我有一个显示和隐藏 div。那是一个添加表单。我想显示表格底部表格中的所有数据。无论表格是否打开，都必须显示数据。当我们添加新帖子时，必须在不刷新的情况下附加到显示数据的顶部。
我的看法
<div class="slidingDiv" style="display:none">
<div class="row">
<div class="col-xs-12">
<div class="box box-primary">
<div class="box-header">
<h3 class="box-title">Add Requirement Item</h3>
</div><!-- /.box-header -->
<!-- form start -->
<?php echo validation_errors(); ?>
<?php
$attributes = array('id' => 'myForm');
echo form_open_multipart(base_url().'moderator/Requirement/add_employee_data',$attributes); ?>
<div class="box-body">
<div class="row">
<div class="col-xs-6">
<div class="form-group">
<div class="row">
<div class="col-xs-12">
<label for="txttitle">Requirement Title (Product/Service) : </label>
</div>
<div class="col-xs-12">
<input type="text" name="txtService" class="form-control" id="txttitle" placeholder="Requirement Title (Product/Service) : " value="" required>
</div>
</div>
</div>
</diV>
<div class="col-xs-6">
<div class="form-group">
<div class="row">
<div class="col-xs-8">
<label for="txtquantity">Estimated Quantity : </label>
</div>
<div class="col-xs-4">
<label for="txtquantity"></label>
</div>
<div class="col-xs-8">
<input type="text" name="txtQuantity" class="form-control" id="txtquantity" placeholder="Estimated Quantity" value="
" required>
</div>
<div class="col-xs-4">
<select class="form-control" name="txtunit" required="required">
<option value="">----Select------</option>
<?php
foreach ($units as $name) {
echo ' <option value="' . $name->id . '">' . $name->name . '</option>';
}
?>
</select>
</div>
</div>
</div>
</diV>
<div class="col-xs-12">
<div class="form-group">
<div class="row">
<div class="col-xs-12">
<label for="txtdetails">Requirement Details : </label>
</div>
<div class="col-xs-12">
<textarea class="textarea" name="txtRequirement" placeholder="Requirement Details" id="txtdetails" style="width: 100%; height: 200px; font-size: 14px; line-height: 18px; border: 1px solid #dddddd; padding: 10px;">
</textarea>
</div>
</div>
</div>
</div>
<div class="col-xs-12">
<div class="form-group">
<div class="row">
<div class="col-xs-6">
<label for="sbUser">Expiry of Requirement : </label><br>
<div class="input-group">
<div class="input-group-addon">
<i class="fa fa-calendar"></i>
</div>
<input id="thedate" type="text" name="txtBidclosing" class="form-control" required="required"/>
</div>
</div>
</div>
</div>
</div>
</div>
</div><!-- /.box-body -->
<div class="box-footer">
<input type="button" class="button" value="submit" />
</form>
</div>
</div>
</div>
</div>
<div id="sort">
<br>
</div>
Ajax 代码
<script>
$(document).ready(function(){
$(".button").click(function(){
$.ajax({
type:"POST",
url: "<?php echo base_url() ?>moderator/Requirement/add_employee_data",
data:$("#myForm").serialize(),
success: function (dataCheck) {
//alert(dataCheck);
$('#sort').append(dataCheck);
//window.location.reload();},
});});});
</script>
型号
public function add_employee_data($data){
$this->db->insert('jil_requirementdetail',$data);
$id = $this->db->insert_id();
$this->db->select('*');
$this->db->from('jil_requirementdetail');
$this->db->where('rqmd_id',$id);
$result = $this->db->get()->result();
return $result;
}
最佳答案
在您的模型上
public function add_employee_data($data){
$this->db->insert('jil_requirementdetail',$data);
$id = $this->db->insert_id();
$this->db->select('*');
$this->db->from('jil_requirementdetail');
// $this->db->where('rqmd_id',$id); // get all data
$this->db->order_by('rqmd_id','DESC');
$result = $this->db->get();
if($result->num_row() > 0)
{
echo json_encode(['status'=>'pass','data'=>$result->result()];
}else{
echo json_encode(['status'=>'fail']);
}
}
关于 Ajax 的成功
<script>
$(document).ready(function(){
$(".button").click(function(){
$.ajax({
type: "POST",
url: "<?php echo base_url() ?>moderator/Requirement/add_employee_data",
data: $("#myForm").serialize(),
success: function(dataCheck){
//alert(dataCheck);
var res = JSON.parse(dataCheck);
if(res.status=='pass')
{
var user_data = res.data;
var html = '';
var length = user_data.length;
for(var i = 0; i < length; i++)
{
//your html
html += '<div>' + user_data[i].name + '</div>'; //add your html structure as you want
}
$('#sort').append(html);
},
}
});
});
});
</script>
关于javascript - 我想显示来自数据库的数据。如果我们新添加一个表单数据，我想使用 codeigniter 在不刷新的情况下附加显示的新帖子，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/36004866/