試圖從我的桌子上的下拉列表中查看信息

我有兩頁。一個包含我的分支下拉列表，而第二個是我的php連接頁面我想從我的下拉列表中選擇一個分支，點擊進程，然後一個表格顯示分支名稱，狀態，金額和日期2 每當單擊過程中，我得到一個空表只用頭這裏是我的代碼：試圖從我的桌子上的下拉列表中查看信息

branch.php

<html> 
<body> 
<br><br><br> 
<form method = "POST" action = "getuser.php" > 
<select name="branch"> 
    <option value="">Select a branch:</option> 
    <option value="1">Maths</option> 
    <option value="3">English</option> 
    <option value="4">Biology</option> 
    <option value="5">Chemistry</option> 
    <option value="6">Economics</option> 
    </select> 
    <br><br> 
    <input type = "submit" name "submit" value ="Process"> 
</form> 
<br>

getuser.php

<?php 
//$q = intval($_GET['q']); 

$con = mysqli_connect('localhost','root','','modem'); 
if (!$con) { 
    die('Could not connect: ' . mysqli_error($con)); 
} 
    if(!isset($_POST['submit'])) { 

     $branch = $_POST['branch']; 

     $result = mysqli_query($con,"SELECT *, DATE_FORMAT(date, '%M %Y')as date2 
     FROM bm_ibadan_division 
     WHERE branch = '$branch' 
     ORDER BY date ASC"); 

     echo "<table align='center' width='400px' border='1'> 
     <tr> 
     <th>Branch</th> 
     <th>Status</th> 
     <th>Amount</th> 
     <th>Date2</th> 
     </tr>"; 

     while($row = mysqli_fetch_array($result)) { 
      echo "<tr>"; 
      echo "<td>" . $row['branch'] . "</td>"; 
      echo "<td>" . $row['status'] . "</td>"; 
      echo "<td>" . $row['amount'] . "</td>"; 
      echo "<td>" . $row['date2'] . "</td>"; 
      echo "</tr>"; 
     } 

     echo "</table>"; 
    } else echo "access denied"; 
     mysqli_close($con); 


?>

來源

2015-03-13 Fola Idowu

$con = mysql_connect('localhost','root','','modem');

如果模具（ '無法連接：'（$騙局！）。 mysql_error（$ CON））;

如果（isset（$ _ POST [ '提交']））{

$branch = $_POST['branch']; 

    $result = mysql_query($con,"SELECT *, DATE_FORMAT(date, '%M %Y')as date2 FROM bm_ibadan_division WHERE branch = $branch ORDER BY date ASC"); 

    echo "<table align='center' width='400px' border='1'> 
    <tr> 
    <th>Branch</th> 
    <th>Status</th> 
    <th>Amount</th> 
    <th>Date2</th> 
    </tr>"; 

    while($row = mysql_fetch_array($result)) { 
     echo "<tr>"; 
     echo "<td>" . $row['branch'] . "</td>"; 
     echo "<td>" . $row['status'] . "</td>"; 
     echo "<td>" . $row['amount'] . "</td>"; 
     echo "<td>" . $row['date2'] . "</td>"; 
     echo "</tr>"; 
    } 

    echo "</table>"; 
} 
else 
{ 
    echo "access denied"; 
} 

mysqli_close($con);

來源

2015-03-13 13:53:49 SneakingOzone

雖然此代碼段可以解決的問題，[包括一個解釋]（http://meta.stackexchange.com/questions/114762 /解釋 - 完全基於代碼的答案）真的有助於提高你的文章的質量。請記住，您將來會爲讀者回答問題，而這些人可能不知道您的代碼建議的原因。 – Bono 2015-03-13 13:58:05

感謝SneakingOzone，但在使用您的建議後，我得到的是拒絕訪問 – 2015-03-13 16:31:37

試圖從我的桌子上的下拉列表中查看信息

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