如何在 codeigniter 中使用活动 class 制作 if else 菜单

Question

我想问一下如何使if else 功能菜单处于活动状态class。这是我在控制器中的代码。

public function index(){

    //i try with this but error 
    $data['activeTab'] == "home";

    $this->load->view('head');
    $this->load->view('header');

    $this->load->view('content',$data);

    $this->load->view('footer');
    $this->load->view('foot');
}

public function about(){

    //i try with this but error
    $data['activeTab'] == "about";

    $this->load->view('head');
    $this->load->view('header');

    $this->load->view('content-about',$data);

    $this->load->view('footer');
    $this->load->view('foot');
}

这是我的观点

<body>  
<div class="container">
    <div class="row">
        <!--HEADER START-->
            <div class="page-header">
            <a href="<?php echo base_url();?>codeig/index.php/site">HEADER</a>
            </div>
        <!--HEADER END-->

        <!--NAVBAR START-->
            <nav class="navbar navbar-default">
                <div class="navbar-header">
                    <button type="button" class="navbar-toggle collapsed" data-toggle="collapse" data-target="#navbar" aria-expanded="false" aria-controls="navbar">
                        <span class="sr-only">Toggle navigation</span>
                        <span class="icon-bar"></span>
                        <span class="icon-bar"></span>
                        <span class="icon-bar"></span>
                    </button>
                </div>
                <div id="navbar" class="collapse navbar-collapse">
                    <ul class="nav navbar-nav">
                        <!--i try with this but error-->
                        <li class="<?php echo ($activeTab=="home")?"active":""; ?>"><a href="<?php echo base_url();?>codeig/index.php/site">Home</a></li>
                        <li><a href="<?php echo base_url();?>codeig/index.php/site/about">About</a></li>
                        <li><a href="#contact">Contact</a></li>
                    </ul>
                </div>
            </nav>
        <!--NAVBAR END-->

我会尝试 activetabs。但是有一个错误。

Answer 1

您需要将 $data 数组传递给您的 viwes 并且 == 是您的比较用途 = 用于分配值

控制器

public function index(){

    //i try with this but error 
    $data['activeTab'] = "home";

    $this->load->view('head',$data);
    $this->load->view('header',$data);

    $this->load->view('content',$data);

    $this->load->view('footer',$data);
    $this->load->view('foot',$data);
}

public function about(){

    //i try with this but error
    $data['activeTab'] = "about";

    $this->load->view('head',$data);
     $this->load->view('header',$data);

    $this->load->view('content-about',$data);

    $this->load->view('footer',$data);
    $this->load->view('foot',$data);
}

Answer 2

将此更改为

$data['activeTab'] == "home";
$data['activeTab'] == "about";

这个

$data['activeTab'] = "home";
$data['activeTab'] = "about";

如果您使用 ==，您会收到此错误

Severity: Notice
Message: Undefined variable: activeTab
Filename: xx/yyy.php
Line Number: xxx

如何在 codeigniter 中使用活动 class 制作 if else 菜单

How to make if else menu with active class in codeigniter

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if-statement

codeigniter