2017-03-07 86 views
2

,这是我的表单.html文件得到连接到adddatabase.php页.having名字和secondname无法从网页数据添加到数据库

 <form action="addtodatabase.php" method="post"> 
     <div class="container"> 
     <form class="form-inline"> 
     <fieldset> 
     <legend>Security Department User Registration</legend> 
     <div class="form-group"> 
     <label for="Firstname">First Name</label> 
     <input type="text" class="form-control" id="Firstname" name="firstname" placeholder="Text input"><br/> 
     </div> 

     <div class="form-group"> 
     <label for="Secondname">Second Name</label> 
     <input type="text" class="form-control" id="Secondname" name="secondname" placeholder="Text input"><br/> 
     </div> 
     </form> 

     <button type="submit" class="btn btn-default">Submit</button> 
     </form> 

addtodatabase.php页面如下。

 if (isset($_POST)) { 
    $firstname = isset($_POST['firstname']) ? $_POST['firstname'] : ''; 
    $secondname = isset($_POST['secondname']) ? $_POST['secondname'] : ''; 
    echo 'Your first name is ' .$firstname. '<br>'; 
    echo 'Your second name is ' .$secondname. '<br>'; 
    } 
    mysqli_query($connect,"INSERT INTO form_details(firstname,secondname) 
      VALUES('$firstname','$secondname')"); 
+0

你的提交按钮必须在表格里面,只要你没有用javascript处理提交内容 – hassan

+0

1. s ubmit按钮必须位于表单内(如果页面刷新用于提交表单)。 '$ connect'代码缺失 –

回答

0

你有一个表格里面的表格是错误的。像这样做:

<form method="POST" action="addtodatabase.php" > 
     <!-- Your inputs --> 
</form> 

而且,把你的提交表单内按钮:

<form method="POST" action="addtodatabase.php" > 
     <!-- Your inputs --> 
     <input type="submit" value="Submit" /> 
</form> 

或者,如果提交按钮外,按钮的form属性设置为formid

<form id="myForm" method="POST" action="addtodatabase.php" > 
     <!-- Your inputs --> 
</form> 

<input type="submit" value="Submit" form="myForm" /> 
+0

非常感谢你的帮助 –