从和SQL数据库创建表包含一个下拉菜单与另一个SQL表的名称列表

问题描述：

我需要创建一个下拉菜单并在每一行中提交按钮的表。下拉菜单包含来自SQL表的顾问列表。当我选择和顾问时，我按下提交按钮，当前行中的项目ID以及选定的顾问ID或名称必须发送到另一页面。在我的情况下，它被发送到delete.php。从和SQL数据库创建表包含一个下拉菜单与另一个SQL表的名称列表

我的代码在下表中显示了一个下拉菜单和一个提交按钮，但是当你按下提交按钮时，只有按下位于表底部的提交按钮，它才会正常工作，如果我按任何其他似乎不从下拉菜单发送信息。

（我知道我的代码看起来凌乱，我正在试验如果有什么不清楚的问我，我会澄清。）非常感谢！

<!DOCTYPE html> 
<html> 
<body> 

<?php 
    //this is he code for the qeue 
// connect to the database udinh sqli 
$con = get_sqli(); 
// get results from database 

if (!$con) { 
    die('Could not connect: ' . mysqli_error($con)); 
} 
//select whole list of students from walk_in 
mysqli_select_db($con,"login"); 
$sql="SELECT * FROM walk_in"; 
$result = mysqli_query($con,$sql); 

if (!$result) { 
    printf("Error: %s\n", mysqli_error($con)); 
    exit(); 
} 
mysqli_close($con); 
//Table to dispaly qeueu of students 
echo "<table border='1' cellpadding='10'>"; 

echo "<tr> <th>ID</th> <th>First Name</th> <th>Last Name</th><th>Advisor Student wants to see</th><th>P ID</th><th>Select Advisor to notify on send</th><th>Send Student</th><th> </tr>"; 

echo "<tr>"; 

//create a table of students by displaying all the data from result and adding a button 
while($row = mysqli_fetch_array($result)) { 
    echo "<tr>"; 
    echo "<td>" . $row['id'] . "</td>"; 
    echo "<td>" . $row['FirstName'] . "</td>"; 
    echo "<td>" . $row['LastName'] . "</td>"; 
    echo "<td>" . $row['Advisor'] . "</td>"; 
    echo "<td>" . $row['pid'] . "</td>"; 
    // echo '<form action="delete.php?id2=' . $row['id'] . '" method="post">'; 
    // drop down menu for selecting advisor as a form submission 

    // used to name each submit button with the id from walk_in 
    $formId = $row['id'] ; 

echo "<td>" ; 
//create a form to submit the sleected advisor and the seelcted student to be removed from the queue 
echo '<form action="delete.php?id=' . $row['id'] . '" method="post">'; 

//another query used to retreive the list of advisors to pupulate the drop down menu 
//create a drop down menu with advisors resulting from the queue 
echo '<select name="formStatus">'; 
$con = get_sqli(); 
      mysqli_select_db($con,"login"); 
$sql="SELECT * FROM login_details WHERE level = 0 AND logged = 1"; 
$result2 = mysqli_query($con,$sql); 

if (!$result2) { 
    printf("Error: %s\n", mysqli_error($con)); 
    exit(); 
} 

      //loops through all advisors for drop down menu creation 
       while ($row2 = mysqli_fetch_array($result2)) { 
        $id = $row2['id']; 
    echo '<option value="'.$id.'">'.$id.'</option>'; 
} 
echo'<option selected="selected"></option>'; 

echo '</select>'; 
echo '<td><input type="submit" name="formSubmit" value= "'.$formId.'" /><td>'; 
//echo '<td><input type="submit" name="formSubmit" value= /><td>'; 
//echo '<td><a href="delete.php?id=' . $row['id'] . '&advisor='. "lol" .'">Send</a></td>'; 
echo "</tr>"; 

} 
// close table> 
echo "</table>"; 


?> 

<p><a href="new.php">Add a new record</a></p> 

    </body> 
</html>

下面是我使用的表格：

login_details含表ADVISER细节

我没有看到结尾的''标签。 –

如果没有javascript/jquery的相关代码，请勿垃圾标签。 –

一些明智的代码缩进将是一个好主意。它可以帮助我们阅读代码，更重要的是，它可以帮助您**调试您的代码** [快速浏览编码标准]（http://www.php-fig.org/psr/psr-2/ ）为了您自己的利益。您可能会被要求在几周/几个月内修改此代码，最后您会感谢我。 – RiggsFolly

答

我忘了关的形式，该问题已得到修复。谢谢你们！

从和SQL数据库创建表包含一个下拉菜单与另一个SQL表的名称列表

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