相关疑难解决方法(0)

嗨有谁知道如何避免/停止重复插入的PHP和HTML？每当我刷新时,相同的数据重复,这不是我想要的.我可以从这里改变什么吗？我听说添加UNIQUE INDEX可以避免/阻止它.也许它有效,但我错误地放置它使它无法工作.有帮助吗？提前致谢!!!真的很感激!

<?php
session_start();
include("Validation.php");

$connect=mysqli_connect("localhost","root","","jailbird");
if(mysqli_error($connect))
{
die("Could not connect.");
}

if(isset($_POST["insert_click"]))
{
//Bookingid is auto increment, therefore no need
//$Bookingid=$_POST["BookingID"];
$Prisonerid=$_SESSION['Prisonerid'];
$Visiting_method=$_POST["VisitingMethod"];
$Visiting_location=$_POST["VisitingLocation"];
$Date=$_POST["Date"];
$Time=$_POST["Time"];

$query=$connect->prepare("insert into Booking(PrisonerID, VisitingMethod, VisitingLocation, Date, Time) values (?,?,?,?,?)");
$query->bind_param('sssss', $Prisonerid, $Visiting_method, $Visiting_location, $Date, $Time);
$query->execute();
}

$query=$connect->prepare("select * from booking WHERE Prisonerid=?");
$query->bind_param('s',$_SESSION['Prisonerid']);
$query->execute();
$query->bind_result($Bookingid, $Prisonerid, $Visiting_method, $Visiting_location, $Date, $Time);

while($query->fetch())
{
echo "<tr>";
//echo "<td width=60>".$Bookingid."</td>";
echo "<td>$Prisonerid</td>";
echo "<td>$Visiting_method</td>";
echo "<td>$Visiting_location</td>";
echo "<td>$Date</td>";
echo "<td>$Time</td>";
echo "</tr>";
}   
?>

因此，我遇到了一个问题，即如果我单击足够快的提交按钮，我的表单将被提交几次。我该如何预防？令牌是自动添加的，但我猜对它没有帮助。表格范例：

  <div class="row padding-10">
    {!! Form::open(array('class' => 'form-horizontal margin-top-10')) !!}
    <div class="form-group">
      {!! Form::label('title', 'Title', ['class' => 'col-md-1 control-label padding-right-10']) !!}
      <div class="col-md-offset-0 col-md-11">
      {!! Form::text('title', null, ['class' => 'form-control']) !!}
      </div>
    </div>
    <div class="form-group">
      {!! Form::label('body', 'Body', ['class' => 'col-md-1 control-label padding-right-10']) !!}
      <div class="col-md-offset-0 col-md-11">
      {!! Form::textarea('body', null, ['class' => 'form-control']) !!}
      </div>
    </div>
    <div class="col-md-offset-5 col-md-3">
      {!! Form::submit('Submit News', ['class' => 'btn btn-primary form-control']) !!}
    </div>
    {!! Form::close() !!}
  </div>

我的NewsController存储方法：

    public function store()
{
$validator = …

如何让PHP打印内联错误而不是更改整个页面？

我希望它能够定位#errors和填充,而不是改变一切.

我目前使用的代码是 die ("Incorrect username or password.");

我是PHP的新手,很抱歉,如果这是一件非常简单的事情.