Question

我正在创建一个类Web页面，允许您通过单击与该类关联的链接来查看类详细信息。我有一个数据库，里面装有班级，注册的学生以及指定的教授。下面我在每个循环中有一个查询，从结果中创建一个表，但是我想创建指向该类页面的条目URL。例如：“localhost / class？class_id = cs120”。我不能只在下面添加一个链接，并将其参数化为cs120，因为所有链接都会去那里。我对SQLITE3，PHP和HTML的理解是有限的，下面是我对此页面的代码。

    <!DOCTYPE html>
    <?php include("secure.php") ?>
    <html lang= "en">
    <head>
        <title> Class Index </title>
        <meta charset= "utf-8" />
        <link rel="stylesheet" href="class_coverpage.css" />
    </head>

    <?php
      //echo("Hello Retrieving Table");
      $username = $_SESSION['user'];
      //echo 'Hello: ' . $username;
      $classquery = "SELECT DISTINCT Classes.Class_id, Classes.Section_id ,Classes.className, Classes.Subject, Classes.Location FROM Classes,StudentClasses,Students WHERE StudentClasses.Class_id = Classes.Class_id AND StudentClasses.Student_id = Students.Student_id AND StudentClasses.Section_id = Classes.Section_id AND Students.Username = '$username'; ";
      //$teacherquery = "SELECT TeacherClasses.Teacher_id, TeacherClasses.Class_id FROM TeacherClasses";
      //echo("Trimming Query");
      trim($classquery);
      //echo("Stripping Slashes");
      $classquery = stripslashes($classquery);
      //echo("Setting up Query");

      $results = $db->query($classquery);
      if (!$results){
        echo("<h2>Error: The query could not be executed.</h2>");
        $error = $db->lastErrorMsg();
        echo("<p>$error<p>");
        exit;
      }

      echo "<table><tr>";
      echo "<td>Course ID:</td>";
      echo "<td>Section ID:</td>";
      echo "<td>Subject:</td>";
      echo "<td>Location:</td></tr>";
      echo("<tr>");
      for($i=0;$i<$num_cols;$i++){
        $head = $results->columnName($i);
        echo("<th>$head</th>");
      }
      echo ("</tr>");
      // Write rows into table
      $ct = 0;
      while ($row = $results->fetchArray(SQLITE3_ASSOC)) {
        echo ("<tr>");
        foreach($row as $v){
            // I want to Make this section down here create a link hopefully.
            //echo ("<td><a href='class_detail.php?Class_id=$row['id']'> $v </a></td>");
        }
        $ct = $ct + 1;
        echo ("</tr>");
      }
      echo 

("</table>");

?>

从数据库结果制作的表中创建可点击链接

0 个答案: