Question

您好我正在开发一个从数据库中提取记录的php下拉列表。我有第一次和第二次下降工作。但是我需要帮助实现第3次下拉。这是我到目前为止所拥有的。前2个工作正常。我真的需要让第三个工作。它依赖于第一个和第二个的输入。

的index.php

   <?php
require_once("dbcontroller.php");


$query ="SELECT * FROM campus";

?>
<html>
<head>
    <TITLE>Campus and Faculty Select</TITLE>
<head>
<style>
        body{width:610px;}
        .frmDronpDown {border: 1px solid #F0F0F0;background-color:#C8EEFD;margin: 2px 0px;padding:40px;}
        .demoInputBox {padding: 10px;border: #F0F0F0 1px solid;border-radius: 4px;background-color: #FFF;width: 50%;}
        .row{padding-bottom:15px;}
</style>
    <script src="jquery.min.js" type="text/javascript"></script>
<script>
    function getcampus_id(val) {
        $.ajax({
        type: "POST",
        url: "get_faculty.php",
        data:'campus_id='+val,
        success: function(data){
            $("#faculty-list").html(data);
        }
        });
    }

    function getfaculty_id(val) {
        alert(val)
        $.ajax({
        type: "POST",
        url: "get_course.php",
        data:'faculty_id='+val,
        success: function(data){
            $("#course-list").html(data);
        }
        });
    }

    function selectcampus_id(val) {
    $("#search-box").val(val);
    $("#suggesstion-box").hide();
    }
</script>
</head>
<body>
<form id="rightcol" action="form.php" method="post" class="container">
    <div class="frmDronpDown">
        <div class="row">
            <label>Campus:</label><br/>
                <select name="campus" id="campus-list" class="demoInputBox" onChange="getcampus_id(this.value);">
                    <option value="">Select Campus</option>
                    <?php
                                $query ="SELECT * FROM campus";
                                $result = mysqli_query($con, $query);
                                //loop
                                while($campus = mysqli_fetch_assoc($result)) :
                                //foreach ($result as $campus){
                            ?>
                        <option value="<?php echo $campus["campus_id"]; ?>"> <?php echo $campus["name"]; ?></option>
                        <?php
                        endwhile;
                        ?>
                </select>
        </div>
        <div class="row">
            <label>Faculty:</label><br/>
                <select name="faculty" id="faculty-list" class="demoInputBox" onChange="getfaculty_id(this.value);>
                    <option value="">Select Faculty</option>
                </select>
        </div>
        <div class="row">
            <label>Course:</label><br/>
                <select name="course" id="course-list" class="demoInputBox" >
                    <option value="">Select Course</option>
                </select>
        </div>
    </div>
</body>
</html>

get_faculty.php

<?php

error_reporting(E_ALL); 
ini_set('display_errors', 1);

require_once("dbcontroller.php");

if(!empty($_POST["campus_id"])) {
    $campus_id = $_POST["campus_id"];
    $query ="SELECT * FROM faculty WHERE campus_id = $campus_id";
    $result = mysqli_query($con, $query);

    if ( $result === false ) {
        echo mysqli_error($con);
        exit;
    }

    echo '<option value="">Select Campus</option>';

    while ( $faculty = mysqli_fetch_assoc($result) ) :  
        echo '<option value="' . $faculty['faculty_id'] . '">';
        echo $faculty['faculty_name']; 
        echo '</option>';
    endwhile;
    }
?>

最后是get_course.php

<?php

error_reporting(E_ALL); 
ini_set('display_errors', 1);

require_once("dbcontroller.php");

if(!empty($_POST["campus_id"])) {
    $course_id = $_POST["campus_id"].$POST_["faculty_id"];
    $query ="SELECT * FROM `course` WHERE `campus_id`and `faculty_id` = $course_id";
    $result = mysqli_query($con, $query);

    if ( $result === false ) {
        echo mysqli_error($con);
        exit;
    }

    echo '<option value="">Select Campus</option>';

    while ( $course = mysqli_fetch_assoc($result) ) :   
        echo '<option value="' . $course['course_id'] . '">';
        echo $faculty['course_name']; 
        echo '</option>';
    endwhile;
    }
?>

Answer 1

您错过了第二个onchange的{{1}}的结束语。

使用此

select 的修改
我试图复制你想要做的事情并让它发挥作用。以下是包含我在此gist

中使用的数据库的


    <select name="faculty" id="faculty-list" class="demoInputBox" onChange="getfaculty_id(this.value);">
       <option value="">Select Faculty</option>
    </select>

编辑的文件

我编辑的文件是

的index.php
在post请求中包含mysqldump以获取course_list，以便我们可以将其用于外键
```
campus_id
```
get_course.php
更正您访问function getfaculty_id(val) { $.ajax({ type: "POST", url: "get_course.php", data:{ 'faculty_id':val, 'campus_id':$('#campus-list').val() }, success: function(data){ $("#course-list").html(data); } }); }数据的方式，不应该$_POST 正确的$POST_["faculty_id"]正如@sean建议的那样在while循环中使用$query代替$course来创建$faculty
```
select
```

将另一个下拉列表添加到动态php，Mysql表单以及其他2个下拉列表

1 个答案: