如何在服务器端处理模式下使用JOIN进行数据库查询

我在我的视图列表中使用jQuery DataTables。 我使用服务器端处理模式，特别适用于大型数据集。 但我的问题是我只能使用单个数据库表来执行此操作。

如何使用自定义查询使用多个表和JOIN而不会改变太多是我的代码？

所以我有这个：

HTML

  <!--  --> 
 
 
 
 Customer Information
 
 Actions
 
 
 
ID
 
First Name
 
Last Name
 
Gender
 
Phone Number
 
Country
 
Postcode
 
Edit
Edit
 
Delete
 
 
 
  

阿贾克斯

  $(document).ready(function() { $.fn.dataTable.ext.legacy.ajax = true; var table = $('#CustomerList').DataTable( { "processing": true, "serverSide": true, "ajax": "api/customer/all", "columnDefs": [ { "targets": 7, "render": function(data, type, row, meta){ // return 'Edit'; return "<a class='btn btn-small btn-info' href='"+row[0]+"/edit'>"; } } ] }); var tt = new $.fn.dataTable.TableTools( $('#CustomerList').DataTable() ); $( tt.fnContainer() ).insertBefore('div.dataTables_wrapper'); }); 

调节器

 public function apiGetCustomers() { /*=================================================================*/ /* * Script: DataTables server-side script for PHP and PostgreSQL * Copyright: 2010 - Allan Jardine * License: GPL v2 or BSD (3-point) */ /* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Easy set variables */ /* Array of database columns which should be read and sent back to DataTables. Use a space where * you want to insert a non-database field (for example a counter or static image) */ $aColumns = array('id', 'firstname', 'lastname', 'gender', 'phone_num', 'country', 'postcode' ); /* Indexed column (used for fast and accurate table cardinality) */ $sIndexColumn = "phone_num"; /* DB table to use */ $sTable = "customers"; /* Database connection information */ $gaSql['user'] = "postgres"; $gaSql['password'] = "postgres"; $gaSql['db'] = "qms"; $gaSql['server'] = "localhost"; /* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * If you just want to use the basic configuration for DataTables with PHP server-side, there is * no need to edit below this line */ /* * DB connection */ $gaSql['link'] = pg_connect( " host=".$gaSql['server']. " dbname=".$gaSql['db']. " user=".$gaSql['user']. " password=".$gaSql['password'] ) or die('Could not connect: ' . pg_last_error()); /* * Paging */ $sLimit = ""; if ( isset( $_GET['iDisplayStart'] ) && $_GET['iDisplayLength'] != '-1' ) { $sLimit = "LIMIT ".intval( $_GET['iDisplayLength'] )." OFFSET ". intval( $_GET['iDisplayStart'] ); } /* * Ordering */ if ( isset( $_GET['iSortCol_0'] ) ) { $sOrder = "ORDER BY "; for ( $i=0 ; $i<intval( $_GET['iSortingCols'] ) ; $i++ ) { if ( $_GET[ 'bSortable_'.intval($_GET['iSortCol_'.$i]) ] == "true" ) { $sOrder .= $aColumns[ intval( $_GET['iSortCol_'.$i] ) ]." ".($_GET['sSortDir_'.$i]==='asc' ? 'asc' : 'desc').", "; } } $sOrder = substr_replace( $sOrder, "", -2 ); if ( $sOrder == "ORDER BY" ) { $sOrder = ""; } } /* * Filtering * NOTE This assumes that the field that is being searched on is a string typed field (ie. one * on which ILIKE can be used). Boolean fields etc will need a modification here. */ $sWhere = ""; if ( $_GET['sSearch'] != "" ) { $sWhere = "WHERE ("; for ( $i=0 ; $i<count($aColumns) ; $i++ ) { if ( $_GET['bSearchable_'.$i] == "true" ) { if($aColumns[$i] != 'id') // Exclude ID for filtering { $sWhere .= $aColumns[$i]." ILIKE '%".pg_escape_string( $_GET['sSearch'] )."%' OR "; } } } $sWhere = substr_replace( $sWhere, "", -3 ); $sWhere .= ")"; } /* Individual column filtering */ for ( $i=0 ; $i intval($_GET['sEcho']), "iTotalRecords" => $iTotal, "iTotalDisplayRecords" => $iFilteredTotal, "aaData" => array() ); while ( $aRow = pg_fetch_array($rResult, null, PGSQL_ASSOC) ) { $row = array(); for ( $i=0 ; $i<count($aColumns) ; $i++ ) { if ( $aColumns[$i] == "version" ) { /* Special output formatting for 'version' column */ $row[] = ($aRow[ $aColumns[$i] ]=="0") ? '-' : $aRow[ $aColumns[$i] ]; } else if ( $aColumns[$i] != ' ' ) { /* General output */ $row[] = $aRow[ $aColumns[$i] ]; } } $output['aaData'][] = $row; } echo json_encode( $output ); // Free resultset pg_free_result( $rResult ); // Closing connection pg_close( $gaSql['link'] ); } 

在我的控制器中，您可以看到$aColumns ，其中包含我希望在表customers获得的表列

如果我想要一个自定义查询来获取如下数据，该怎么办？

 $query = "SELECT a.id as crmid, b.name, a.title, a.firstname, a.surname, a.disposition, a.gross, a.created_at, a.phone_num FROM forms a INNER JOIN users b ON a.agent_id = b.id;"; 

所以我有内部联接而不是只有一个表。

Answer 1

有一个技巧可以使用JOIN而不会过多地修改代码。

改变这一行：

 $sTable = "customers"; 

至：

 $sTable = "( SELECT a.id AS crmid, b.name FROM forms a INNER JOIN users b ON a.agent_id = b.id ) table"; 

我为了代码清晰起见简化了上面的查询。 只需确保所有列名都是唯一的，否则请在需要时使用别名。

然后在$aColumns变量中使用列名/别名。 对于上面的查询，它将是

 $aColumns = array('crmid', 'name'); 

Answer 2

您需要使用关系创建视图。

1. 在MySql中创建VIEW：

    CREATE VIEW 'the_view' AS SELECT a.id, a.num_factura_proveedor, b.nombre_comercial FROM compras a INNER JOIN terceros b ON a.tercero_id = b.id 

2. 在服务器端脚本：

    $sTable = "the_view"; //the columns of the view $aColumns = array('id' , 'num_factura_proveedor', 'nombre_comercial'); $sIndexColumn = "id";