Question

<!doctype html>
<html>
  <head>
    <title>Post Answer</title>
  </head>
  <body>
    <?php
    include ("connection.php");

      $result = mysqli_query($con, "SELECT * FROM student_table WHERE SID='$_POST[sid]'") or die('Query failed');
      $tmp = mysqli_fetch_array($result);
      $nn = $tmp['nickname'];
      $a = $_POST['answer'];
      $sid = $tmp['SID'];
      $c = $tmp['course'];
      $q_id = $_POST['q_id'];
      $sql = mysqli_connect("127.0.0.1", "root", "", "project");
      $sql = "INSERT INTO answer_table (nickname, answer, SID, vote, course) VALUES ('$nn', '$a', '$sid', '', '$c')";

      if (!mysqli_query($con, $sql)) {
        die('Error: ' . mysqli_error($con));
      }

      echo ("<SCRIPT LANGUAGE='JavaScript'>
                window.location.href='question.php';
                window.alert('Thank You! Your answer is online now.')
                  </SCRIPT>");
    ?>
  </body>
</html>

at＆＃34; question.php＆＃34;文件，我想收到＆＃34; $ q_id＆＃34;作为"$_GET['q_id'];"

我可以通过＆＃34; $ q_id＆＃34;第17行到＃34; question.php＆＃34;第26行的页面？

Answer 1

只需将其添加为javascript中的查询字符串

即可

echo ("<script language='JavaScript'>
         window.location.href='question.php?q_id=$q_id';
         window.alert('Thank You! Your answer is online now.')
       </script>");

Answer 2

虽然已经晚了4年，但是@Seth的答案可能不正确。将PHP变量放在Javascript中无效，必须将其回显。

echo ("<script language='JavaScript'>
     window.location.href='question.php?q_id=<?php echo $q_id ?>';
     window.alert('Thank You! Your answer is online now.')
   </script>");

“ <？php echo $ q_id？>”会将$ q_id中包含的值传递给JS函数。

通过“window.location.href”传递参数

2 个答案: