通过ajax将多个变量值传递给php文件，运行mysql查询并检索数据

Question

更新:我想从mysql表中检索学生的全名，根据用户输入的学号、课程和学期的值。

用户在文本字段中键入学号，同时从选择列表中选择课程和学期。当用户在输入学号并选择课程后从选择列表中选择学期时，我希望在文本字段中检索学生姓名。

我现在可以在文本框中检索学生的名字了。有没有什么办法可以检索学生的中间名和姓氏，并将它们显示在两个不同的文本框中？请帮帮忙。

下面是我使用的代码：

HTML代码：

    <form id="formname1" action = "" method= "post">

  <dd>
        <table>


            <tr>
                <td>Roll No.<font color = "red"><B>*</B></font></td>
                <td><input type = "text" id = "roll_number" name = "roll_number"></td>


                <td>Course:<font color = "red"><B>*</B></font></td>
                <td>
                    <select name ="course_name" id = "course">
                        <option value = "">Select</option>
                        <option value = "B.Ed">B.Ed</option>
                        <option value = "M.Ed">M.Ed</option>
                    </select>
                </td>

                <td>Semester:<font color = "red"><B>*</B></font></td>
                <td>
                    <select name ="sem" id = "sem">
                        <option value = "">Select</option>
                        <option value = "1">1</option>
                        <option value = "2">2</option>
                        <option value = "3">3</option>
                        <option value = "4">4</option>

                    </select>
                </td>

            </tr>

            <tr>
                <td>Student Name:<font color = "red"><B>*</B></font></td>
                <td><input type = "text" id = "f_name" name = "s_name"></td>

                <td><input type = "text" id = "m_name" name = "m_name"></td>

                <td><input type = "text" id = "l_name" name = "l_name"></td>

            </tr>

        </table>

    </form>

AJX代码：

    <script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

<script>

            $(document).ready(function () {

            $("#sem").change(function(){

             var roll_no = $("#roll_number").val(); 
             var course = $("#course").val();
             var sem = $("#sem").val();
             $.post("retrieve_name.php",{
             r_no:roll_no,cname:course,semester:sem},

             function(data){
             $("#f_name").val((data[0].first_name));
             $("#m_name").val((data[0].middle_name));
             $("#l_name").val((data[0].last_name));
             });
             }); 
        });


</script>    

PHP文件：

      <?php

        $con=mysqli_connect("$host", "$username", "$password", "$db_name");


$rows1 = array[]; 

$sql = "select first_name, middle_name, last_name from students s, students_in_courses sin
where s.roll_number = sin.roll_number
AND s.roll_number = '$roll' 
AND sin.course_name = '$course'
AND sin.semester = '$sem' ";

$result = mysqli_query($con,$sql);

if(!$result)
    {
        echo "No records matching your query were found.";  
    }
    else
    {
        while ($rows = mysqli_fetch_array($result))
        {
            //echo $rows[first_name];
            //echo $rows[middle_name];
            //echo $rows[last_name];
            $rows1[]= $rows;
        }
    }
    ?>

Answer 1

输入不能是内容父标记。

更改：

$("#s_name").html(data);

至：

$("#s_name").val(data);

Answer 2

以下是正确的脚本代码：

$(document).ready(function () {

            $("#sem").change(function(){

             var roll_no = $("#roll_number").val(); 
             var course = $("#course").val();
             var sem = $("#sem").val();
             $.post("test.php",{
             r_no:roll_no,cname:course,semester:sem},
             function(data){
             $("#s_name").val(data);
             });
             });
 });