PHP，HTML JS MySQL動態下拉列表

Question

我有一個表格，我可以進行所有工作並提交，除了...用戶應該選擇部門，然后在4個缺陷字段中填充與該部門匹配的缺陷。 部門和缺陷4件作品。 如果我刪除4則3個作品，則刪除4＆3和2作品，依此類推。 我認為它正在丟失listindex，但我迷路了。 我也將各種變量更改為在每個Defect中也是唯一的，並且結果相同。 我只是腦子里凍結，語法有些我想念的東西。 以下是部門1和2的代碼。

    <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

修復。 我在Department中調用一個函數，需要添加單個函數調用並更改被調用函數以匹配。 哇，抽筋。