSON.parse(fs.readFileSync('config.json'));
var host = config.host;
var port = config.port;
var connection = mysql.createConnection({
host : 'localhost',
user : 'root',
password : 'root',
database : 'innovation_one'
});
function connectToDb(){
connection.connect(function(err){
if (err){
console.log('error: ' + err.stack);
return;
}
return console.log('Connected established!');
});
}
var server = http.createServer(function(request,response){
var parsed = url.parse(request.url);
var mimetypeLookup = mime.lookup(request.url);
if(request.method == "POST") {
// POST
} else if (request.method == "GET") {
// GET
if (request.url == '/get-servers/'){
connectToDb();
connection.query('SELECT * FROM servers', function(err,rows,fields) {
if (err) {
console.log(err);
return
}
var data = [];
for( i = 0 ; i < rows.length ; i++){
data.push(rows[i].name);
data.push(rows[i].client);
data.push(rows[i].type);
data.push(rows[i].host);
data.push(rows[i].ssh);
data.push(rows[i].mysql);
}
response.writeHead(200, {'Content-Type': 'text/html'});
response.end(JSON.stringify(data,fields));
});
}
}
}).listen(port, host);
console.log('Server running at http://127.0.0.1:4114/');
。 （但你必须稍微分叉以适应你的问题）
答案 2 :(得分：3)
订购
$page = Input::get('page', 1);
$paginate = 5;
$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
->where("user_id", "=", $id)
->union($recipes)
->orderBy('created_at','desc')
->get();
$slice = array_slice($items, $paginate * ($page - 1), $paginate);
$result = Paginator::make($slice, count($items), $paginate);
return View::make('yourView',compact('result'))->with( 'result', $result );
查看页面：
@foreach($result as $data)
{{ $data->your_column_name;}}
@endforeach
{{$result->links();}} //for pagination
它对更多人的帮助..因为没有人无法理解显示数据 查看页面联合与分页和命令..谢谢你
答案 3 :(得分：2)
使用雄辩
我改编了jdme的answer以便与Eloquent一起使用。我创建了一个类，扩展了默认的Eloquent Builder，并覆盖了union方法来解决分页问题。
创建app\Builder\BuilderWithFixes.php：
<?php
namespace App\Builder;
use Illuminate\Database\Eloquent\Builder;
class BuilderWithFixes extends Builder
{
/**
* Add a union statement to the query.
*
* @param \Illuminate\Database\Query\Builder|\Closure $query
* @param bool $all
* @return \Illuminate\Database\Query\Builder|static
*/
public function union($query, $all = false)
{
$query = parent::union($query, $all);
$querySql = $query->toSql();
return $this->model->from(\DB::raw("($querySql) as ".$this->model->table))->select($this->model->table.'.*')->addBinding($this->getBindings());
}
}
在您的模型中（例如app\Post.php），包括下面的方法newEloquentBuilder，用\App\Builder\BuilderWithFixes替换默认的Eloquent Builder：
<?php
namespace App;
use Eloquent as Model;
class Post extends Model
{
// your model stuffs...
public function newEloquentBuilder($query)
{
return new \App\Builder\BuilderWithFixes($query);
}
}
现在，您可以正常地在模型中同时使用union + paginate（在这种情况下为Post），例如：
$recipes = Recipe::select("id", "title", "user_id", "description", "created_at")
->where("user_id", "=", $id);
$items = Post::select("id", "title", "user_id", "content", "created_at")
->where("user_id", "=", $id)
->union($recipes)
->paginate(5);
答案 4 :(得分：1)
我遇到了同样的问题，很遗憾我无法获得与{{ $result->links() }}的页面链接，但我找到了另一种编写分页部分并显示页面链接的方法
Custom data pagination with Laravel 5
//Create a new Laravel collection from the array data
$collection = new Collection($searchResults);
//Define how many items we want to be visible in each page
$perPage = 5;
//Slice the collection to get the items to display in current page
$currentPageSearchResults = $collection->slice($currentPage * $perPage, $perPage)->all();
//Create our paginator and pass it to the view
$paginatedSearchResults= new LengthAwarePaginator($currentPageSearchResults, count($collection), $perPage);
return view('search', ['results' => $paginatedSearchResults]);
答案 5 :(得分：1)
使用Illuminate\Database\Query\Builder中更优雅的方法重申jdme answer。
$recipes = DB::table("recipes") ..
$items = DB::table("posts")->union($recipes) ..
$query = DB::query()
->fromSub($items, "some_query_name");
// Let's paginate!
$query->paginate(5);
我希望这有帮助！
答案 6 :(得分：1)
对于分页收集，请执行以下操作：
将此添加到\ app \ Providers \ AppServiceProvider中的启动功能
/**
* Paginate a standard Laravel Collection.
