2017-05-21 44 views
-1

我想使用php在数据库表中保存输入选择选项。我想将它与现有的属性 我知道如何使用单独的表它时添加它保存在一个表中,但我想它与其他几个属性添加到现有表我想保存在数据库表中的输入选择选项使用php

PS的第一个问题可以给我

<?php 

if(isset($_POST['signup'])) { 

    $name = $_POST['name']; 
    $email = $_POST['email']; 
    $phone = $_POST['phone']; 
    $password = $_POST['password']; 
    $department = $_POST['deptlist']; 

    echo $name; 

    $connection = mysqli_connect('localhost', 'root', '', 'filetracking');  

    if($connection) { 
     echo "We are connected"; 
    } else { 
     die("Database connection failed"); 
    } 

    $query = "INSERT INTO faculty(name,email,contactno,password,office) "; 
    $query .= "VALUES ('$name', '$email', '$phone', '$password', '$department')"; 

    $result = mysqli_query($connection, $query); 
    if(!$result) { 
     die('Query FAILED' . mysqli_error($connection)); 
    } else { 
     echo "Record Create"; 
    } 
} 

?> 

<form method="post" action="signup.php"> 
    <div class="form-group"> 
     <label for="name" class="control-label">Name</label> 
     <input type="text" name="name" id="Name" class="inputmd textinput form-control required" placeholder="Full Name" style="margin-bottom: 10px" required="true"> 
    </div> 

    <div class="form-group"> 
     <label for="email" class="control-label">Email</label> 
     <input type="email" name="email" id="Email" class="inputmd textinput form-control required" placeholder="Email" style="margin-bottom: 10px" required="true"> 
    </div> 

    <div class="form-group"> 
     <label for="phone" class="control-label">Phone</label> 
     <input type="number" name="phone" id="Phone" class="inputmd textinput form-control required" placeholder="Phone No" style="margin-bottom: 10px" required="true"> 
    </div> 

    <div class="form-group"> 
     <label for="password" class="control-label">Password</label> 
     <input type="password" name="password" id="Password" class="inputmd textinput form-control" placeholder="Password" style="margin-bottom: 10px" required="true"> 
    </div> 

    <div class="form-group">       
     <label for="deptlist" class="control-label">Department</label> 
     <select name="deptlist" form="deptform" class="inputmd form-control" style="margin-bottom: 10px"> 
      <option value="DCSE">Department of Computer Systems Engineering</option> 
      <option value="DME">Department of Mechanical Engineering</option> 
      <option value="DEE">Department of Electrical Engineering</option> 
      <option value="DCS">Department of Computer Science</option> 
     </select> 

    <div class="form-group">      
     <input type="submit" name="signup" id="Singup" value="Sign Up" class="btn btn-info col-md-4 col-md-offset-4"> 
    </div> 

</div> 

+0

你究竟试过了什么? –

+0

看看?到图片..你会吗? –

+2

*“看看图片..你会吗?”* - 将您的代码发布为需要操作的格式。另外,也是HTML。你来寻求帮助,那么你也需要帮助我们。 –

回答

0

您错误地提到了一个名为deptform <select name="deptlist" form="deptform"的表格,但没有这样的表格。这可以防止脚本接收这些数据。

+0

请不要链接新手到W3Schools。该网站被认为是有害的。 –

+0

已删除。为什么它被认为是有害的? (我没有太注意它) –

+0

因为他们的PHP和JS代码都充满了漏洞。而有关HTML和CSS的文档有错误和/或过时的倾向。 –

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