我正在开发一个下拉菜单,该菜单将HTML optgroup用于员工所属的群组名称。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 | 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)
唯一的问题是,我正在努力弄清楚如何让optgroup正常工作。我已经尝试了无数次,而且它真的开始让我感到沮丧。 以下几乎是我希望输出为(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,但实际上并不确定。这是我的功能,但它并没有完全奏效:
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功能截至美国中部时间凌晨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 $name; ?></option>
<?php endforeach; ?>
</optgroup>
<?php endforeach;
}
答案 0 :(得分:5)
两个for循环没有嵌套在你的代码中:
foreach ($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
和$opt
在循环结束后也可用。)
你必须嵌套循环(using alternative syntax for control structures):
<?php foreach($groups as $label => $opt): ?>
<optgroup label="<?php echo $label; ?>">
<?php foreach ($opt as $id => $name): ?>
<option value="<?php echo $id; ?>"><?php echo $name; ?></option>
<?php endforeach; ?>
</optgroup>
<?php endforeach; ?>
此外,我认为在创建数组时必须使用emp_id
,而不是grp_id
:
while ($qa = $employees->GetRows()) {
$groups[$qa['groupname']][$qa['emp_id']] = $qa['empname'];
}