我试图在查询执行后立即刷新页面。该页面能够显示弹出消息“成功”。但是，它无法正确刷新页面，只能无限显示弹出消息。当前网页的网址是http://localhost/test/test.php?post=1PHP刷新页面有'？var = value'

<form method="post" role="form" action="reply.php"> 
    <fieldset> 
    <div class="form-group"> 
    <textarea name="reply" class="form-control" rows="3" placeholder="Comment" required autofocus=""></textarea> 
    </div>   
    <button name="post" type="submit" class="[ btn btn-success ]" data-loading-text="Loading...">Post reply</button> 
     </fieldset> 
</form>

reply.php

if (isset($_POST['post'])) {    
       $description =$_POST['reply']; 
       $stmt = "INSERT INTO reply (comments) VALUES (:description)"; 
       $p = $MySQLi_CON -> prepare($stmt); 

       $results = $p -> execute(array(
       ":description" => $description 
       )); 
       echo '<script language = "javascript">'; 
       echo 'alert("Successful")'; 
       echo '</script>'; 
       echo "<script> location.reload(true); </script>"; 

       if(!$results){ 
         echo '<script language = "javascript">'; 
         echo 'alert("Fail")'; 
         echo '</script>'; 
         echo "<script> location.reload(true); </script>";    
       }   
    }

来源

2017-02-28 fypforstack

JavaScript的JavaScript头？只需使用：header（'location：URL to same page'）;或者你可以使用jQuery与Ajax，这也是一个更好的解决方案，然后在PHP中使用JavaScript！ – Soheyl

我试图找到相同的页面，它是http：//localhost/test/test.php？post = 1。如果我尝试了这种方法，它只会去http：//localhost/test/test.php – fypforstack

谁说的？！你可以在标题的URL中使用任何你想要的东西！只需使用：header（'location：localhost/test/test.php？post = 1'）; – Soheyl

你为什么不使用使用此代码

<?php 
if (isset($_POST['post'])) { 
      $description =$_POST['reply']; 
      $stmt = "INSERT INTO reply (comments) VALUES (:description)"; 
      $p = $MySQLi_CON -> prepare($stmt); 

      $results = $p -> execute(array(
      ":description" => $description 
      )); 
     header('Location: http://localhost/test/test.php?post=1'); exit(); 
} 
?>

您也可以使用此代码

<?php 
if (isset($_POST['post'])) { 
$description = $_POST['reply']; 
$stmt = "INSERT INTO reply (comments) VALUES (:description)"; 
$p = $MySQLi_CON->prepare($stmt); 

$results = $p->execute(array(
    ":description" => $description 
)); 
header("Location: " . $_SERVER['HTTP_REFERER']); 
exit(); 
} 
?>

来源

2017-02-28 13:45:58

我试过header（'Location：http：//localhost/test/test.php '），但test.php？post = 1是'？var = value'。该URL不是固定的。 1是post_ID – fypforstack

@fypforstack的价值，这基本上是我说的:) – Soheyl

@soheyl 但我可能有例如。我只想刷新当前页面我在 test.php？post = 1 test.php？post = 2 test.php？post = 3 – fypforstack

PHP刷新页面有'？var = value'

reply.php

回答

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