我试图制作一个将图书添加到图书馆书籍数据库的表格,但对于我的生活,我无法让它发挥作用。这是我的代码:
<div class="container-fluid">
<div class="row">
<div class="col-lg-12">
<h2 class="page-header">Add Books</h2>
<br>
<?php
if (isset($_POST['submit'])){
$host="*****"; // Host name
$username="*****"; // Mysql username
$password="******"; // Mysql password
$db_name="******"; // Database name
$conn = mysqli_connect($host, $username, $password, $db_name);
$name=$_POST['Name'];
$author=$_POST['Author'];
$published=$_POST['Published'];
$isbn=$_POST['ISBN'];
$category=$_POST['Category'];
$sql="INSERT INTO Library (Name, Author, Published, ISBN, Category)
VALUES('$name', '$author', '$published', '$isbn', '$category')";
$query = mysqli_query($conn,$sql);
if($query){
echo "Successfully Added!";
}else{
echo "Error"."<br />".$sql.mysqli_error($con);
}
}
?>
<br>
<form name="addBookForm" action="" method="post">
<fieldset class="form-group">
<label for="frmName">Book Name</label>
<input type="text" class="form-control" id="frmName" placeholder="Name" name="Name">
</fieldset>
<fieldset class="form-group">
<label for="frmAuthor">Author</label>
<input type="text" class="form-control" id="frmAuthor" placeholder="Author" name="Author">
</fieldset>
<fieldset class="form-group">
<label for="frmPublished">Published</label>
<input type="text" class="form-control" id="frmPublished" placeholder="Published" name="Published">
</fieldset>
<fieldset class="form-group">
<label for="frmISBN">ISBN</label>
<input type="text" class="form-control" id="frmISBN" placeholder="ISBN" name="ISBN">
</fieldset>
<fieldset class="form-group">
<label for="frmCategory">Category</label>
<select class="form-control" id="frmCategory" name="Category">
<option value="Politics/Philosophy">Politics/Philosophy</option>
<option value="History / Labour History">History / Labour History</option>
</select>
</fieldset>
<button type="submit" class="btn btn-secondary">Add Book</button>
</form>
</div>
</div>
</div>
我整天都在撞墙,是否有人能告诉我自己做错了什么?当我按下提交按钮时,它只是回发到页面而没有任何成功或错误消息,当我检查数据库时,没有插入数据。它就像刷新页面一样。
答案 0 :(得分:3)
您已收到if声明:
if (isset($_POST['submit'])){
$ _ POST [&#39;提交&#39;] ...好吧,它是一个表单输入,名称为submit。也许一个按钮? = P
<button type="submit" class="btn btn-secondary">Add Book</button>
GASP!没有名字!
<button type="submit" name="submit" class="btn btn-secondary">Add Book</button>
我们走了。