【问题标题】:using JQuery to change a second drop down selection box from mysql DB使用 JQuery 从 mysql DB 更改第二个下拉选择框
【发布时间】:2013-08-27 04:03:53
【问题描述】:

当第一个框选择了汽车制造时,我试图让第二个下拉框更改和填充。它可以很好地连接到数据库,如果我在第二个下拉列表中输入“Honda”而不是“carmake3”,那么它会给我一个所有本田汽车的列表。我只需要根据用户的选择工作。任何帮助将不胜感激!

<script src="http://code.jquery.com/jquery-1.9.1.js"></script>
<form action="step-3" method="post">
Car 3
</div>

<?php
mysql_connect('localhost', '**********', '**********');
mysql_select_db('**********');
$sql = "SELECT Make FROM CarMakes";
$result = mysql_query($sql);
echo "<select name='carmake3'>";
 echo "<option value='Make'>Make</option>"; 
while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['Make'] . "'>" . $row['Make'] . "</option>";}
echo "</select>";
?>

<script type="text/javascript">
jQuery(document).ready(function() {
    jQuery('carmodel3').change(function() {
        jQuery('carmake3').change();
    });
});
</script>

<?php
mysql_connect('localhost', '**********', '**********');
mysql_select_db('**********');
$sql1 = "SELECT * FROM myTable WHERE Make='carmake3'";
$result1 = mysql_query($sql1);
echo "<select name='carmodel3'>";
  echo "<option value='Model'>Model</option>"; 
while ($row = mysql_fetch_array($result1)) {
  echo "<option value='" . $row['Model'] . "'>" . $row['Model'] . "</option>";}
echo "</select>";
?>
<input type="text" maxlength="4" name="car3year" placeholder="year" class="WriteInBox"/>

<input type="text" maxlength="6" name="car3miles" placeholder="miles" class="WriteInBox"/>

</div>

</form>

【问题讨论】:

  • 很多事情需要纠正..
  • 请帮帮我。一些技巧。或者如果您想为我重做,我将不胜感激
  • 我很乐意,但不是为了那个...你需要通过elance联系我。
  • 我刚刚发布了一份工作并邀请了你。如果你不明白,请告诉我。我需要尽快完成这项工作
  • 你也有答案...

标签: jquery mysql select


【解决方案1】:

为 Make 执行下面给出的 html

$sql = "SELECT Make FROM CarMakes";
$result = mysql_query($sql);
<select id="carmake3" name='carmake3' onchange="get_makes();">
<option value='Make'>Make</option>
<?php while ($row = mysql_fetch_array($result)) {
    echo "<option value='" . $row['Make'] . "'>" . $row['Make'] . "</option>";}
?>
</select>
<div id="get_car_make"></div> // Sub will be appended here using ajax

写一个ajax函数get_makes();

<script type="text/javascript">
function get_makes() { // Call to ajax function
    var carmake3 = $('#carmake3').val();
    var dataString = "carmake3="+carmake3+"&redirecturl="+redirecturl;
    $.ajax({
        type: "POST",
        url: "getcarmakenames.php", // Name of the php files
        data: dataString,
        success: function(html)
        {
            $("#get_car_make").html(html);
        }
    });
}
</script>

文件 getcarmakenames.php - 将从下面的文件中获取子文件,该文件将附加到 div

if ($_POST) {
    $carmake3 = $_POST['carmake3'];
    if ($carmake3 != '') {
       $sql1 = "SELECT * FROM myTable WHERE Make='carmake3'";
       $result1 = mysql_query($sql1);
       echo "<select name='carmodel3'>";
       echo "<option value='Model'>Model</option>"; 
       while ($row = mysql_fetch_array($result1)) {
          echo "<option value='" . $row['Model'] . "'>" . $row['Model'] . "</option>";}
       echo "</select>";
    }
    else
    {
        echo  '';
    }
}

【讨论】:

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