PHP、HTML JS MySQL 动态下拉菜单

Question

我有一个表格，除了... 用户应该选择部门并且 4 个缺陷字段应该填充与部门匹配的缺陷之外，我所有的东西都在工作和提交。部门和缺陷 4 工作。如果我删除 4 然后 3 作品，删除 4&3 和 2 作品等等。我认为它正在丢失 listindex 但我迷路了。我已将各种变量更改为在每个缺陷中也是唯一的并且结果相同。我只是大脑冻结，语法有些东西我错过了一些东西。以下是 Dept、Defect 1 和 2 的代码。

    <strong>Department Name:</strong>
        <select name=Department_Nam class="required-entry" id="Department_Nam" onchange="javascript: dynamicdropdown(this.options[this.selectedIndex].value);">
            <option value="">Select Department</option>
            <?php if ($resultdep->num_rows > 0) {
              while($row = mysqli_fetch_assoc($resultdep)) {?>
                <option value="<?php echo $row['Department_Nam']; ?>"> <?php echo $row['Department_Nam']; ?></option>

          <?php        }
             } ?>

        </select>
    </div>

    <strong>Defect:</strong><br>
        <script>
            document.write('<select name="DefectDescrip" id="DefectDescrip"> <option value="">Please select defect</option></select>')
        </script>
        <noscript>
            <select name="DefectDescrip" id="DefectDescrip">
                <option value="">Please select defect</option>
            </select>
        </noscript>
    </div>
    <script>
        var rowFrameworkResultInJs =<?php echo json_encode($rowFrameworkResult);?>;
        function dynamicdropdown(listindex)
        {
            document.getElementById("DefectDescrip").length = 0;
            document.getElementById("DefectDescrip").options[0]=new Option("Please select defect","");
            if (listindex) {
                var lookup = {};
                var j = 1;
                for (var i = 0, len = rowFrameworkResultInJs.length; i < len; i++) {
                    if (rowFrameworkResultInJs[i].Reject_Code_Department == listindex) {
                        document.getElementById("DefectDescrip").options[j]=new Option(rowFrameworkResultInJs[i].Reject_Code_Descrip,rowFrameworkResultInJs[i].Reject_Code_Descrip);
                        j = j+1;
                    }
                }
            }
         return true;   
        }
   </script>


   <br><strong>Defect 2:</strong><br>
        <script>
            document.write('<select name="DefectDescrip2" id="DefectDescrip2"><option value="">Please select defect</option></select>')
        </script>
        <noscript>
            <select name="DefectDescrip2" id="DefectDescrip2">
                <option value="">Please select defect</option>
            </select>
        </noscript>
    </div>
    <script>
        var rowFrameworkResultInJs =<?php echo json_encode($rowFrameworkResult);?>;
        function dynamicdropdown(listindex)
        {
            document.getElementById("DefectDescrip2").length = 0;
            document.getElementById("DefectDescrip2").options[0]=new Option("Please select defect","");
            if (listindex) {
                var lookup = {};
                var j = 1;
                for (var i = 0, len = rowFrameworkResultInJs.length; i < len; i++) {
                    if (rowFrameworkResultInJs[i].Reject_Code_Department == listindex) {
                        document.getElementById("DefectDescrip2").options[j]=new Option(rowFrameworkResultInJs[i].Reject_Code_Descrip,rowFrameworkResultInJs[i].Reject_Code_Descrip);
                        j = j+1;
                    }
                }
            }

            return true;
        }
   </script>

Answer 1

修复它。我正在调用部门中的一个函数，需要添加单个函数调用并更改被调用函数以匹配。哇，头抽筋。