2017-05-08 41 views
-1

我有以下代码,它显示了我的书籍表中的所有书籍。使用mySQL从“select”语句中删除数据库列

<select id="booksToDelete"> 
    <?php 
     foreach($bookTitles as $bookTitle) { 
     echo '<option>'; 
     echo $bookTitle['Title']; 
     echo '</option>'; 
     } 
    ?> 
    </select> 

<input type="submit" name="deleteBook" id="deleteBook" value="Delete Book" /> 
    <?php 

我希望能够根据用户在html页面上选择的选项删除任何图书,但我不知道如何完成它。这是我到目前为止:

if(isset($_POST['deleteBook'])) 
     $deleteQuery= 'DELETE FROM book WHERE Title=**I don't know what goes here**; 
    ?> 

请帮忙!!

+1

标题出现在引号中,但“标题”不是唯一的,你应该使用ID。如果问题是如何在选项中添加值,则需要添加属性“值”。 – chris85

回答

0

正如chris85在评论部分指出的那样,您应该使用ID s来确定要在表格中删除的内容。如果有书籍具有相同的标题,则使用title会导致多重删除。

让我们修复您的option标签。我们将使用value属性将ID作为值,而不是title。让我们也为您的select字段使用name属性。

这是在一个form标记里面吗?

<form method="POST" action=""> <!-- PUT THE PAGE WHERE YOU WILL PROCESS THE REQUEST TO THE ACTION ATTRIBUTE --> 
    <select name="booksToDelete" id="booksToDelete"> 
<?php 
    foreach($bookTitles as $bookTitle) { 
     echo '<option value="'.$bookTitle['id'].'">'; /** REPLACE CORRESPONDING INDEX WITH THE UNIQUE ID COLUMN OF THE book TABLE **/ 
     echo $bookTitle['Title']; 
     echo '</option>'; 
    } 
?> 
    </select> 
    <input type="submit" name="deleteBook" id="deleteBook" value="Delete Book" /> 
</form> 

然后一旦提交。顺便说一下,你用什么mysql API来连接你的数据库或执行查询?对于这个例子,让我们使用mysqli_*'s prepared statement

/** ESTABLISH YOUR CONNECTION **/ 
$con = new mysqli("host", "user", "password", "database"); /** REPLACE NECESSARY DATA **/ 

/* CHECK CONNECTION */ 
if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 

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

    $deleteQuery = "DELETE FROM book WHERE id = ?"; /** REPLACE NECESSARY UNIQUE ID COLUMN **/ 
    /*         ^  */ 

    $stmt = $con->prepare($deleteQuery); 
    $stmt->bind_param("i", $_POST["booksToDelete"]); 
    $stmt->execute(); 
    $stmt->close(); 

}