【发布时间】:2015-11-07 13:35:37
【问题描述】:
我需要帮助在 Codeigniter 中使用 ajax 将数据传递给模态引导程序
当我将数据传递给模态时,模态引导程序有空白输入。我知道我的 ajax 工作,因为如果我传递静态数据,模态将显示它。但是如果我从 GET 传递数据,模式输入总是空白
我是 ajax 新手,所以我认为我的 JSON 解析有问题。
型号:
public function getEditContent($table,$field,$key)
{
return $this->db->get_where($table, array($field => $key))->result();
}
控制器:
public function Test()
{
$this->load->library('Datatables');
$this->_header();
$this->load->view('Front/v_test',$data);
$this->_footer();
}
public function testdata()
{
$data['result'] = $this->Model->getEditContent('arsip','idsurat','1');
echo json_encode($data);
}
查看:
<div class="col-xs-12 col-sm-12">
<!-- Button HTML (to Trigger Modal) -->
<a href="javascript:void(0)" onclick="openModal()">
<span class="glyphicon glyphicon-pencil"></span></a>
</div>
<!-- Modal HTML -->
<div id="form-content" class="modal fade">
<div class="modal-dialog">
<div class="modal-content">
<div class="modal-header">
<a class="close" data-dismiss="modal">×</a>
<h3>Input Data Surat Keluar</h3>
</div>
<div class="modal-body">
<?php echo form_open('Tambah/insert_suratkeluar'); ?>
<div class="form-group">
<label for="inputEmail">No Surat</label>
<input type="text" class="form-control" name="nosurat" placeholder="No Surat" id="nosurat">
</div>
<div class="form-group">
<label for="tglkirim">Tanggal Kirim</label>
<div class='input-group date' id='datetimepicker1'>
<input type='text' class="form-control" name="tglkirim" placeholder="Masukkan Tanggal Kirim" />
<span class="input-group-addon">
<span class="glyphicon glyphicon-calendar"></span>
</span>
</div>
</div>
<script type="text/javascript">
$(function ()
{
$('#datetimepicker1').datetimepicker(
{
format: 'YYYY/MM/DD'
});
});
</script>
<div class="form-group">
<label for="inputPassword">Alamat</label>
<input type="text" class="form-control" name="alamat" placeholder="Alamat" id="alamat" value="">
</div>
<div class="form-group">
<label for="notelp">No Telp</label>
<input type="text" class="form-control" name="notelp" placeholder="No Telp" value="">
</div>
<div class="form-group">
<label for="inputPassword">Email</label>
<input type="email" class="form-control" name="email" placeholder="Email" value="">
</div>
<div class="form-group">
<label for="inputPassword">Nama Perusahaan</label>
<input type="text" class="form-control" name="nmusaha" placeholder="Nama Perusahaan" value="">
</div>
<div class="form-group">
<label for="inputPassword">Deskripsi</label>
<input type="text" class="form-control" name="deskripsi" placeholder="Deskripsi" value="">
</div>
<div class="modal-footer">
<button type="submit" class="btn btn-primary">Submit</button>
<a href="#" class="btn" data-dismiss="modal">Cancel</a>
</div>
<?php echo form_close(); ?>
</div>
</div>
</div>
</div>
jQuery :
function openModal()
{
$.ajax(
{
url:"testdata/",
type : 'GET',
dataType : 'json',
success:function (result)
{
$('#nosurat').val(result.nosurat);
$('#alamat').val("Test Alamat");
$('#form-content').modal('show');
},
});
}
我在 firebug 中得到 JSON:
【问题讨论】:
-
试试
$('#nosurat').val(result[0].nosurat); -
^ 是的,这项工作,谢谢,你应该回答,这样我就可以投票给你的答案
标签: jquery ajax twitter-bootstrap codeigniter