如何在服务器端处理模式下使用 JOIN 进行数据库查询
Posted
技术标签:
【中文标题】如何在服务器端处理模式下使用 JOIN 进行数据库查询【英文标题】:How to use database query with JOIN in server-side processing mode 【发布时间】:2015-12-30 18:13:47 【问题描述】:我正在为我的视图列表使用 jQuery DataTables。我使用了服务器端处理模式,该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做到这一点。
如果我的代码不做太多更改,如何使用多个带有JOIN
的表进行自定义查询?
所以我有这个:
HTML
<table id="CustomerList" class="table table-striped table-bordered" cellspacing="0" >
<thead>
<tr>
<th colspan="7"> <center>Customer Information<center></th>
<th colspan="1"> <center>Actions<center></th>
</tr>
<tr>
<th>ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Gender</th>
<th>Phone Number</th>
<th>Country</th>
<th>Postcode</th>
<th>Edit</th>
<!-- <th>Edit</th>
<th>Delete</th> -->
</tr>
</thead>
<tbody>
</tbody>
</table>
阿贾克斯
<script type="text/javascript">
$(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 '<a href="/qms/public/customer/' + row[0] + '/edit">Edit</a>';
return "<a class='btn btn-small btn-info' href='<?php echo URL::to('customer').'/';?>"+row[0]+"/edit'><span class='glyphicon glyphicon glyphicon-edit' aria-hidden='true'></span></a>";
]
);
var tt = new $.fn.dataTable.TableTools( $('#CustomerList').DataTable() );
$( tt.fnContainer() ).insertBefore('div.dataTables_wrapper');
);
控制器
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 ; $i<count($aColumns) ; $i++ )
if ( $_GET['bSearchable_'.$i] == "true" && $_GET['sSearch_'.$i] != '' )
if ( $sWhere == "" )
$sWhere = "WHERE ";
else
$sWhere .= " AND ";
$sWhere .= $aColumns[$i]." ILIKE '%".pg_escape_string($_GET['sSearch_'.$i])."%' ";
$sQuery = "
SELECT ".str_replace(" , ", " ", implode(", ", $aColumns))."
FROM $sTable
$sWhere
$sOrder
$sLimit
";
$rResult = pg_query( $gaSql['link'], $sQuery ) or die(pg_last_error());
$sQuery = "
SELECT $sIndexColumn
FROM $sTable
";
$rResultTotal = pg_query( $gaSql['link'], $sQuery ) or die(pg_last_error());
$iTotal = pg_num_rows($rResultTotal);
pg_free_result( $rResultTotal );
if ( $sWhere != "" )
$sQuery = "
SELECT $sIndexColumn
FROM $sTable
$sWhere
";
$rResultFilterTotal = pg_query( $gaSql['link'], $sQuery ) or die(pg_last_error());
$iFilteredTotal = pg_num_rows($rResultFilterTotal);
pg_free_result( $rResultFilterTotal );
else
$iFilteredTotal = $iTotal;
/*
* Output
*/
$output = array(
"sEcho" => intval($_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
,其中包含我想在表格customers
中获取的表格列
如果我想要一个自定义查询来获取如下数据怎么办:
$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。
【问题讨论】:
您是在询问分页数据吗?我对 PHP 或 Laravel 不熟悉,但通过 Google 快速发现:https://laracasts.com/discuss/channels/general-discussion/jquery-datatables-and-laravel-server-side-implementation 【参考方案1】:我知道这篇文章已经很老了,但它给了我线索。我遇到了同样的挑战,偶然发现了@Óscar Sánchez 的回应。
仅针对可能遇到相同问题并需要帮助的任何人,这就是我所做的。
-
首先我创建了一个视图
CREATE VIEW economic AS select tbl_pay.idno, tbl_pay.name, tbl_pay.amount, tbl_pay.cat, tbl_pay.item, tbl_pay.code, tbl_rev.t_id, tbl_sect.s_name, tbl_pay.bnk, DATE_FORMAT(tbl_pay.paydate, '%d %M %Y') AS PayDate, tbl_pay.reno, tbl_pay.slip, tbl_pay.cStatus, tbl_pay.cvdate, YEAR(paydate) AS PayYr
FROM tbl_pay
INNER JOIN tbl_rev ON tbl_pay.cat = tbl_rev.revid
INNER JOIN tbl_sect ON tbl_rev.t_id = tbl_sect.t_id
WHERE tbl_rev.t_id = 1 AND YEAR(paydate)=YEAR(NOW()) ORDER BY tbl_pay.cat, tbl_pay.name
-
创建名为“经济”的视图后,我现在只在服务器端文件中添加了该表名,如下所示:
<?php
include('dd_connection.php');
$column = array('code', 'cat', 'name', 'item', 'amount', 'bnk', 'reno', 'cvdate', 'PayDate');
$query = "
SELECT * FROM economic
";
if(isset($_POST['filter_cat']) && $_POST['filter_cat'] != '')
$query .= '
WHERE cat = "'.$_POST['filter_cat'].'"
';
if(isset($_POST['order']))
$query .= 'ORDER BY '.$column[$_POST['order']['0']['column']].' '.$_POST['order']['0']['dir'].' ';
else
$query .= 'ORDER BY name ASC ';
and other lines of code below....
现在根据您的 mysql 版本,在创建视图时,您可能需要将视图名称保留为 经济,不带标点符号,或者像这样添加打开和关闭标点符号 '经济'。
现在,当您在线上传服务器端文件时,不要忘记重复创建在线视图的相同过程,否则您的页面将不显示任何记录。
【讨论】:
【参考方案2】:您需要创建一个包含关系的视图。
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
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";
【讨论】:
【参考方案3】:在不过多修改代码的情况下使用JOIN
有一个技巧。
改变这一行:
$sTable = "customers";
到:
$sTable =
"(
SELECT a.id AS crmid, b.name
FROM forms a
INNER JOIN users b ON a.agent_id = b.id
) table";
我只是为了代码清晰起见简化了上面的查询。只需确保所有列名都是唯一的,否则在需要时使用别名。
然后在$aColumns
变量中使用列名/别名。对于上面的查询,它将是
$aColumns = array('crmid', 'name');
【讨论】:
嗨。我尝试了您的建议,但出现错误,查询失败:错误:FROM 中的子查询必须具有别名 LINE 3: FROM ( ^ HINT: For example, FROM (SELECT ...) [AS] foo. 这指向我行 $rResult = pg_query( $gaSql['link'], $sQuery ) or die(pg_last_error()); 非虚拟机。解决了这个问题。谢谢! 我必须通过将“$table
”替换为“$table”来编辑库中的 ssp.class.php 文件。它现在按预期工作。
这可能已经快 6 年了,但在 2021 年完美运行!谢谢!以上是关于如何在服务器端处理模式下使用 JOIN 进行数据库查询的主要内容,如果未能解决你的问题,请参考以下文章
如何在不使用 Join 的情况下处理引用其他表的相关子查询的问题