Question

在处理数据和填充方面需要帮助：

我正在创建Employee数据库，有两个页面： 1.添加员工 2.修改员工

添加员工，获取EMP名称，EMP ID，经理名称（下拉列表），位置（选择它们的无线电选项）等输入所有这些输入都写入数据库。

员工ID

        <div class="fb-input-number">
          <input id="item12_number_1" name="emp_number" min="0" max="999999999" autocomplete="off" data-hint="" required="" step="1" type="number">
        </div>
      </div>
       <select id="item13_select_1" name="Manager" data-hint="">
            <option id="item13_0_option" value="Tom" >
              TOM
            </option>
            <option id="item13_1_option" value="Harry" selected="selected">
              HARRY
            </option>
      </select>
       <label id="item16_0_label">
            <input id="item16_0_radio" checked="checked" name="radio_location" data-hint="" value="IND" type="radio">
            <span id="item16_0_span" class="fb-fieldlabel">
              INDIA
            </span>
        </label>
        <label id="item16_1_label">
            <input id="item16_1_radio" name="radio_location" value="EMEA" type="radio">
            <span id="item16_1_span" class="fb-fieldlabel">
              EMEA
            </span>
        </label>

在“修改员工”页面中，基于emplyee ID，我想要使用db到html页面的值来呈现页面。我正在使用类似的页面来修改条目。我可以按值= $ EMPID显示文本字段。如何为下拉列表和复选框执行相同操作？

员工ID

        <div class="fb-input-number">
          <input id="item12_number_1" name="emp_number" min="0" max="999999999" autocomplete="off" data-hint="" required="" step="1" type="number" value="11222">
        </div>
      </div>
       <select id="item13_select_1" name="Manager" data-hint="">
            <option id="item13_0_option" value="Tom" >
              TOM
            </option>
            <option id="item13_1_option" value="Harry" selected="selected">
              HARRY
            </option>
      </select>
       <label id="item16_0_label">
            <input id="item16_0_radio"  name="radio_location" data-hint="" value="IND" type="radio">
            <span id="item16_0_span" class="fb-fieldlabel">
              INDIA
            </span>
        </label>
        <label id="item16_1_label">
            <input id="item16_1_radio" name="radio_location" value="EMEA" type="radio" checked="checked">
            <span id="item16_1_span" class="fb-fieldlabel">
              EMEA
            </span>
        </label>

你能帮我吗？

基本上我想出了修改页面，它将从数据库中检索数据并以HTML格式显示。下面的代码没有帮助显示所选的值：

<?php
mysql_connect($SERVER_IP, $USERNAME, $PWD) or die(mysql_error());
mysql_select_db($DBNAME) or die(mysql_error());

$result=mysql_query("SELECT * from `Data` where Emp_ID = $MyEmpID ");
$info = mysql_fetch_array( $result );
$Location=$info['Location'];
?>

<html>
<body>
        <div class="fb-dropdown">
              <select id="item13_select_1" name="location" data-hint="">
                <option id="item13_0_option" value="BLR" <? if ($Location=="BLR") echo "selected"; ?> >
                  Bangalore
                </option>
                <option id="item13_1_option" value="EMEA" <? if ($Location=="EMEA") echo "selected"; ?> >
                  EMEA
                </option>
              </select>
            </div>
</body>
</html>

我该如何解决？

Answer 1

当您的脚本呈现选择列表时，如果其值与DB值匹配，则必须检查每个选项。如果是，请将所选属性添加到该选项。

Answer 2

以下解决了问题：

<?php
mysql_connect($SERVER_IP, $USERNAME, $PWD) or die(mysql_error());
mysql_select_db($DBNAME) or die(mysql_error());

$result=mysql_query("SELECT * from `Data` where Emp_ID = $MyEmpID ");
$info = mysql_fetch_array( $result );
$Location=$info['Location'];
$locations[0]="BLR";
$locations[1]="Provo";
$locations[2]="EMEA";
$locations[3]="APAC";
?>

<html>
<body>
        <div class="fb-dropdown">
              echo "
        <select name='location'>";

     for ($i=0; $i<=4; $i++)
     {
        if($Location == $locations[$i]){
          echo "<option selected value='$locations[$i]'>$locations[$i]</option>";
        }else {
          echo "<option value='$locations[$i]'>$locations[$i]</option>";
        }
     }

echo" </select>
            </div>
</body>
</html>

HTML / PHP将数据从DB渲染到HTML表单

2 个答案: