問題描述

我正在開發一個下拉菜單，它使用 HTML optgroups 作為員工所屬的組名稱.這是 MySQL 查詢和輸出:

唯一的問題是，我最難弄清楚如何讓 optgroup 正常工作.我已經嘗試了無數次，它真的開始讓我感到沮喪.以下幾乎是我想要的輸出(示例):

<optgroup label="組 1"><option name="20">員工 2</option><option name="19">員工 1</option></optgroup><optgroup label="第 2 組"><option name="21">員工 3</option></optgroup></選擇>
基本上，optgroup 需要是groupname"，選項name"應該是emp_id"，動作option"(下拉項)是empname".
我希望這是可以做到的，但真的不確定.這是我擁有的功能，但效果不佳:
function getDynGrpList() {全球 $db;//$query = "SELECT * FROM員工 ORDER BY grp_id desc;";$query = "選擇雇員.emp_id,employees.empname,employees.grp_id,groups.groupname FROM 雇員離開加入組雇員.grp_id = groups.grp_id 按組名 asc 排序;";$employees = $db->GetAll($query);$groups = array();而 ($qa = $employees->GetRows()) {$groups[$qa['groupname']][$qa['grp_id']] = $qa['empname'];}foreach ($groups as $label => $opt) { ?><optgroup label="<?php echo $label; ?>"><?php }foreach ($groups[$label] as $id => $name) { ?><option value="<?php echo $id; ?>"><?php echo $name;?></option><?php } ?></optgroup><?php }
getDynGrpList 函數截至美國中部標準時間凌晨 3:15 (2/27):
function getDynGrpList() {全球 $db;//$query = "SELECT * FROM員工 ORDER BY grp_id desc;";$query = "選擇雇員.emp_id,employees.empname,employees.grp_id,groups.groupname FROM 雇員離開加入組雇員.grp_id = groups.grp_id 按組名 asc 排序;";$employees = $db->GetAll($query);$groups = array();而 ($qa = $employees->GetRows()) {$groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];}var_export($groups);foreach($groups as $label => $opt): ?><optgroup label="<?php echo $label; ?>"><?php foreach ($opt as $id => $name): ?><option value="<?php echo $id; ?>"><?php echo $name;?></option><?php endforeach;?></optgroup><?php endforeach;}
最終解決方案(在 Felix Kling 的幫助下)
function getDynGrpList() {全球 $db;$query = "選擇雇員.emp_id,employees.empname,employees.grp_id,groups.groupname FROM 雇員離開加入組雇員.grp_id = groups.grp_id 按組名 asc 排序;";$employees = $db->GetAll($query);$groups = array();foreach ($employees 作為 $employee) {$groups[$employee['groupname']][$employee['emp_id']] = $employee['empname'];}foreach($groups as $label => $opt): ?><optgroup label="<?php echo $label; ?>"><?php foreach ($opt as $id => $name): ?><option value="<?php echo $id; ?>"><?php echo $name;?></option><?php endforeach;?></optgroup><?php endforeach;}
 解決方案 
這兩個 for 循環沒有嵌套在你的代碼中:
foreach ($groups as $label => $opt) { ?><optgroup label="<?php echo $label; ?>"><?php } <-- 這里錯了foreach ($groups[$label] as $id => $name) { ?><option value="<?php echo $id; ?>"><?php echo $name;?></option><?php } ?>
結果是首先創建了所有 opt 組，然后添加了最后一組的員工(因為 $label 和 $opt 在循環后也可用完成).
你必須嵌套循環(使用替代語法控制結構):
 $opt): ?><optgroup label="<?php echo $label; ?>"><?php foreach ($opt as $id => $name): ?><option value="<?php echo $id; ?>"><?php echo $name;?></option><?php endforeach;?></optgroup><?php endforeach;?>
此外，我認為您必須在創建數組時使用 emp_id，而不是 grp_id:
while ($qa = $employees->GetRows()) {$groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];}
I am developing a drop down menu that uses HTML optgroups for group names that employees are a part of.  Here is the MySQL query and output:
mysql> SELECT employees.emp_id,employees.empname,employees.grp_id,groups.groupname FROM employees left join groups on employees.grp_id = groups.grp_id order by groupname asc;
+--------+------------+--------+-----------+
| emp_id | empname    | grp_id | groupname |
+--------+------------+--------+-----------+
|     20 | Employee 2 |     13 | Group 1   |
|     19 | Employee 1 |     13 | Group 1   |
|     21 | Employee 3 |     14 | Group 2   |
+--------+------------+--------+-----------+
3 rows in set (0.00 sec)
The only issue is, I'm having the hardest time figuring out how to get the optgroup to work correctly.  I've tried countless times, and it's really starting to frustrate me. The following is pretty much I want the output to be (example):
<select name="dropdownmenu">
    <optgroup label="Group 1">
        <option name="20">Employee 2</option>
        <option name="19">Employee 1</option>
    </optgroup>
    <optgroup label="Group 2">
        <option name="21">Employee 3</option>
    </optgroup>
</select>
Basically, the optgroup needs to be the "groupname", the option "name" should be the "emp_id", and the action "option" (drop down item) is the "empname".

