我的数据库连接是正确的,我可以选择和回显我的表的行。但是,我的表单数据没有更新我的数据库。我试图在同一页面上提交表单,所以我将该操作留空。这里是我的PHP:表单数据正在提交,但不会进入mySQL
<?php
if (isset($_POST['submit'])) {
$con=mysqli_connect("localhost","hey","password!","dbname");
$sql="INSERT INTO StudentList (StudentNum, LastName, FirstName, Address, City, State, Zip, Balance, FirstTermAttended)
VALUES('$_POST[inputID]','$_POST[inputLast]','$_POST[inputFirst]','$_POST[inputAddress]','$_POST[inputCity]','$_POST[inputState]','$_POST[inputZip]','$_POST[inputBalance]','$_POST[inputTerm]')";
echo "It worked";
}
?>
,在这里我的形式:
<form class="form-horizontal" role="form" method="post" action="">
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputID" placeholder="Student ID Number">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputLast" placeholder="Last Name">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputFirst" placeholder="First Name">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputAddress" placeholder="Street Address">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputCity" placeholder="City">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputState" placeholder="State">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputZip" placeholder="Zip">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputBalance" placeholder="Current Balance">
</div>
</div>
<div class="form-group">
<div class="col-sm-offset-1 col-sm-10">
<input type="text" class="form-control" id="inputTerm" placeholder="First Term Attended">
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<input type="submit" value="Save Student Information" id="submit" name="submit" class="btn btn-primary">
</div>
</form>
我也有这个在我的头,可以把它搞砸的事情了?
<?php
// Create connection
$con=mysqli_connect("localhost","hey","password!","dbname");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
你是不是错过了'mysqli_query()'?同时请清理从'$ _POST'收到的值 –
您提供的唯一命名输入是您的提交按钮。 –
[**如何防止PHP中的SQL注入?**](http://stackoverflow.com/q/60174/1415724)<这是一条消息/线索。 ;-) –