PHP MySqli上的重复数据内部连接3表 [英] Duplicate Data Inner Join 3 Tables on PHP MySqli

问题描述

我在php中使用mysqli ext程序样式显示3个不同表的数据时出现重复问题。我的桌子是：



分类：id_kat，nama_kat



subkategori：id_sub，id_kat，nama_sub



supersubkategori：id_supersub，id_sub，id_kat，nama_supersub，kategori_seo



这是我的代码：

 <？  php   
   
           require    ../ config.php;   
   
           $ sql     =  SELECT * FROM supersubkategori INNER JOIN kategori on kategori.id_kat = supersubkategori.id_kat 
 INNER JOIN subkategori ON subkategori.id_kat = supersubkategori.id_kat ORDER BY id_supersub ASC ;   
   
           $ result     =   mysqli_query（$ conn，  $ sql）;   
   
           if    （mysqli_num_rows（$ result）   >  0）
 {
 while（$ data = mysqli_fetch_array（$ result））
 {
 echo< span class =code-keyword><   tr     class   =   \ 表    table-stripedd \  >  
 <   td     align   =   \  left \ > 。$ data ['id_supersub']。<   / td  >  
 <   td    对齐  =   \  left \  > 。$ data ['nama_supersub']。<   / td  >  
 <   td     align   =   \  left\ > 。$ data ['nama_kat']。<   / td  >  
  <   td     align   =   \  left \ > 。$ data ['nama_sub']。<   / td  >  
 <   td     align   =   \  left\ >  
 <   a     href   =   \  supersub_ubah.php？id_supersub   =   $ data [id_supersub] \ > 编辑<   / a  >  
 <   a     href   =   \  supersub_hapus.php？id_supersub   =   $ data [id_supersub] \ >  Hapus < span class =code-keyword><   / a  >  
 <   / td  >  
 <   / tr  > ; 
} 
} 
其他
 {
 echoBelum ada data。; 
} 
？> 

解决方案

sql = SELECT * FROM supersubkategori INNER JOIN kategori ON kategori.id_kat = supersubkategori.id_kat
INNER JOIN subkategori ON subkategori.id_kat = supersubkategori.id_kat ORDER BY id_supersub ASC ;

< span class =code-summarycomment>

result = mysqli_query（

conn，

I have duplicate problem when showing data with 3 different table in php with mysqli ext procedural style. My tables are:

kategori: id_kat, nama_kat

subkategori: id_sub, id_kat, nama_sub

supersubkategori: id_supersub, id_sub, id_kat, nama_supersub, kategori_seo

This is my code:

<?php

    require "../config.php";

    $sql = "SELECT * FROM supersubkategori INNER JOIN kategori ON kategori.id_kat = supersubkategori.id_kat
                        INNER JOIN subkategori ON subkategori.id_kat = supersubkategori.id_kat ORDER BY id_supersub ASC";

    $result = mysqli_query($conn, $sql);

    if (mysqli_num_rows($result) > 0)
    {
        while ($data = mysqli_fetch_array($result))
        {
            echo "<tr class=\"table table-striped\">
                            <td align=\"left\">".$data['id_supersub']."</td>
                            <td align=\"left\">".$data['nama_supersub']."</td>
                            <td align=\"left\">".$data['nama_kat']."</td>
                            <td align=\"left\">".$data['nama_sub']."</td>
                            <td align=\"left\">
                                <a href=\"supersub_ubah.php?id_supersub=$data[id_supersub]\">Edit</a>
                                <a href=\"supersub_hapus.php?id_supersub=$data[id_supersub]\">Hapus</a>
                            </td>
                        </tr>";
        }
    }
        else
        {
            echo "Belum ada data.";
        }
    ?>

解决方案

sql = "SELECT * FROM supersubkategori INNER JOIN kategori ON kategori.id_kat = supersubkategori.id_kat INNER JOIN subkategori ON subkategori.id_kat = supersubkategori.id_kat ORDER BY id_supersub ASC";

result = mysqli_query(

conn,

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