php代码在脚本标记下无法正常运行

时间:2014-04-01 10:58:08

标签: javascript php

我有以下PHP代码独立运行,但问题是它也在表单加载上运行 所以我把它放在标签里面

       <script>
    $(document).ready(function() {
    <?php
     $allowedExts = array("gif", "jpeg", "jpg", "png");
     $temp = explode(".", $_FILES["file"]["name"]);
     $extension = end($temp);
if ((($_FILES["file"]["type"] == "image/gif")
|| ($_FILES["file"]["type"] == "image/jpeg")
|| ($_FILES["file"]["type"] == "image/jpg")
|| ($_FILES["file"]["type"] == "image/pjpeg")
|| ($_FILES["file"]["type"] == "image/x-png")
|| ($_FILES["file"]["type"] == "image/png"))
&& ($_FILES["file"]["size"] < 20000)
&& in_array($extension, $allowedExts))
{
if ($_FILES["file"]["error"] > 0)
{
echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
}
else
{
if (file_exists("upload/" . $_FILES["file"]["name"]))
  {
  echo '<script>alert(" Logo with this name already exists.")</script>';
  }
else
  {
  move_uploaded_file($_FILES["file"]["tmp_name"],
  "upload/" . $_FILES["file"]["name"]);
  echo '<script>alert("project logo uploaded successfully")</script>';
  }
  }
  }
  else
  {
  echo '<script> alert("Invalid file")</script>';
       }
     ?>         
    });
    </script>

但问题是它正在执行上传文件但不生成警报 请帮忙!!

2 个答案:

答案 0 :(得分:2)

代码的输出;

 <script>
    $(document).ready(function() {
 <script> alert("Invalid file")</script>
    });
    </script>

应该是;

<script>
        $(document).ready(function() {
     alert("Invalid file");
        });
        </script>

答案 1 :(得分:1)

您在另一个<script>元素中不能拥有<script>元素。 通过validator运行生成的HTML。