2017-12-27 182 views
0

你好,我有一個基本的HTML表單的Index.htmlLAMP堆棧 - 更新HTML表 - 不映射到/從DB

<html> 
<body> 
<form action="update.php" method="POST"> 
    Department: <input type="text" name="department"><br><br> 
    Subname: <input type="text" name="subname"><br><br> 
    Labels: <input type="text" name="labels"><br><br> 
    Pagerduty: <input type="text" name="pagerduty"><br><br> 
    Description: <input type="text" name="description"><br><br> 
    <input type="submit" value="Submit" name="submit"> 
</form> 
</body> 
</html> 

在提交以下PHP腳本執行與細節來更新我的DB 。 update.php

<?php 
$hostname = "localhost"; 
$username = "root"; 
$password = "xxxxxxxxxxx"; 
$db = "dora"; 
$dbconnect=mysqli_connect($hostname,$username,$password,$db); 
if ($dbconnect->connect_error) { 
die("Database connection failed: " . $dbconnect->connect_error); 
} 
if(isset($_POST['submit'])) { 
$department=$_POST['department']; 
$subname=$_POST['subname']; 
$labels=$_POST['labels']; 
$pagerduty=$_POST['pagerduty']; 
$description=$_POST['description']; 
$query = "INSERT INTO dora (department, subname, labels, pagerduty, description) 
VALUES ('$department', '$subname', '$labels', '$pagerduty', '$description')"; 
if (!mysqli_query($dbconnect, $query)) { 
     die('An error occurred when submitting your review.'); 
    } else { 
    echo "Thanks for your review."; 
    } 
    } 
?> 

然後我想另一個頁面顯示來自DB main.html中

<html> 
<body> 
<?php 
$hostname = "localhost"; 
$username = "root"; 
$password = "xxxxxxxx"; 
$db = "dora"; 
$dbconnect=mysqli_connect($hostname,$username,$password,$db); 
if ($dbconnect->connect_error) { 
    die("Database connection failed: " . $dbconnect->connect_error); 
} 
?> 
<table border="1" align="center"> 
<tr> 
    <td>Department</td> 
    <td>Subname</td> 
    <td>Labels</td> 
    <td>Pagerduty</td> 
    <td>Description</td> 
</tr> 
<?php 
$query = mysqli_query($dbconnect, "SELECT * FROM dora") 
    or die (mysqli_error($dbconnect)); 

while ($row = mysqli_fetch_array($query)) { 
    echo 
    "<tr> 
    <td>{$row['department']}</td> 
    <td>{$row['subname']}</td> 
    <td>{$row['labels']}</td> 
    <td>{$row['pagerduty']}</td> 
    <td>{$row['description']}</td> 
    </tr>\n"; 
} 
?> 
</table> 
</body> 
</html> 

一切內容的HTML表格由兩列除了正常工作不映射即Subname & Labels

我已經仔細檢查了我的代碼,但我無法發現錯誤,任何人都可以幫我嗎?

我已經重新啓動HTTPS等,但似乎沒有任何工作:(

DORA SCHEMA

MariaDB的從多拉[多拉]>顯示列;

+-------------+--------------+------+-----+---------+----------------+ 
| Field  | Type   | Null | Key | Default | Extra   | 
+-------------+--------------+------+-----+---------+----------------+ 
| TAB_ID  | int(11)  | NO | PRI | NULL | auto_increment | 
| department | varchar(200) | YES |  | NULL |    | 
| subname  | varchar(200) | YES |  | NULL |    | 
| labels  | varchar(200) | YES |  | NULL |    | 
| pagerduty | varchar(200) | YES |  | NULL |    | 
| description | varchar(200) | YES |  | NULL |    | 
+-------------+--------------+------+-----+---------+----------------+ 

從結果我的表

Department Subname Labels Pagerduty Description 
1        4    5 
1        4    5 
sdfsdf      adfasdfad  adfadfadfa 
+0

愚蠢的問題,但...你證明:對於子名稱和標籤值確實得到(使用,也就是說,phpMyAdmin的)插入? – Rushikumar

+0

+ --------- + |子名稱| + --------- + | | | | + --------- +是的,我檢查了,沒有任何進展:( – user3236169

+0

只適用於這兩個領域,其他所有人都在工作 – user3236169

回答

1

我測試了你的代碼它工作正常。唯一的區別是即時通訊使用InnoDB應該是相同的。請嘗試以下操作:

1更改這些列名稱。

2重新創建您的數據庫。

3確保沒有trigers設置兩個列木板

+0

是的,這是我的下一步,現在我已經用盡了幫助。我會盡力讓你知道。謝謝您的幫助! – user3236169