Mysql 如何在单行中显示具有相似ID的多行记录？

我有两张桌子，用户和工作。工作表有多个用户的工作经验。我必须根据开始日期检索最近的两个公司

条件：

1.如果工作表有多个用户Id的记录，则检索最近的2条记录并显示在单行中，例如user_Id=1

2.如果工作表有一条记录，则该记录是最近的公司记录，第二条记录应为空。用户id=3

    user
    user_id  First_name  Last_name
      1        AAA         BBB
      2        PPP         QQQ
      3        SSS         RRR

    work
    user_id recent_company position start_year end_year
        1        ABC          CCC     2014       2015
        1        PQR          DDD     2013       2014
        1        MNO          EEE     2012       2013
        1        MNO          EEE     0000       0000
        2        XYZ          TTT     2008       2009
        2        IJK          MMM     2005       2008
        3        QRS          ZZZ     2001       2002

我尝试了最新的公司记录1：

select u.user_id,u.first_name,u.last_name,uw.recent_company1,uw.position1,uw.start_year,uw.end_year from muser u
left join
(SELECT user_id,recent_company,position,MAX(start_year) as start_year,end_year 
FROM work 
group by user_id 
order by user_id) uw ON u.user_id =uw.user_id

根据起始年份，我对最近第一家公司的查询结果：

user_id  First_name  Last_name     recent_company1   
  1        AAA         BBB            ABC                          
  2        PPP         QQQ            XYZ                           
  3        SSS         RRR            QRS             

position1    start_year1    end_year1
   CCC         2014           2015
   TTT         2008           2009
   ZZZ         2001           2002

user_id  First_name  Last_name   recent_company1   position1    
  1        AAA         BBB          ABC             CCC                 
  2        PPP         QQQ          XYZ             TTT                
  3        SSS         RRR          QRS             ZZZ                 

start_year1    end_year1
    2014         2015
    2008         2009
    2001         2002

   recent_company2   position2   start_year2    end_year2
      PQR               DDD         2013          2014
      IJK               MMM         2005          2008
      NULL              NULL        NULL          NULL

最近第一家和第二家公司在起始年的所需产量：

user_id  First_name  Last_name     recent_company1   
  1        AAA         BBB            ABC                          
  2        PPP         QQQ            XYZ                           
  3        SSS         RRR            QRS             

position1    start_year1    end_year1
   CCC         2014           2015
   TTT         2008           2009
   ZZZ         2001           2002

user_id  First_name  Last_name   recent_company1   position1    
  1        AAA         BBB          ABC             CCC                 
  2        PPP         QQQ          XYZ             TTT                
  3        SSS         RRR          QRS             ZZZ                 

start_year1    end_year1
    2014         2015
    2008         2009
    2001         2002

   recent_company2   position2   start_year2    end_year2
      PQR               DDD         2013          2014
      IJK               MMM         2005          2008
      NULL              NULL        NULL          NULL

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我用php脚本做了同样的事情。 我也在用sql查询寻找答案。。这样就可以在sql查询中完成

  function tag_listing() {
    echo 'Row Number' . ',' . 'TagID' . ',' . 'Status' . ',' . 'UserName' . ',' . 'Country' . ',' . 'PostCode' . ',';
    $array = array('First', 'Second', 'Third', 'Fourth', 'Fifth', 'Sixth', 'Seventh', 'Eighth', 'Ninth', 'Tenth');
    $entityName = "Allocation Entity";
    $entity_type = "Entity Type";
    $allocDate = "Allocation Date";
    $sqlCount = "SELECT count(*) as total from tag_allocation  group by id_tag HAVING COUNT(*) > 1
                 ORDER BY COUNT(*) DESC";
    $countQuery = $this->db->query($sqlCount);
    foreach ($countQuery->result() as $NewTotal) {
        $resRows = $NewTotal->total;
        $item = array_slice($array, 0, $resRows);
        foreach ($item as $items) {
            echo $items . ' ' . $entityName . ',';
            echo $items . ' ' . $entity_type . ',';
            echo $items . ' ' . $allocDate . ',';
        }
        break;
    }
    $query = $this->db->query("select  id_tag_id, IT.id_tag, ITS.status,
    CASE WHEN firstname IS NULL THEN '' ELSE CONCAT(firstname,' ',lastname) END as Username,
    CASE WHEN U.country IS NULL THEN '' ELSE U.country END as Country,
    CASE WHEN U.postcode IS NULL THEN '' ELSE U.postcode END as postcode,
    CASE WHEN E.entity_name IS NULL THEN '' ELSE E.entity_name END as entity_name,
    ET.entity_type, TA.allocation_date, TA.allocation_type from id_tag IT
    INNER JOIN id_tag_status ITS ON IT.id_tag_status_id = ITS.id_tag_status_id
    RIGHT JOIN tag_allocation TA ON IT.id_tag = TA.id_tag
    INNER JOIN entity_type ET ON ET.entity_type_id = TA.allocation_type
    LEFT JOIN allocation_entity E ON E.entity_id = TA.entity_id
    LEFT JOIN user U ON U.user_id = IT.user_id
    order by IT.id_tag, TA.allocation_date asc        ");
    $IDTAG = '';
    $rowCount = 0;
    foreach ($query->result() as $row) {
        $newID = $row->id_tag;
        if ($IDTAG != $newID) {
            echo "\r\n";
        }
        if ($IDTAG == $newID) {
            echo $entityName = $row->entity_name . ',' . $entityType = $row->entity_type . ',' . $allocationDate = $row->allocation_date . ',';
        } else {
            $rowCount = $rowCount + 1;
            $IDTAG = $row->id_tag;
            echo $rowCount . ',' . $idTag = $row->id_tag . ',' . $status = $row->status . ',' . $Username = $row->Username . ',' . $Country = $row->Country . ',' . $postcode = $row->postcode . ',' . $entityName = $row->entity_name . ',' . $entityType = $row->entity_type . ',' . $allocationDate = $row->allocation_date . ',';
        }
    }
    exit();
}

希望能有帮助。。试试这个，你就会得到它。。

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谢谢…

是否有可能重复结果？