2的foreach與MySQL查詢

Question

我有2個MySQL表工具欄和subtoolbar

工具欄

| id | toolbar | status | 

| 1 | Usuarios | ACTIVE | 

| 2 | Bancas | ACTIVE | 

| 3 | Balance | ACTIVE | 

subtoolbar循環

| id | subtoolbar | toolbarid | status | 


| 1 | Editar  | 1   | ACTIVE | 

| 2 | Agregar | 1   | ACTIVE | 

| 3 | Bancas  | 2   | ACTIVE | 

| 4 | Listeros | 2   | ACTIVE | 

| 5 | Agencias | 2   | ACTIVE | 

| 6 | Taquillas | 2   | ACTIVE | 

| 7 | Layout  | 3   | ACTIVE | 

我想列出每個報告其報表，所以我嘗試了以下

$res=mysql_query("select * FROM toolbar"); 

$num_rows = (mysql_num_rows($res) -1); 

while ($x = mysql_fetch_assoc($res)){ 

    $toolid[] = $x['id']; 

    $toolbar[] = $x['toolbar']; 

    $toolstatus[] = $x['status']; 

} 

在PHP

foreach (range(0, $num_rows) as $number) { 

$res1=mysql_query("select * FROM subtoolbar where 

toolbarid='$toolid[$number]'"); 

$num_rows2 = (mysql_num_rows($res1) -1); 

while ($y = mysql_fetch_assoc($res1)){ 

    $subtoolid[] = $y['id']; 

    $subtoolbar[] = $y['subtoolbar']; 

    $toolbarid[] = $y['toolbarid']; 

    $subtoolstatus[] = $y['status']; 

} 


echo $toolbar[$number]; 

foreach (range(0, $num_rows2) as $number2) { 

echo $subtoolbar[$number2]; 

}} 

，結果走出這樣

USUARIOS

• Editar

• Agregar

Bancas

• Editar

• Agregar

• Bancas

• Listeros

餘額

• Editar

Answer 1

請試試這個方法： 但是一定要記住：

1. 必須使用mysqli_ （），而不是mysql_（）

2. 檢查數組是否不爲空&值是否設置。

   $res=mysql_query("select t.id,t.toolbar,st.* FROM toolbar t left join 
       subtoolbar st on t.id=toolbarid where t.status='ACTIVE' and 
       st.status='ACTIVE'"); 
   
   if(mysql_num_rows($res)>0) 
   { 
       while ($rs = mysql_fetch_assoc($res)){ 
        $retArr[$rs['toolbarid']][] = array("toolbar"=>$rs["subtoolbar"], "subtoolbar"=>$rs["subtoolbar"]); 
       } 
   } 
   
   foreach($retArr as $rs) 
   { 
   $cnt=0; 
   
       foreach($rs as $toolbarName => $data) 
       { 
       if($cnt==0) 
       { 
        echo $data["toolbar"]."<br>"; 
       } 
       $cnt++; 
       echo $data["subtoolbar"]."<br>"; 
   
       } 
   
   } 
   

Answer 2

我分享我自己的代碼，在正常工作我的定點

<div id="navbar" class="navbar-collapse collapse"> 

          <ul class="nav navbar-nav"> 
          <?php 
          $Clsmenu = new Menus; 
          $menu_res = mysql_query("SELECT * FROM `nav_menus`");(); 
          while($menu_details=mysql_fetch_array($menu_res)) 
          {        
           $sub_menu_query=mysql_query("SELECT * FROM `category` WHERE `Menu_id`='$menu_details['id']'"); 
           if(mysql_num_rows($sub_menu_query)) 
           {?> 
           <li class="dropdown"> 
            <a href="category.html" class="dropdown-toggle" data-toggle="dropdown"><?php echo $menu_details['name']; ?></a> 
            <ul class="mega_dropdown dropdown-menu" style="width: 830px;"> 
             <?php 
             while($submenu_relation_details=mysql_fetch_array($sub_menu_query)) 
             { 
            ?> 
             <li class="block-container col-sm-2"> 
              <ul class="block"> 
               <li class="link_container group_header"> 
                <a href="#"><?php echo $submenu_relation_details['name']; ?></a> 
               </li> 
               <?php 
                $sub_menu=mysql_query("SELECT * FROM `sub_category` WHERE `cat_id`='$submenu_relation_details['id']'"); 
                $i=1; 
                while($data=mysql_fetch_array($sub_menu)) 
                { 
                ?> 
                <li class="link_container"><a href="Selectedproduct.php?id=<?php echo base64_encode($data['id']); ?>"><?php echo $data['name']; ?></a></li> 
                <?php 
                $i++; 
                } 
               ?> 
              </ul> 
             </li> 
             <?php 
             } 
             ?> 
             </ul> 

           </li> 

           <?php 
           } 
           else 
           { 
           ?> 
           <li> 
           <a href="category.html" ><?php echo $menu_details['name']; ?></a></li> 
           <?php 
           } 

          } 
          ?> 
          </ul> 

         </div> 

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