*
* @param int $perPage
* @param int $total
* @param int $page
* @param string $pageName
* @return array
*/
Collection::macro('paginate', function($perPage, $total = null, $page = null, $pageName = 'page') {
$page = $page ?: LengthAwarePaginator::resolveCurrentPage($pageName);
return new LengthAwarePaginator(
$this->forPage($page, $perPage),
$total ?: $this->count(),
$perPage,
$page,
[
'path' => LengthAwarePaginator::resolveCurrentPath(),
'pageName' => $pageName,
]
);
});
此后，对于所有集合，您都可以像您的代码一样进行分页
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at")
->where("user_id", "=", $id)
->union($recipes)
->paginate(5)
答案 7 :(得分：0)
我知道这个答案为时已晚。但是我想分享我的问题和解决方案。
※这是样品。 MariaDB，约146,000条记录。
SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM customers A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
UNION ALL
SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM employees A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
Reference from www.tech-corgi.com（やり方２），我更新了PHP代码以在查询中进行过滤，然后正常调用分页。
在获取大记录之前，我必须添加一个条件（过滤器）。在此示例中为 organization_id 。
$query = "
SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM customers A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
WHERE 1 = 1
AND B.organization_id = {ORGANIZATION_ID}
UNION ALL
SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM employees A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
WHERE 1 = 1
AND B.organization_id = {ORGANIZATION_ID}
";
$organization_id = request()->organization_id;
$query = str_replace("{ORGANIZATION_ID}", $organization_id, $query);
但是它仍然不能在 paginate（）中使用。有一个技巧可以解决这个问题。见下文。
技巧：将查询放入()中。例如：(SELECT * FROM TABLE_A)。
原因： paginage（）将生成并运行计数查询SELECT count(*) FROM (SELECT * FROM TABLE_A)，如果我们没有将放在方括号内，则计数查询将不是正确的查询
$query = "
( SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM customers A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
WHERE 1 = 1
AND B.organization_id = {ORGANIZATION_ID}
UNION ALL
SELECT A.a_id
, A.a_name
, B.organization_id
, B.organization_name
FROM employees A
LEFT JOIN organizations B ON (A.organization_id = B.organization_id)
WHERE 1 = 1
AND B.organization_id = {ORGANIZATION_ID}
) AS VIEW_RESULT
";
$organization_id = request()->organization_id;
$query = str_replace("{ORGANIZATION_ID}", $organization_id, $query);
$resultSet = DB::table(DB::raw($query))->paginate(20);
现在我可以正常使用了：
希望对您有帮助！
答案 8 :(得分：0)
可接受的答案非常适合查询生成器。
但是这是我的Laravel Eloquent Builder方法。
假设我们指的是同一型号
$q1 = Model::createByMe(); // some condition
$q2 = Model::createByMyFriend(); // another condition
$q2->union($q1);
$querySql = $q2->toSql();
$query = Model::from(DB::raw("($querySql) as a"))->select('a.*')->addBinding($q2->getBindings());
$paginated_data = $query->paginate();
我正在使用Laravel 5.6
答案 9 :(得分：0)
获取分页总数是这里的问题。这是我在使用$builder->paginate()
"SQLSTATE[21000]: Cardinality violation: 1222 The used SELECT statements have a different number of columns (SQL: (select count(*) as aggregate from `institute_category_places` where `status` = approved and (`category_id` in (3) or `name` LIKE %dancing class% or `description` LIKE %dancing class% or `address_line1` LIKE %dancing class% or `address_line2` LIKE %dancing class% or `city` LIKE %dancing class% or `province` LIKE %dancing class% or `country` LIKE %dancing class%) and `institute_category_places`.`deleted_at` is null) union (select * from `institute_category_places` where `status` = approved and (`category_id` in (3, 4) or `name` LIKE %dancing% or `description` LIKE %dancing% or `address_line1` LIKE %dancing% or `address_line2` LIKE %dancing% or `city` LIKE %dancing% or `province` LIKE %dancing% or `country` LIKE %dancing% or `name` LIKE %class% or `description` LIKE %class% or `address_line1` LIKE %class% or `address_line2` LIKE %class% or `city` LIKE %class% or `province` LIKE %class% or `country` LIKE %class%) and `institute_category_places`.`deleted_at` is null))"
如果您要在没有总数的情况下分页，可以使用
$builder->limit($per_page)->offset($per_page * ($page - 1))->get();
仅获取页面中的行集。
获取所有行并计算总数会降低内存效率。所以我用下面的方法来获取总数。
$bindings = $query_builder->getBindings();
$sql = $query_builder->toSql();
foreach ($bindings as $binding) {
$value = is_numeric($binding) ? $binding : "'" . $binding . "'";
$sql = preg_replace('/\?/', $value, $sql, 1);
}
$sql = str_replace('\\', '\\\\', $sql);
$total = DB::select(DB::raw("select count(*) as total_count from ($sql) as count_table"));
然后我们必须手动对结果进行分页。
$page = Input::get('page', 1);
$per_page = 15;
$search_results = $query_builder->limit($per_page)->offset($per_page * ($page - 1))->get();
$result = new LengthAwarePaginator($search_results, $total[0]->total_count, $per_page, $page, ['path' => $request->url()]);
如果可以使用原始sql查询，则CPU和内存效率更高。
答案 10 :(得分：0)
对于那些仍在寻找答案的人，我已经尝试union和paginate并在laravel 5.7.20 下获得了正确的结果 >。这将比合并集合然后进行分页更好，后者将无法处理大量数据。
一些演示代码（在我的情况下，我将处理具有相同表名的多个数据库）：
$dbs=["db_name1","db_name2"];
$query=DB::table("$dbs[0].table_name");
for($i=1;$i<count($log_dbs);$i++){
$query=DB::table("$dbs[$i].table_name")->union($query);
}
$query=$query->orderBy('id','desc')->paginate(50);
我还没有尝试过其他更高版本的laravel。但至少它现在可以正常工作！
更多信息
我的laravel的先前版本是5.7.9，它将报告Cardinality violation错误。因此，laravel团队在5.7.x的某些版本中解决了此问题。
答案 11 :(得分：-1)
$page = Input::get('page', 1);
$paginate = 5;
$recipes = DB::table("recipes")->select("id", "title", "user_id", "description", "created_at")
->where("user_id", "=", $id);
$items = DB::table("posts")->select("id", "title", "user_id", "content", "created_at") ->where("user_id", "=", $id)->union($recipes)->get()->toArray();
$slice = array_slice($items, $paginate * ($page - 1), $paginate);
$result = new Paginator($slice , $paginate);