【问题标题】:Change elements of a form in php based on the selection from a drop-down list using javascript根据使用 javascript 从下拉列表中的选择更改 php 中表单的元素
【发布时间】:2020-01-22 00:17:37
【问题描述】:

根据部门选择,我已经预先定义了要在表格中使用的设备列表。当我选择一个部门时,设备清单需要相应地改变。该脚本似乎根本没有被调用。

在 HTML 中:

<select name="department_id" onchange="updateEquipmentList()" >
    <option value="2">Department 2</option>
    <option value="3">Department 3</option>
    <option value="4">Department 4</option>
</select>

在 Javascript 中:

<script>
    function updateEquipmentList()
    {
        var myDepartment = document.getElementById("department_id").value;
        switch(myDepartment){
            case 2:
                equipment_list = <?php echo $equipment_2; ?>
                break;
            case 3:
                equipment_list = <?php echo $equipment_3; ?>
                break;
            case 4:
                equipment_list = <?php echo $equipment_4; ?>
                break;
        }
        <?php $str_equipment_list = "<script>document.write(equipment_list);</script>"; ?>
    }
</script>

【问题讨论】:

    标签: javascript php dropdown


    【解决方案1】:

    这是完整的代码

    <?php
    $equipment_2 = 'Hammer';
    $equipment_3 = 'Tape';
    $equipment_4 = 'Cutter';
    ?>
    <!DOCTYPE html>
    <html>
    <head></head>
    
    <body>
    <div >
        <select id="department_id" name="department" onchange="updateEquipmentList()" >
        <option value="2">Department 2</option>
        <option value="3">Department 3</option>
        <option value="4">Department 4</option>
    </select>
    </div>
    
    
    <script>
        function updateEquipmentList()
        {
            var myDepartment = document.getElementById("department_id").value;
            var equipment_list = '';
            switch(myDepartment){
                case '2':
                    equipment_list = '<?php echo $equipment_2; ?>';
                    break;
                case '3':
                    equipment_list = '<?php echo $equipment_3; ?>';
                    break;
                case '4':
                    equipment_list = '<?php echo $equipment_4; ?>';
                    break;
                default:
                    equipment_list = 'Nothing';
    
            }
            document.write(equipment_list);
    
        }
    </script>
    </body>
    </html>
    

    【讨论】:

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