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Php POST数据未进入控制器代码点火器_Php_Codeigniter - Fatal编程技术网

Php POST数据未进入控制器代码点火器

Php POST数据未进入控制器代码点火器,php,codeigniter,Php,Codeigniter,我发布了一个包含一些数据的表单,但是我得到了一个空的POST数组。表格是 <form name="opening-statistics" action ="<?php echo base_url()?>statistic/openings"> <select name="titles" multiple="multiple"> <?php foreach($titles as $title) { ?> &l

我发布了一个包含一些数据的表单,但是我得到了一个空的POST数组。表格是

<form name="opening-statistics" action ="<?php echo base_url()?>statistic/openings">
   <select name="titles" multiple="multiple">
       <?php foreach($titles as $title) { ?> 
          <option value="<?php echo $title['idtitles']?>"><?php echo $title['title']?></option>
       <?php } ?>
    </select>
<input type="submit" name="submit" value="Submit">
在我的路线上,我有

$route['statistic/openings-statistics'] = 'statistic/openings';
它总是出现在else语句中

谁能告诉我原因


提前感谢

试着像这样抓住帖子:

 public function openings(){
    $data['boxes'] = $this->Statistic_model->getBoxes();
    $data['titles'] = $this->Statistic_model->getTitles();
    if($_POST){
        echo 'I am here';
        $form = $this->input->post();
        var_dump($form);
        exit();
    }
    else{
        echo 'I am here';
        $this->load->view('statistic/openings-statistics', $data);
    }
}
是的,CI确实有:

$this->input->post(NULL, TRUE); // returns all POST items with XSS filter 
$this->input->post(); // returns all POST items without XSS filter

method=“POST”
添加到表单中…我现在获得了数据,但我的表单中有一个multiselect,我得到了一条记录,$this->input->POST('titles')给了您什么?最后选择的选项的值请参阅CI指南()搜索表单\u multiselect(),有一个关于如何获取值的说明我没有使用form helper:(
$this->input->post(NULL, TRUE); // returns all POST items with XSS filter 
$this->input->post(); // returns all POST items without XSS filter