Question

我试图用我的变量从php调出javascript警报。到目前为止，我的upload.php文件是：

<?php
if(isset($_POST['btn-upload']))
{
    $pic = rand(1000,100000)."-".$_FILES['pic']['name'];
    $pic_loc = $_FILES['pic']['tmp_name'];
    $folder="uploaded_files/";
    if(move_uploaded_file($pic_loc,$folder.$pic))
    {
        ?><script>alert('File successfully uploaded.\n![File Upload]('+window.location.href+')');
</script><?php
    }
    else
    {
        ?><script>alert('Sorry, error while uploading file. Please try again');</script><?php
    }
}
?>
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Photo Uploader for use in markdown</title>
</head>
<body>
<h2>Photo Uploader & Markdown generator</h2>
<p>Click 'Choose file' to choose the file to be uploaded. The filename should appear. Then click 'upload'. <br>A popup should show you what to copy and paste.</p>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="pic" />
<button type="submit" name="btn-upload">upload</button>
</form>
</body>
</html>

然后我的html代码看起来像（只包括相关部分）：

<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="pic" />
<button type="submit" name="btn-upload">upload</button>

此脚本的目的是将图片上传到服务器，然后显示缩减代码以供用户使用该图像。如果文件正确上传，我的目标是输出以下内容： ![Alternative Text](http://www.example.com/folder/photo.jpg)

我尝试了以下内容：

<?php
if(isset($_POST['btn-upload']))
{
    $pic = rand(1000,100000)."-".$_FILES['pic']['name'];
    $pic_loc = $_FILES['pic']['tmp_name'];
    $folder="uploaded_files/";
    if(move_uploaded_file($pic_loc,$folder.$pic))
    {
        ?><script>alert('File successfully uploaded.\n![File Upload]('+window.location.href+.$folder.'/'.$pic.')');</script><?php
    }
    else
    {
        ?><script>alert('Sorry, error while uploading file. Please try again');</script><?php
    }
}
?>
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Photo Uploader for use in markdown</title>
</head>
<body>
<h2>Photo Uploader & Markdown generator</h2>
<p>Click 'Choose file' to choose the file to be uploaded. The filename should appear. Then click 'upload'. <br>A popup should show you what to copy and paste.</p>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="pic" />
<button type="submit" name="btn-upload">upload</button>
</form>
</body>
</html>

这会生成一个上传文件但不显示js警报的工作网页。我也尝试了以下内容：

<?php
if(isset($_POST['btn-upload']))
{
    $pic = rand(1000,100000)."-".$_FILES['pic']['name'];
    $pic_loc = $_FILES['pic']['tmp_name'];
    $folder="uploaded_files/";
    <script>var folder = "<?php echo $folder ?>";</script>
    <script>var pic = "<?php echo $pic ?>";</script>
    if(move_uploaded_file($pic_loc,$folder.$pic))
    {
        ?><script>alert('File successfully uploaded.\n![File Upload]('+window.location.href+folder+'/'+pic+')');</script><?php
    }
    else
    {
        ?><script>alert('Sorry, error while uploading file. Please try again');</script><?php
    }
}
?>
<!DOCTYPE html>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Photo Uploader for use in markdown</title>
</head>
<body>
<h2>Photo Uploader & Markdown generator</h2>
<p>Click 'Choose file' to choose the file to be uploaded. The filename should appear. Then click 'upload'. <br>A popup should show you what to copy and paste.</p>
<form action="" method="post" enctype="multipart/form-data">
<input type="file" name="pic" />
<button type="submit" name="btn-upload">upload</button>
</form>
</body>
</html>

这导致http错误500

有什么建议吗？

非常感谢，

Answer 1

您可以在php中创建一个函数，并在需要调用警报的地方调用它。

功能：

function alertMsg($str) {
    print("<script>alert('$str')</script>");
}

并在php中调用

//string
alertMsg("Success");

//php variable
$alertMsg = "Some alert message";
alertMsg($alertMsg);

//even you can concatenate both
alertMsg("This is an alert. ".$alertMsg);

希望这会有所帮助。感谢。

Answer 2

在这两次尝试中，您都试图将PHP，HTML和JavaScript混合起来，就好像它们都是相同的语言一样。他们不是。从其中任何一个的角度来看，另一个中的代码只不过是一个字符串。他们不能直接分享变量和逻辑。

了解这一行：

alert('File successfully uploaded.\n![File Upload]('+window.location.href+.$folder.'/'.$pic.')');

正在尝试直接在JavaScript中使用PHP代码（变量和语法）。这只会导致JavaScript中出现语法错误，浏览器的开发控制台会向您指出这些错误。相反，将PHP代码括在<?php ?>标记中并回显结果：

alert('File successfully uploaded.\n![File Upload]('+window.location.href+'<?php echo $folder.'/'.$pic ?>'+')');

第二次尝试存在同样的问题，您将HTML / JavaScript直接放在PHP中：

$folder="uploaded_files/";
<script>var folder = "<?php echo $folder ?>";</script>

这导致PHP语法错误，PHP日志会告诉您（以及您正在获取的500内部服务器错误）。

PHP代码需要位于<?php ?>标记中。总是。所以这就像是：

$folder="uploaded_files/";
?>
<script>var folder = "<?php echo $folder ?>";</script>
<script>var pic = "<?php echo $pic ?>";</script>
<?php
if(move_uploaded_file($pic_loc,$folder.$pic))

另请注意，在HTML / JavaScript中，您不需要{em>每行 JavaScript代码的<script>标记。您可以在一个<script>元素中包含多行代码。

Answer 3

在JS中使用变量PHP

<?php

  if (isset($_POST['btn-upload'])) {

    $pic     = rand(1000,100000)."-".$_FILES['pic']['name'];
    $pic_loc = $_FILES['pic']['tmp_name'];
    $folder  = "uploaded_files";

    if (move_uploaded_file($pic_loc, $folder . '/' . $pic)) { ?> 

      <script>
        alert("File successfully uploaded! " + 
              "\n" + 
              location.hostname + 
              "<?php echo '/' . $folder . '/' . $pic; ?>");
      </script>

<?php

    } else { ?>

      <script>alert("Sorry, error while uploading file. Please try again");</script>

<?php

    }

  }

?>

location.hostname = $ _SERVER [＆＃39; HTTP_HOST＆＃39;] // localhost

如何使用PHP变量显示JavaScript警报？

3 个答案: