Question

第一次在这里张贴所以要善良！出于某种原因，当我的PHP脚本到达时 if（$ beds ==＆＃39; nopref＆＃39;）我可以显示消息的唯一方法是使用echo $ message。使用javascript的以下行不会像我的其他页面一样显示它。有任何想法吗？代码：

<?php    
$beds = $_POST['beds'];
$orientation = $_POST['orientation'];
$checkin = $_POST['checkin'];
$checkout = $_POST['checkout'];
$conn = mysqli_connect("localhost", "user", "password", "name"); 
    if(!$conn) 
    { 
       die("Connection failed: " . mysqli_connect_error()); 
    } 
    if ($beds == 'nopref')
    {
        $message = "Please select how many beds you'd like";
        echo $message;
        echo "<script type='text/javascript'>alert('$message');</script>";
    }
 ?>

Answer 1

由于'you'd，您的文字已损坏。试试这个：

$message = htmlspecialchars("Please select how many beds you'd like", ENT_QUOTES);

Answer 2

你可以逃避＆＃39;在JavaScript中，如\＆＃39;

 if ($beds == 'nopref')
    {
        $message = "Please select how many beds you\'d like";
        echo $message;
        echo '<script>alert("'.$message.'");</script>';
    }

Answer 3

试试这个

 print "<script type='text/javascript'>window.alert('$message');</script>";

或

print "<script type='text/javascript'>window.alert('Please select how many beds you had like');</script>";

Answer 4

试试这段代码：

<?php
   $message=htmlspecialchars("Hello World's",ENT_QUOTES);
   echo "<script>alert('$message')</script>";
?>

在PHP中没有显示Javascript错误消息

4 个答案: