如何从Codeigniter表单创建链接?

时间:2015-08-06 23:40:56

标签: php codeigniter

我需要将用户输入的两个变量传输到表单中,但我需要将所有数据一起发送到数据库。我无法使用$this->input->post$_REQUEST,我需要创建www.xxx.com/controler?var1=xx*&var2=xx之类的链接。

这是我的代码:

    <?php if($this->input->get_post('tip', TRUE) =='3') {?>
<input type="hidden" name="var_plati" id="var_plati" value="https://www.example.com/beta/tesst/credit-card-reservation-form?tour=<?php echo $_REQUEST['tour'] ?>&tip=<?php echo $_REQUEST['tip'];?>&c=<?php echo $this->input->post("var_charter"); ?>&ad=<?php echo THIS I NEED?>&dep=<?php echo $_REQUEST['var_dep'] ?>" > 

1 个答案:

答案 0 :(得分:0)

方法1:您可以将变量用作href链接 例如:

<a href="approve.php?id=<?php echo $sub->id ?>" role="button" class="btn">Add Topic</a>

方法2:使用ajax调用将代码发送到codeigniter控制器

 var string_array="&tour="+<?php echo $_REQUEST['var1'] ?>+"&tip="+<?php echo $_REQUEST['var2'] ?>+"&c="
                                +<?php echo $this->input->post("var_charter"); ?>+"&ad="+<?php echo 'THIS I NEED'?>+"&address="+address+"&short_address="+short_address
                                +"&dep="+<?php echo $_REQUEST['var_dep'] ?>;
                             $.ajax({
                                        type: "POST",
                                        url: "<?php echo site_url('https://www.example.com/beta/learning/reservation-form');?>",
                                        data:string_array,
                                        success: function(msg) {
                                            console.log(msg);
                                        }
                                        });

                                }