2017-04-21 48 views
0

我设计了一个表单来将数据插入本地主机上的数据库。在单个页面表单应用程序中插入数据库

<form action='' method='post'> 
<input type='submit' name='CRUD' value='New Data'> 
<br><br> 
<input type='submit' name='CRUD' value='Retrieve Data'> 
<br> 

<hr> 
</form> 

<?php 
error_reporting(0); 
$x = $_POST['CRUD']; 

if ($x == "New Data") { 
    require 'part1.php'; 
} 
?> 

然后,我做了一个表单,将数据插入另一个文件。

<form method='post'> 
<label for='site'>Name: </label> 
<input type='text' name='site'> 
<br><br> 

<label for='date'>Date: </label> 
<input type='date' name='time'> 
<br><br> 

<label for='page'>Web URL: </label> 
<input type='url' name='page'> 
<br><br> 

<label for='desc'>Description: </label> 
<input type='text' name='desc'> 
<br><br> 

<input type='submit' name='finish' value='Go'><input type="reset"> 
</form> 

<?php 

if (!empty($_POST)){ 

    $servername = "localhost"; 
    $username = "root"; 
    $password = ""; 
    $dbname = "assignment5"; 

    $resource = $_POST['site']; 
    $date = $_POST['time']; 
    $url = $_POST['page']; 
    $explain = $_POST['desc']; 

    // Create connection 
    $conn = new mysqli('localhost', 'root', $password, $dbname) or 
    die("Unable to connect"); 

    // Check connection 
    if ($conn->connect_error) { 
     die("Connection failed: " . $conn->connect_error); 
    } 

    $sql = "INSERT INTO thedata (date, Name, URL, Description) 
    VALUES ('$date', '$resource', '$url', '$explain')"; 

    if (mysqli_query($conn, $sql)) { 
     echo "New record created successfully"; 
    } else { 
     echo "Error: " . $sql . "<br>" . mysqli_error($conn); 
    } 

    $conn->close(); 
} 
?> 

在有自己的,他们按计划工作,但我需要的是具有相同的页面两种形式。这样做会在插入默认数据时出错,而不是表单的输入。

回答

0

如果你想把2个表单放在同一个页面中,你必须给每个表单一个提交按钮..注意使用相同的提交按钮到相同的表格

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