Question

我有一个表单，用户可以在其中输入部门名称，然后是多个选择框，其中包含在该部门中工作的多个员工职位/职位。然后通过foreach（）循环将记录保存到我的SQL表中，这样就可以得到结果：

我现在正在寻找一种方法来允许用户编辑他们提交的部门和位置，因此想要在一个字段中再次显示存储的记录，这使我无法将数组数据恢复到选择字段。 ..

我试图使用另一个foreach来填充<select>字段但到目前为止没有运气。我不确定问题是否只是错误的代码或者Select2插件是否也给我带来了问题......

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>";

有人能指出我正确的方向吗？谢谢， A2K

编辑：更新代码以反映下面海报建议的更改

Answer 1

我会给你两个解决方案

首先：

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>$dept_name</td>
            <td>
                <select id=\"editDeptPositions\" name=\"editDeptPositions\" class=\"form-control select2\">
                    <option value='".$dept_positions."'>".$dept_positions."</option>";
                    $position_query = ("SELECT distinct Positions FROM qci_departments");
                    if(!$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>";
}

在while内进行第二次查询以获取位置并填充选择选项。

第二：//我更喜欢这个

$positions_arr = array();
$position_query = ("SELECT distinct Positions FROM qci_departments");
if(!$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)){
        $positions_arr[] = $row_positions['Positions']; 
    }
}
while($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>$dept_name</td>
            <td>
                <select id=\"editDeptPositions\" name=\"editDeptPositions\" class=\"form-control select2\">
                    <option value='".$dept_positions."'>".$dept_positions."</option>";
                    foreach($positions_arr as $value){
                         echo "<option value='".$value."'>".$value."</option>";
                    }
                echo "</select>
            </td>
        </tr>";
}

制作一个包含所有位置的数组，然后在其中制作一个foreach以填充选择选项

PHP / MySQLi - 显示表中的数组数据

1 个答案: