Question

我有4个表，我想使用一个查询从这些表中选择信息。

我有一个表名

tbl_marketing

db_maid db_client db_status db_process
4        test      done     ddd

tbl_meeting

db_meetingid db_meetingsubject db_mid
1               test            4

tbl_phonecall

db_phid db_subject db_mid
1       ggg          4
2       fff          4

tbl_email

db_eid  db_email db_mid

1        xxx       4

备注：这是我的4个表的示例。我的表包含更多列 db_mid是tbl_marketing

的ID

我想在不重复的情况下选择所有信息

我使用此查询：

select
      marketing.*,
 meeting.db_meetingid,meeting.db_meetingsubject,meeting.db_sd,meeting.db_dd,meeting.db_duration as meetingDuration,meeting.db_place,meeting.db_mom,meeting.db_momattache,meeting.db_status as meetingStatus,meeting.db_nextmeeting,meeting.db_lastmeeting,meeting.db_attendees,meeting.db_note as meetingNote,meeting.db_mid,
      phonecall.db_phid,phonecall.db_subject as phoneSubject,phonecall.db_desc as phoneDesc,phonecall.db_nextdate,phonecall.db_doc,phonecall.db_duration as phoneDuration,phonecall.db_phstatus,phonecall.db_pnote,phonecall.db_client as phoneClient,phonecall.db_crf,phonecall.db_callto,phonecall.db_phone as phonecallPhone,phonecall.db_logs as phoneLogs,phonecall.db_mid,phonecall.db_dateedit as phoneEdit,phonecall.db_phdate,
   email.db_eid,email.db_edate,email.db_esubject,email.db_edesc,email.db_erf,email.db_emailto,email.db_email as eEmail,email.db_dos,email.db_dor,email.db_estatus,email.db_logs as eLogs,email.db_note as eNote,
   location.db_location as location,
   location.db_lid,
   company.db_company as com,
   company.db_coid,
   subcompany.db_subid,
   subcompany.db_subcompany as sub,
   user.db_uid,
   concat(user.db_fname,' ' ,user.db_lname) as tr,
   subj.db_subjid,
   subj.db_subject as s
      from tbl_marketing as marketing
      left join tbl_location as location
      on
      marketing.db_location=location.db_lid
      left join tbl_meeting as meeting
      on
      marketing.db_maid=meeting.db_mid
      left join tbl_phonecall as phonecall
      on
      marketing.db_maid=phonecall.db_mid
      left join tbl_email as email
      on
      marketing.db_maid=email.db_mid
      left join tbl_company as company
      on
      marketing.db_companyname=company.db_coid
      left join tbl_subcompany as subcompany
      on
      marketing.db_subcompany=subcompany.db_subid
      left join tbl_user as user
      on
      marketing.db_transfered=user.db_uid
      left join tbl_subject as subj
      on
      phonecall.db_subject=subj.db_subjid
      where
      marketing.db_maid='$read'
      and
      meeting.db_mid='$read'
      and
      email.db_mid='$read'
      and
      phonecall.db_mid='$read'

我使用此查询来选择信息，但当我有两行时，示例中的db_mid与tbl_phonecall相同，则信息重复如下：

表格营销

1 test 4

表电话

1 ggg 4

2 fff 4

表格会议

1 test 4

表格电子邮件

1 xxx 4

如上所见，所有表都有重复的值没有phonecall，因为他有2行我想显示没有重复的所有信息

如果我在中途使用群组，我将无法收到tbl_phonecall

中的第二个信息

$rows=array();
while($rows=mysqli_fetch_array($sql)){
$row[]=$rows;}
<table>
foreach($rows as row){
<tr>
<td><?php echo $row['phID'];?></td>
</tr>
} 
</table>

输出

# subject mid

1  test   4

2  ggg    4

不喜欢这个

1,2测试，ggg 4,4

Answer 1

如果您在phonecall表中只有重复项，则以下内容应该有效。如果其他左连接表可以有重复项，只需根据需要添加GROUP_CONCAT。

select
    marketing.*,
    meeting.db_meetingid,
    meeting.db_meetingsubject,
    meeting.db_sd,
    meeting.db_dd,
    meeting.db_duration as meetingDuration,
    meeting.db_place,
    meeting.db_mom,
    meeting.db_momattache,
    meeting.db_status as meetingStatus,
    meeting.db_nextmeeting,
    meeting.db_lastmeeting,
    meeting.db_attendees,
    meeting.db_note as meetingNote,
    meeting.db_mid,
    GROUP_CONCAT(phonecall.db_phid) as phID,
    GROUP_CONCAT(phonecall.db_subject) as phoneSubject,
    GROUP_CONCAT(phonecall.db_desc) as phoneDesc,
    GROUP_CONCAT(phonecall.db_nextdate) as phNextDate,
    GROUP_CONCAT(phonecall.db_doc) as phDoc,
    GROUP_CONCAT(phonecall.db_duration) as phoneDuration,
    GROUP_CONCAT(phonecall.db_phstatus) as phStatus,
    GROUP_CONCAT(phonecall.db_pnote) as phPNote,
    GROUP_CONCAT(phonecall.db_client) as phoneClient,
    GROUP_CONCAT(phonecall.db_crf) as phCrf,
    GROUP_CONCAT(phonecall.db_callto) as phCallTo,
    GROUP_CONCAT(phonecall.db_phone) as phonecallPhone,
    GROUP_CONCAT(phonecall.db_logs) as phoneLogs,
    GROUP_CONCAT(phonecall.db_mid) as phMid,
    GROUP_CONCAT(phonecall.db_dateedit) as phoneEdit,
    GROUP_CONCAT(phonecall.db_phdate) as phdate,
    email.db_eid,
    email.db_edate,
    email.db_esubject,
    email.db_edesc,
    email.db_erf,
    email.db_emailto,
    email.db_email as eEmail,
    email.db_dos,
    email.db_dor,
    email.db_estatus,
    email.db_logs as eLogs,
    email.db_note as eNote,
    location.db_location as location,
    location.db_lid,
    company.db_company as com,
    company.db_coid,
    subcompany.db_subid,
    subcompany.db_subcompany as sub,
    user.db_uid,
    concat(user.db_fname,' ' ,user.db_lname) as tr,
    subj.db_subjid,
    subj.db_subject as s
from tbl_marketing as marketing
left join tbl_location as location
    on marketing.db_location=location.db_lid
left join tbl_meeting as meeting
    on marketing.db_maid=meeting.db_mid
left join tbl_phonecall as phonecall
    on marketing.db_maid=phonecall.db_mid
left join tbl_email as email
    on marketing.db_maid=email.db_mid
left join tbl_company as company
    on marketing.db_companyname=company.db_coid
left join tbl_subcompany as subcompany
    on marketing.db_subcompany=subcompany.db_subid
left join tbl_user as user
    on marketing.db_transfered=user.db_uid
left join tbl_subject as subj
    on phonecall.db_subject=subj.db_subjid
where marketing.db_maid='$read' and
    meeting.db_mid='$read' and
    email.db_mid='$read' and
    phonecall.db_mid='$read'
GROUP BY meeting.db_mid

MySQL连接4个表而不重复

1 个答案: