Ajax Wird nicht im div-Tag angezeigt

Question

Ich bin in der Lage, die Ergebnisse von der Abfrage zu erhalten, indem Sie zwar Ajax, aber anstelle der Ergebnisse im div-Tag, das auf mainInstructor2.php angezeigt wird, es auf einer anderen Seite anzeigt, die in Aktion aufgelistete InstructorStudent.php. Jede mögliche Hilfe würde geschätzt :)

Hier ist mein HTML:

<p id ="byStudent"> Search all the tutoring sessions by student</p> 
     <form id="SearchByStudent" action ="InstructorStudent.php" method="post"> 
      Student Username <input = "textbox" type="text" name ="studentSearch" id="studentSearch" value="" required /> 
      <button class ="button" type="submit" name ="seachStudentSubmit" id ="searchStudentSubmit" value="submit"> Submit </button> 
     </form> 

     <div id ="searchByStudentContainer"> 
     </div> 

Hier ist mein javascript:

<script> 

      $(SearchByStudent).submit(function(event){ 
       event.preventDefault(); 
       var form = $(this).serialize(); 
       var url = $form.attr('action'); 

       var posting = $.post(url, { studentSearch: $('input[name="studentSearch"]').val() }); 

       posting.done(function(data) { 
        var content = data; 
        $(searchByStudentContainer).html(content); 
       }); 
      }); 


    </script> 

Hier ist meine php:

<?php 

    session_start(); 

    include_once("Connection.php"); 

    $studentSearch = $_POST['studentSearch']; 

    $sql1 = "SELECT Students.StudentUsername 
      FROM Students 
      INNER JOIN CLASS_STUDENT 
      ON Students.StudentUsername = CLASS_STUDENT.StudentUsername 
      INNER JOIN Class 
      ON CLASS_STUDENT.ClassID = Class.ClassID 
      INNER JOIN Instructors 
      ON Class.InstructorID = Instructors.InstructorID 
      WHERE Students.StudentUsername = '$studentSearch'"; 


    $result1= mysqli_query($link, $sql1); 

    $resultarr = mysqli_fetch_assoc($result1); 

    $studentSea = $resultarr['StudentUsername']; 

    if(!$studentSea) { 

     echo " 
     <script> 
       alert ('Could not find student on file!'); 
       window.location ='mainInstructor2.php' 
      </script>"; 
    } 
    else { 
    $sql2 = "SELECT Students.StudentUsername, CLASS_STUDENT.Year, CLASS_STUDENT.Semester, TUTOR_STUDENT.Date, 
    TUTOR_STUDENT.TopicCovered, TUTOR_STUDENT.BeginTime, TUTOR_STUDENT.EndTime, TUTOR_STUDENT.TotalMinutes, 
    TUTOR_STUDENT.Categories, TUTOR_STUDENT.TutorUsername 
    FROM Students 
    INNER JOIN CLASS_STUDENT 
    ON Students.StudentUsername = CLASS_STUDENT.StudentUsername 
    INNER JOIN TUTOR_STUDENT 
    ON CLASS_STUDENT.StudentUsername = TUTOR_STUDENT.StudentUsername 
    WHERE Students.StudentUsername = '$studentSearch'"; 

    $result= mysqli_query($link, $sql2); 
      if ($result->num_rows > 0) 
      { 

       echo " 
        <table> 
         <tr> 
          <th>Student Username</th> 
          <th>Year</th> 
          <th>Semester</th> 
          <th>Date</th> 
          <th>Topic Covered</th> 
          <th>Begin Time</th> 
          <th>End Time</th> 
          <th>Total Minutes</th> 
          <th>Category</th> 
          <th>Tutor Username</th> 

         </tr> 
          "; 

          // output data of each row 
       while($row = $result->fetch_assoc()) 
       { 
         echo " 
          <tr> 
           <td>".$row["StudentUsername"]. "</td> 
         "; 
         echo " 
           <td>".$row["Year"]. "</td> 
         "; 
         echo " 
           <td>".$row["Semester"]. "</td> 
         "; 
         echo " 
           <td>".$row["Date"]. "</td> 
         "; 
         echo " 
           <td>".$row["TopicCovered"]. "</td> 
         "; 
         echo " 
           <td>".$row["BeginTime"]. "</td> 
         "; 
          echo " 
           <td>".$row["EndTime"]. "</td> 
         "; 
         echo " 
           <td>".$row["TotalMinutes"]. "</td> 
         "; 
          echo " 
           <td>".$row["Categories"]. "</td> 
         "; 
         echo " 
           <td>".$row["TutorUsername"]. "</td> 
         "; 

         echo "</tr>"; 
            } 
         echo "</table>"; 

       } 
      else 
       echo "This Student doesn't have any tutoring sessions."; 



     mysqli_close($link); 
    } 

?> 

Answer 1

Schauen Sie sich diese Linie:

<div id ="searchByStudentContainer"></div> 

$(searchByStudentContainer).html(content); 

searchByStudentContainer ist ID und Sie verwenden # nicht, während Sie darauf verweisen.

So diese Zeile ändern:

$('#searchByStudentContainer').html(content); 

Answer 2

versuchen, dies ich denke, u #

posting.done(function(data) { 
 
        var content = data; 
 
        $(#searchByStudentContainer).html(content); 
 
       });

Answer 3

document.getElementById verpasst ("searchByStudentContainer") innerHTML- = Inhalt. ;

können Sie versuchen, diese

anstelle dieses

$ (searchByStudentContainer) .html (Inhalt);