Question

我正在使用对话框在提交时显示表单和回显消息。这很好用。我还有一个下拉菜单，从中选择值填充第二个下拉菜单。这在它自己很好用，但如果我尝试合并这两个，那么div id =“divId”没有显示或填充更改。现在我将是第一个承认我的jQuery知识至少是基本的，所以我很感激专家的帮助。我已经提供了代码供您注意并等待您的意见。感谢

此代码位于单独的function.js

中

// Function to add box

function addbox() {

    $("#boxaddform").dialog({
        autoOpen: false,
        resizable: true,
        modal: true,
        title: 'Submit a box intake request',
        width: 470,
        beforeclose: function (event, ui) {
            $("#addbox").html("");

        }

    });

    $('#boxsubmit').click(function () {

        var box = $('.box').val();
        var service = $('#service').val();
        var authorised = $('.authorised').val();
        var data = 'box=' + box + '&authorised=' + authorised + '&service=' + service;
        $.ajax({
            type: "POST",
            url: "boxesadd.php",
            data: data,
            success: function (data) {
                $("#boxform").get(0).reset();
                $('#addbox').html(data);
                //$("#form").dialog('close');
                $("#flex1").flexReload();

            }
        });
        return false;

    });

    $("#boxaddform").dialog('open');

}

HTML

<script language="javascript" type="text/javascript">
      $(function() {
            $("#company").change(function() {
              if ($(this).val()!="") $.get("getOptions.php?customer=" + $(this).val(), function(data) {
                $("#divId").html(data);
                });
              });
      });
</script>

<!--- Form to add box -->

<div id="boxaddform" style="display:none;">
  <form id="boxform" method="post" class="webform" name="boxform" />

        <label for="company">Select a Company:</label>
                <select name="company" id="company" />
                    <option SELECTED VALUE="">Select a Company</option>
                        <?php
                                do {  
                                ?>
                    <option value="<?php echo $row_Recordsetcust['customer']?>"><?php echo $row_Recordsetcust['customer']?></option>
                        <?php

                    } 
                                while ($row_Recordsetcust = mysql_fetch_assoc($Recordsetcust));
                                $rows = mysql_num_rows($Recordsetcust);
                if($rows > 0)

                    {
                                mysql_data_seek($Recordsetcust, 0);
                                $row_Recordsetcust = mysql_fetch_assoc($Recordsetcust);
                    }

                ?>
            </select><br />
            <div id="divId"></div><br />
        <label for="service">Enter service level:</label>
                <select name="service" id="service">
                    <option SELECTED VALUE="">Select an option</option>
                    <option value="Standard">Standard</option>
                    <option value="Rapid">Rapid</option>
                </select>
                <br />
        <label for="box">Enter a Box#:</label>
                <input id="box" name="box" type="text" class="text ui-widget-content ui-corner-all inputbox box" />

        <label for="authorised">Requested By:</label>
                <input name="authorised" type="text" class="text ui-widget-content ui-corner-all inputbox authorised" id="authorised" value="<?php echo $_SESSION['kt_name_usr']; ?>" /><br />
    <div id="addbox"></div>
      <button id="boxsubmit" class="submit">Submit</button>
  </form>
</div>

getOptions.php

<?php
    $customer = mysql_real_escape_string( $_GET["customer"] ); // not used here, it's the customer choosen

    $con = mysql_connect("localhost","root","");
    $db = "sample";
      if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }

        mysql_select_db($db, $con);
        $query_rs_select_address2 = sprintf("SELECT * FROM company_com where idcode_com = '$customer'");
        $rs_select_address2 = mysql_query($query_rs_select_address2, $con) or die(mysql_error());
        $row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2);
        $totalRows_rs_select_address2 = mysql_num_rows($rs_select_address2);



          echo 'Select delivery address'.'<select name="customeraddress">';
          echo '<option value="">Select delivery address</option>';
          while ($row_rs_select_address2 = mysql_fetch_assoc($rs_select_address2))
          {
          $address=$row_rs_select_address2['address1_com']. " ".
          $row_rs_select_address2['address2_com']. " ".
          $row_rs_select_address2['address3_com']. " ".
          $row_rs_select_address2['town_com']. " ".
          $row_rs_select_address2['postcode_com'];
          echo '<option
          value="address">'.$address.'</option>';
          }
          echo '</select>';
?>

Answer 1

尝试把这个

$("#company").live('change', function() {
  if ($(this).val()!="") $.get("getOptions.php?customer=" + $(this).val(), function(data)  {
    $("#divId").html(data);
  });
});

在您提供的脚本的第一部分中。

我的想法是，更改处理代码可能没有捕获下拉列表而根本没有运行。为了测试那个put alert（'test'）;在你当前的$（“＃company”）里面.change（function（）{...}）;块。

jQuery对话框不显示更改div

1 个答案: