我有codeigniter的问题。我想在下一页中将表单值(从下拉列表)传递为URL段。然而,我搜索了高低,但找不到解决方案。
以下是我的观点:
<?= form_open('admin_gallery_upload/upload_images/'); ?>
<?= form_label('Gallerij', 'gallery_id'); ?><br/>
<?= form_dropdown('gallery_id', $select_dropdown); ?>
<?= form_submit('submit', 'Volgende', 'class="submit"'); ?>
<?= form_close(); ?>
我的控制器:
function upload_images() {
$gallery_id = $this->input->post("gallery_id");
echo $gallery_id;
}
因此,它不应像控制器中那样回显$gallery_id
,而应该成为第三个网址段
答案 0 :(得分:1)
尝试此操作并将选择框ID设为gallery_id
:
function redirectFunction(){
var val = document.getElementById("gallery_id").value;
alert(val); //see if alerts the correct value that is selected from the dropdown
window.location.href = '<?php echo base_url()?>admin_gallery_upload/upload_images/'+val;
}
更改以下行,按照我的说法添加下拉列表的ID,
$js = 'id="gallery_id" onChange="redirectFunction();"';
form_dropdown('gallery_id', $select_dropdown,'',$js);
答案 1 :(得分:1)
您可以默认传递表单open function extra参数,如下所示: -
<? $arr = array('method'=> 'GET');?>
<?= form_open('admin_gallery_upload/upload_images/',$arr); ?>
使用以下方法获取控制器: -
$this->input->get();