我有一个表单,用户可以输入部门名称,然后是多个选择框,其中多个员工职位/职位正在此部门工作。PHP/MySQLi - 显示来自表
我现在正在寻找一种方式,让用户编辑自己提交的部门和:然后将记录通过foreach()循环,这是给我这个结果保存到我的SQL表因此想要在一个字段中再次显示存储的记录,这留下了将数组数据返回到选择字段的问题...
我试图用另一个foreach填充<select>
字段,但是到目前为止没有运气。我不知道,如果问题仅仅是糟糕的代码,或者如果选择二插件在这里给我的问题,以及...
PHP(UPDATE2)
<?php
// Start MySQLi connection
$db = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
// Build basic query, Admins can see all records
$sql = ("SELECT * FROM qci_departments GROUP BY Department ORDER BY Department");
// run the query or show an error message
if(!$result = $db->query($sql)){
echo('There was an error selecting data from the table [' . $db->error . ']');
}
while($row = mysqli_fetch_array($result)){
$dept_id = $row['ID'];
$dept_name = $row['Department'];
$dept_positions = $row['Positions'];
echo "
<tr>
<td>
<input type=\"text\" class=\"form-control\" id=\"editDeptName\" name=\"editDeptName\" value=\"$dept_id\">
</td>
<td>$dept_name</td>
<td>
<select id=\"editDeptPositions\" name=\"editDeptPositions\" class=\"form-control\" multiple>
<option value='".$dept_positions."'>".$dept_positions."</option>";
//$position_query = ("SELECT distinct Positions FROM qci_departments");
$position_query = ("SELECT distinct Positions FROM qci_departments where Department = '".$row['Department']."'");
if(!$result_positions = $db->query($position_query)){
echo('There was an error selecting data from the table [' . $db->error . ']');
} else {
while($row_positions = mysqli_fetch_assoc($result_positions)){
echo "<option value='".$row_positions['Positions']."'>".$row_positions['Positions']."</option>";
}
}
echo "</select>
</td>
</tr>";
任何人都可以点我在正确的方向,请? 谢谢 A2K
编辑:更新的代码,以反映海报下方
似乎确定?检查你有'echo'
“;' –您期望的值你在一段时间内在这种情况下foreach是无用的。 –
@RafaelShkembi确实如此,应该是 'echo“”;' –