<?php
$options = array(
500 => 500,
1000 => 1000,
2500 => 2500,
5000 => 5000,
);
echo form_dropdown('units',$options);
?>
$db_num = 25; # a random number
$post_num = $set_num * (units-selected)
如果单击提交,我如何将单位下拉列表的选定值与我的$db_num
相乘?我认为这可能需要javascript,但是没有吗?
在我的控制器上我尝试了这个:
$db_num = $this->input->post('db_num');
$units = $this->input->post('units');
$answer = $db_num * $units;
...
echo $answer;
(结果是没有错误,没有插入。我做错了什么?)
答案 0 :(得分:0)
类似的东西:
<form ... onsubmit="return functionwhereyouwilldothemath(this)" >
</form>
<script type="text/javascript">
function functionwhereyouwilldothemath(el) { //el = javascript form object
var db_num = <?=$db_num?>; //yup this is legit.
//do your math
return true; //will cause the page to reload uppon submit.
}
</script>
答案 1 :(得分:0)
传递我的变量的数据库字段需要设置为INT
- 问题已解决。
答案 2 :(得分:0)
使用php implode函数存储和爆炸显示功能
$ itm = $ _POST ['hospital_department']; $ comma_separated = implode(“ - ”,$ itm);
$data['hospital_department'] = $comma_separated;