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灯堆栈-更新HTML表-不映射到DB或从DB映射

lamp html php

user3236169 · 技术社区 · 6 年前

你好,我有一个基本的HTML表单 指数html

<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>

在提交时,将执行以下PHP脚本以使用详细信息更新我的DB。 使现代化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.";
  }
  }
?>

然后,我希望另一个页面显示一个HTML表,其中包含来自DB的内容 主要的html

<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>

除了两列没有相互映射之外,其他一切都正常工作,即 Subname & Labels

我已经仔细检查了我的代码,但我发现不了错误,有人能帮我吗?

我重新启动了https等,但似乎没有任何效果:(

DORA架构

MariaDB[朵拉]>显示dora中的列;

+-------------+--------------+------+-----+---------+----------------+
| 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    |                |
+-------------+--------------+------+-----+---------+----------------+

我的表格中的结果

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

1 回复 | 直到 6 年前

USER249 6 年前

我测试了你的代码,效果很好。唯一的区别是我使用的InnoDB应该是相同的。请尝试以下操作:

1更改这些列的名称。

2重新创建数据库。

3确保没有触发器将两列设置为支架