<div class="panel-body">
<script type="text/javascript">
$.ajax({
type : 'POST',
url : '<?php echo site_url('welcome/displayallquestionsandoptions')?>',
data : '' // query string
success : function(formagain){
$('.panel-body').html(formagain);
//this will replace the content of div with new form
}
});
</script>
</div>
控制器:
public function displayallquestionsandoptions() {
if($this->input->is_ajax_request()){
$this->load->database();
$this->load->library('table');
$query = $this->db->query("SELECT QuestionId,Name,SurveyId,CreatedOn from question");
$table = $this->table->generate($query);
}else{
}
}
请帮助我是新的编码器和php
在您的控制器中,您需要将$this->table->generate($query);
生成的表echo
返回给ajax调用。此外,您需要将数组或数据库结果对象传递给generate
函数。因此,您需要在控制器
$table = $this->table->generate($query->result_array()); //Pass an array
echo $table; //Added this
然后,AJAX调用中的success函数将捕获从控制器回显的数据,您可以对它进行任何处理。success : function(formagain){ //foragain contains the table HTML from the controller
$('.panel-body').html(formagain);
//this will replace the content of div with new form
}