我在codeigniter中有一个程序,在登录后会有一个搜索框来搜索(ajax搜索)数据库中的数据,它将以链接的形式存在。但是,如果我有任何不在数据库中的单词,它应显示“找不到数据”,并且应该有一个“请求按钮”。单击请求按钮时,搜索框中键入的单词应与登录的用户名一起保存在数据库中,并应显示在屏幕上发送的请求。
这是我的ajaxsearch控制器:
if(is_null($this->input->get('id')))
{
$this->load->view('data');
}
else
{
$this->load->model('Info_model');
$data['Infotable']=$this->Info_model->Infotable($this->input->get('id'));
$this->load->view('view_result',$data);
}
这是我的看法:
1.数据页面
<form action="" method="get">
<div class="row">
<div class="col-lg-10 col-lg-offset-1">
<div class="input-group">
<span class="input-group-addon" >SEARCH</span>
<input autocomplete="off" id="search" type="text" class="form-control input-lg" placeholder="Search " >
</div>
</div>
</div>
<div class="space"></div>
</form>
提前致谢
<form action="<?php echo base_url();?>Test/ajaxsearch" method="post">
<div class="row">
<div class="col-lg-10 col-lg-offset-1">
<div class="input-group">
<span class="input-group-addon" >SEARCH</span>
<input autocomplete="off" id="search" name="search" type="text" class="form-control input-lg" placeholder="Search " >
<button type="submit">Search</button>
</div>
</div>
</div>
<div class="space"></div>
</form>
然后在你的ajaxsearch()
public function ajaxsearch() {
if(@$this->input->post('search') != "") {
$data['Infotable']=$this->Info_model->Infotable($this->input->post('search'));
if(count($data['Infotable']) > 0) {
$this->load->view('view_result',$data);
}else {
$data['dataNotFound'] = "No data Found";
$this->load->view('view_result',$data);
}
}
}
然后在view_result中
<?php
if(isset($dataNotFound)) {
echo $dataNotFound;
}else {
foreach($Infotable as $objects) {
echo $objects;
}
}
?>