在CodeIgniter中將數據加載到模態

Question

我正在嘗試將數據填充到我的模式形式的下拉選項中，但似乎無法正常工作。 該列表未顯示在我的下拉列表中。 我不確定這是怎么回事。 感謝您的建議。 謝謝

這是我的模式形式視圖（view_timeline.php）

<div class="modal fade" id="myTimeModal" tabindex="-1" role="dialog" aria-labelledby="myTimeModalLabel" aria-hidden="true">
        <div class="modal-dialog">
            <div class="modal-content">
                <div class="modal-header">
                <button type="button" class="close" data-dismiss="modal" aria-hidden="true">&times;</button>                                            
                <h4 class="modal-title" id="myTimeModalLabel">Add New Timeline</h4>
            </div>                                                                                      
        <form role="form" id="TimeForm" action="<?php echo base_url().'admin/addTimeline'; ?>" method= "POST">
            <div class="modal-body">
            <div class="form-group">
                <label>Select Bank *</label>
                <select name="fiid" id="fiid" class="form-control" onChange="fnLoadBank(this)" required=TRUE>
                    <option disabled selected>SELECT BANK</option>
                </select>
            </div>
            // some other item
            </div> <!-- end modal-body -->
            <div class="modal-footer">
                <button type="button" class="btn btn-default" id="reset" data-dismiss="modal">Cancel</button>
                <button type="submit" value="submit" form="TimeForm" class="btn btn-primary"> Add Timeline</button>
            </div> <!-- end modal-footer -->
        </form>

這是我的JavaScript

<script type="text/javascript">
        function fnLoadBank(me) {
        $('.fiid').empty();
        $('.fiid').html('<option disabled selected>SELECT BANK</option>');
        $.ajax({
            type:'POST',
            url: '<?php echo base_url() ?>/admin/getBank()',
            dataType: 'json',
            success: function(dataBank) {
                var str = '';
                if (dataBank) {
                    $.each(dataBank, function(key, row) {
                    str += '<option value="' + row.bankID + '">' + row.bankName + '</option>';
                });
            $('.fiid').append(str);
            }
        }
        });
        }                   
</script>

這是我的admin / getBank控制器

public function getBank()
{
    $dataBank = $this->model_bank->bank_getAll();
    return $dataBank;
}

這是我的控制器來加載view_timeline.php

public function index()
{
    $data['dataBank'] = $this->model_bank->bank_getAll();
    $this->load->view(pages/view_timeline,$data);
}

我的輸出示例

++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+ BANK ID + BANK NAME + DATE/TIME   + TRANSACTION DETAILS      +
++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
+ 0001    + BANK A    + 1111/101210 + Transaction 1            +
+--------------------------------------------------------------+
+ 0002    + BANK B    + 1111/103050 + Transaction 2            +
+--------------------------------------------------------------+
+ 0001    + BANK A    + 1111/110105 + Transaction 3            +
+--------------------------------------------------------------+

                                          +ADD MORE TRANSACTION+

因此，當單擊“添加更多交易”時，我需要選擇銀行（從可用銀行列表中），填寫詳細信息並保存（這將添加到我的表格中而不刷新我的頁面）。 因此，我使用模態彈出表單。

Answer 1

這樣嘗試

<script type="text/javascript">
$(document).ready(function(e) 
{
   fnLoadBank();
});
    function fnLoadBank() 
    {
    $.ajax({
        type:'POST',
        url: '<?php echo base_url("admin/getBank");?>',
        success: function(dataBank) 
        {
            $('#fiid').html(dataBank);
        }
    }
    });
 }                   
</script>

<?php 
public function getBank()
{
    $dataBank = $this->model_bank->bank_getAll();

    $option .= '<option value="" disabled selected >SELECT BANK</option>';
    foreach ($dataBank as $value) {

        $option .= "<option value=".$value['bankID'].">".$value['bankName']."</option>";
    }
    echo $option;

}
?>

然后嘗試將此選項綁定到您選擇的銀行ID

$('#fiid').html(dataBank);