Question

我有2个下拉菜单（projectnumnmer和projectnaam），它们搜索表“ projecten”并显示可以选择的添加数据，效果很好！，尽管我希望这些下拉菜单可以找到彼此的匹配项，并且将相应的（匹配）数据填写到另一个字段中，反之亦然。

希望有人能帮助我，帮助将不胜感激。：-）

这是我现在拥有的代码：

<!-- ***************** Projectnummer ***************** -->

            <tr valign="top"> 
                <td><?php echo "$language019"; ?></td>
                <td>

<!-- ***************** Projectnummer selecteren uit database ***************** -->
<?php $result = mysqli_query($mysqli, "SELECT * FROM projecten ORDER BY pnummer ASC"); // using mysqli_query instead ?>
<img src="images/projectnummer.png">&nbsp;<select id="projectnummerselect" name="projectnummer2" onchange="myFunction2()">
<option selected><?php echo "$language176"; ?>.</option>

    <?php 
    while($res = mysqli_fetch_array($result)) {         

        echo "<option value='".$res['pnummer']."'>".$res['pnummer']."</option>";
    }
    ?>
</select>
<br />
<script>
function myFunction2() {
  var x = document.getElementById("projectnummerselect").value;
  document.getElementById("projectnummer").value = "" + x;
}
</script>
<input type="text" id="projectnummer" size="4" maxlength="6" name="projectnummer" onfocus="this.value='19'" value="<?php echo $_COOKIE["cprojectnummer"]; ?>">

<!-- ***************** /Projectnummer selecteren uit database ***************** -->
</td>
            </tr>
<!-- ***************** /Projectnummer ***************** -->


<!-- ***************** Projectnaam ***************** -->
            <tr valign="top"> 
                <td><?php echo "$language020"; ?></td>
                <td>

<!-- ***************** Projectnaam selecteren uit database ***************** -->
<?php $result = mysqli_query($mysqli, "SELECT * FROM projecten ORDER BY pnaam ASC"); // using mysqli_query instead ?>
<img src="images/projectnaam.png">&nbsp;<select id="projectnaamselect" name="projectnaam2" onchange="myFunction3()">
<option selected><?php echo "$language177"; ?>.</option>

    <?php 
    while($res = mysqli_fetch_array($result)) {         

        echo "<option value='".$res['pnaam']."'>".$res['pnaam']."</option>";
    }
    ?>
</select>
<br />
<script>
function myFunction3() {
  var x = document.getElementById("projectnaamselect").value;
  document.getElementById("projectnaam").value = "" + x;
}
</script>
<input type="text" id="projectnaam" size="40" name="projectnaam" value="<?php echo $_COOKIE["cprojectnaam"]; ?>">

<!-- ***************** /Projectnaam selecteren uit database ***************** -->


</td>
            </tr>
<!-- ***************** /Projectnaam ***************** -->

匹配并显示2个下拉菜单中的数据

0 个答案: