我正在尝试在执行查询后立即刷新页面。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。当前页面网址为http://localhost/test/test.php?post=1
<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>
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>";
}
}
答案 0 :(得分:3)
您可以使用此代码
<?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" => $description
));
header('Location: http://localhost/test/test.php?post=1'); exit();
}
?>
您也可以使用此代码
<?php
if (isset($_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();
}
?>