I hope this is something that can be done, but really not sure.  Here's the function I have, but it doesn't exactly work well:
function getDynGrpList() {
    global $db;

    // $query = "SELECT * FROM employees ORDER BY grp_id desc;";
    $query = "SELECT employees.emp_id,employees.empname,employees.grp_id,groups.groupname FROM employees left join groups on employees.grp_id = groups.grp_id order by groupname asc;";
    $employees = $db->GetAll($query);
    $groups = array();

    while ($qa = $employees->GetRows()) {
        $groups[$qa['groupname']][$qa['grp_id']] = $qa['empname'];
    }
    foreach ($groups as $label => $opt) { ?>
        <optgroup label="<?php echo $label; ?>">
<?php   }
        foreach ($groups[$label] as $id => $name) { ?>
            <option value="<?php echo $id; ?>"><?php echo $name; ?></option>
<?php } ?>
        </optgroup>
<?php }
getDynGrpList function as of 3:15AM CST (2/27):
function getDynGrpList() {
    global $db;

    // $query = "SELECT * FROM employees ORDER BY grp_id desc;";
    $query = "SELECT employees.emp_id,employees.empname,employees.grp_id,groups.groupname FROM employees left join groups on employees.grp_id = groups.grp_id order by groupname asc;";
    $employees = $db->GetAll($query);
    $groups = array();
    while ($qa = $employees->GetRows()) {
        $groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];
    }
    var_export($groups);
    foreach($groups as $label => $opt): ?>
        <optgroup label="<?php echo $label; ?>">
    <?php foreach ($opt as $id => $name): ?>
        <option value="<?php echo $id; ?>"><?php echo $name; ?></option>
    <?php endforeach; ?>
    </optgroup>
<?php endforeach;
}
Final Solution (with the help of Felix Kling)
function getDynGrpList() {
    global $db;

    $query = "SELECT employees.emp_id,employees.empname,employees.grp_id,groups.groupname FROM employees left join groups on employees.grp_id = groups.grp_id order by groupname asc;";
    $employees = $db->GetAll($query);
    $groups = array();
    foreach ($employees as $employee) {
        $groups[$employee['groupname']][$employee['emp_id']] = $employee['empname'];
    }
    foreach($groups as $label => $opt): ?>
        <optgroup label="<?php echo $label; ?>">
    <?php foreach ($opt as $id => $name): ?>
        <option value="<?php echo $id; ?>"><?php echo $name; ?></option>
    <?php endforeach; ?>
    </optgroup>
<?php endforeach;
}

 解決方案 
The two for loops are not nested in your code:
foreach ($groups as $label => $opt) { ?>
    <optgroup label="<?php echo $label; ?>">
<?php   } <-- wrong here
    foreach ($groups[$label] as $id => $name) { ?>
        <option value="<?php echo $id; ?>"><?php echo $name; ?></option>
<?php } ?>
The result is that first all opt groups are created and then the employees for the last group are added (because $label and $opt are also available after the loop finished).

You have to nest the loops (using alternative syntax for control structures):
<?php foreach($groups as $label => $opt): ?>
    <optgroup label="<?php echo $label; ?>">
    <?php foreach ($opt as $id => $name): ?>
        <option value="<?php echo $id; ?>"><?php echo $name; ?></option>
    <?php endforeach; ?>
    </optgroup>
<?php endforeach; ?>
Furthermore, I think you have to use the emp_id, not the grp_id when creating the array:
while ($qa = $employees->GetRows()) {
    $groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];
}


                        
這篇關于PHP:使用 optgroup 動態下拉的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！
                
【網站聲明】本站部分內容來源于互聯網,旨在幫助大家更快的解決問題，如果有圖片或者內容侵犯了您的權益，請聯系我們刪除處理，感謝您的支持！
                
                

             

                  

                  上一篇：使用復選框、PHP 和 MySQL 刪除多行 
                  下一篇：如何在下拉更改事件上調用 AJAX 請求? 
                  
                
              
            
          
          
          

            

              

                

                  
相關文檔推薦
                

                

                  

                    PHP、MySQL PDOException 的死鎖異常代碼?
                  
                  
Deadlock exception code for PHP, MySQL PDOException?(PHP、MySQL PDOException 的死鎖異常代碼?)
                
                

                

                  

                    PHP PDO MySQL 可滾動游標不起作用
                  
                  
PHP PDO MySQL scrollable cursor doesn#39;t work(PHP PDO MySQL 可滾動游標不起作用)
                
                

                

                  

                    PHP PDO ODBC 連接
                  
                  
PHP PDO ODBC connection(PHP PDO ODBC 連接)
                
                

                

                  

                    使用 PDO::FETCH_CLASS 和魔術方法
                  
                  
Using PDO::FETCH_CLASS with Magic Methods(使用 PDO::FETCH_CLASS 和魔術方法)
                
                

                

                  

                    php pdo 只從 mysql 獲取一個值；等于變量的值
                  
                  
php pdo get only one value from mysql; value that equals to variable(php pdo 只從 mysql 獲取一個值；等于變量的值)
                
                

                

                  

                    MSSQL PDO 找不到驅動程序
                  
                  
MSSQL PDO could not find driver(MSSQL PDO 找不到驅動程序)
                
                

            
          
        
        
<tbody id='xs2XF'></tbody>
<small id='xs2XF'></small><noframes id='xs2XF'>
<tfoot id='xs2XF'></tfoot>
<bdo id='xs2XF'></bdo><ul id='xs2XF'></ul>
<i id='xs2XF'><tr id='xs2XF'><dt id='xs2XF'><q id='xs2XF'><span id='xs2XF'><b id='xs2XF'><form id='xs2XF'><ins id='xs2XF'></ins><ul id='xs2XF'></ul><sub id='xs2XF'></sub></form><legend id='xs2XF'></legend><bdo id='xs2XF'><pre id='xs2XF'><center id='xs2XF'></center></pre></bdo></b><th id='xs2XF'></th></span></q></dt></tr></i><div   class="l0qxevy"   id='xs2XF'><tfoot id='xs2XF'></tfoot><dl id='xs2XF'><fieldset id='xs2XF'></fieldset></dl></div>
<legend id='xs2XF'><style id='xs2XF'><dir id='xs2XF'><q id='xs2XF'></q></dir></style></legend>