Question

嘿，我有多个下拉选项，我想抓住它的值。一般的想法是从每个单独的下拉列表中获取值以通过ajax调用。下面是3个代码片段。前两个是我想要收集的值，第三个是通过调用。任何帮助表示赞赏。谢谢。

    <div id="year"  class="dropdownmenu styled-select">
    <label>Year:</label>
        <select name="year" id="yearlist" onchange="getId(this.value);">
            <option value="">Select Year</option>

        <?php       
        $query = "select distinct(Year) from websitemasterlist order by Year ASC";
        $results = mysqli_query($conn, $query);

        foreach($results as $info) {
        ?>
            <option value="<?php echo $info[Year]; ?>"><?php echo $info[Year]; ?></option>

        <?php
        }
        ?>
        </select>
    </div>    

    <div id="make"  class="dropdownmenu styled-select">
    <label>Make:</label>
        <select name="make" id="makelist" onchange="getId2(this.value);">
            <option value="">Select Make</option>
        </select>
    </div>

    <div id="model"  class="dropdownmenu styled-select">
    <label>Model:</label>
        <select name="model" id="modellist" onchange="getId3(this.value);">
            <option value="">Select Model</option>
        </select>
    </div>

继承人收集这些信息的代码我想知道如何收集年份并在1更改时生成。

function getId(val){
    jQuery.ajax({
                method: "POST",
                url: "http://comugg.com/getdata.php",
                data: "year="+val,
                success:function(data){
                    jQuery("#makelist").html(data);
                }
    });
}

Answer 1

您可以像这样更改功能以发布更多数据。更多信息和jquery示例http://api.jquery.com/jquery.ajax/#example-0

function getId(val){
    var modelVariable = $( '#model:selected' ).val();
    jQuery.ajax({
                method: "POST",
                url: "http://comugg.com/getdata.php",
                data: {
                    year : val,
                    make : makeVariable
                },
                success:function(data){
                    jQuery("#makelist").html(data);
                }
    });
}

试图弄清楚如何通过onchange =“”传递多个值

1 个答案: