6

我正在开发一个下拉菜单，该菜单使用 HTML optgroups 作为员工所属的组名。这是 MySQL 查询和输出：

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 | 员工名 | grp_id | 组名 |
+--------+------------+--------+------------+
| 20 | 员工 2 | 13 | 第 1 组 |
| 19 | 员工 1 | 13 | 第 1 组 |
| 21 | 员工 3 | 14 | 第 2 组 |
+--------+------------+--------+------------+
3 行一组（0.00 秒）

唯一的问题是，我很难弄清楚如何让 optgroup 正常工作。我已经尝试了无数次，这真的让我感到沮丧。以下是我非常希望输出为（示例）：

<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>

基本上，optgroup 需要是“groupname”，选项“name”应该是“emp_id”，动作“option”（下拉项）是“empname”。

我希望这是可以做到的，但真的不确定。这是我拥有的功能，但它并不能很好地工作：

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 函数截至美国中部标准时间凌晨 3:15（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;
}

最终解决方案（在 Felix Kling 的帮助下）

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;
}
4

1 回答 1

5

两个 for 循环没有嵌套在您的代码中：

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 } ?>

结果是首先创建所有 opt 组，然后添加最后一个组的员工（因为$label并且$opt在循环完成后也可用）。

您必须嵌套循环（使用控制结构的替代语法）：

<?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; ?>

此外，我认为您必须在创建数组时使用emp_id，而不是：grp_id

while ($qa = $employees->GetRows()) {
    $groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];
}
于 2011-02-27T08:23:25.113 回答