在codeigniter中，我通过form_open提交表单，所有按钮都由我的form_open（'Home / page'）链接
[英]In codeigniter i submit a form by form_open there all button linked by my form_open('Home/page')
问题描述
<?php echo form_open('Home/OtherDetail'); ?> <div class="row" style="margin-bottom:5%;" > <div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600; font-size: 100%;">Name:</div> <div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="name" style="width:100%" value=""></div> </div> <div class="row" style="margin-bottom:5%;"> <div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600;font-size: 100%;">Father Full Name:</div> <div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="FatherName" style="width:100%" value=""></div> </div> <div class="row" style="margin-bottom:5%;"> <div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600;font-size: 100%;">Address:</div> <div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="Address" style="width:100%" value=""></div> </div> <a href="Home/new"><button >add</button></a> <a href="Home/new"><button >negative</button></a> <?php form_submit('name',value); ?> //so all button which are comes in between from open and form close linked with form_open <?php form_close(); ?>
//因此所有从open和form close之间出现的按钮都与form_open链接
3 个解决方案
解决方案1
0 2018-08-23 12:42:38
请检查以下代码，它可以完美工作。
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8">
<title>Simple CodeIgniter App - Todos</title>
<link rel="stylesheet"
href="//maxcdn.bootstrapcdn.com/bootstrap/3.2.0/css/bootstrap.min.css">
</head>
<body>
<?php echo form_open('Home/OtherDetail'); ?>
<div class="row" style="margin-bottom:5%;" >
<div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600; font-size: 100%;">Name:</div>
<div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="name" style="width:100%" value=""></div>
</div>
<div class="row" style="margin-bottom:5%;">
<div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600;font-size: 100%;">Father Full Name:</div>
<div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="FatherName" style="width:100%" value=""></div>
</div>
<div class="row" style="margin-bottom:5%;">
<div class="col col-lg-4 col-md-12 col-sm-12 col-xs-12" style="font-weight: 600;font-size: 100%;">Address:</div>
<div class="col col-lg-8 col-md-12 col-sm-12 col-xs-12"><input type="text" name="Address" style="width:100%" value=""></div>
</div>
<a href="Home/new">
<?php echo form_button('add','add'); ?>
<!-- <button type>add</button> -->
</a>
<a href="Home/new">
<?php echo form_button('negative','negative'); ?>
<!-- <button type>negative</button> -->
</a>
<?php echo form_submit('name','btn_name'); ?>//so all button which are comes in between from open and form close linked with form_open
<?php echo form_close(); ?>
</body>
</html>
解决方案2
0 2018-08-24 16:18:00
如果您不希望HTML BUTTON像Submit那样起作用，则需要添加标记type =“ button”。
<form>
<button>This button submit</button>
<button type="button">This button DO NOT submit</button>
<!-- Default attribute type value is "submit" -->
</form>
解决方案3
0 2018-08-27 05:12:17
我弄清楚为什么会这样...那是由于按钮标记位于form_open标记和form_close标记之间，因为我将按钮标记更改为div而不是正常工作....所以有结论您提交的表格永远不要使用任何按钮标签，而只能使用div标签，span标签等。
单击表单中的“提交”按钮后，如何打开特定的网页？
[英]How to open a specific web page once submit button is clicked in the form?
如何使用 JavaScript 函数根据单选按钮选择让我的表单提交按钮打开不同的 URL？
[英]How can I make my form submit button open a different URL based on radio button selection using JavaScript function?
声明:本站的技术帖子网页，遵循CC BY-SA 4.0协议，如果您需要转载，请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.