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Php 如何将值从控制器传输到查看codeigniter到dropdownlist_Php_Codeigniter - Fatal编程技术网

Php 如何将值从控制器传输到查看codeigniter到dropdownlist

Php 如何将值从控制器传输到查看codeigniter到dropdownlist,php,codeigniter,Php,Codeigniter,我想将国家/地区转换为视图codeigniter中的输入类型选择: 假设我通过上述查询得到两个国家,比如印度和美国,我想将其传递给视图 我是通过ajax实现的吗 请帮助我,我是codeigniter的新手因此在您的控制器中您希望为您的构建一系列选项。差不多 if (is_array($jsonhome)) { foreach ($jsonhome as $query) { foreach ($query['results

我想将国家/地区转换为视图codeigniter中的输入类型选择:

假设我通过上述查询得到两个国家,比如印度和美国,我想将其传递给视图

我是通过ajax实现的吗


请帮助我,我是codeigniter的新手

因此在您的控制器中您希望为您的
构建一系列选项。差不多

  if (is_array($jsonhome)) {
                foreach ($jsonhome as $query) {
                    foreach ($query['results']['place'] as $places) {


                        if (is_array($places['country'])) {
                            echo "Country:\n";
                            echo "Content: " . $places['country']['content'] . "\n\n";
                        }

                        if (is_array($places['admin1'])) {
                            echo "State:\n";
                            echo "Content: " . $places['admin1']['content'] . "\n\n";
                        }

                        if (is_array($places['admin2'])) {
                            echo "District/City:\n";
                            echo "Content: " . $places['admin2']['content'] . "\n\n";
                        }
                    }
                }
            }

    $location_data['user_current_country'] = $places['country']['content'];

            $this->load->view('header_register', $location_data);
            $this->load->view('body_complete_register', $location_data);
            $this->load->view('footer_register');
然后在您的视图中您可以使用
表单下拉列表
功能,前提是您已加载表单助手

//Build array of country options
$aData['countryOptions'] = array();
foreach ($jsonhome as $query) {
    foreach ($query['results']['place'] as $places) {
        $aData['countryOptions'][] = $places['country']['content'];
    }
}

$this->load->view('body_complete_register', $aData);
当然,如果你不想使用助手,你可以这样做

$this->load->helper('form');
echo form_dropdown('countries', $countryOptions);


这是视图代码还是控制器?要加载的代码视图在哪里?必须在视图中放置上述代码
$places['country']['content']
将只返回最后一条记录。是否将整个$jsonhome传递给查看?是。您必须将其传递给视图,并在视图中而不是控制器中执行上述代码。非常感谢。是的,收到了:)@vini没问题!:)如果答案有帮助,请接受(谢谢你已经接受了)
<select name="countries" id="countries">
    <?php foreach($countryOptions as $key => $countryName) { ?>
        <option value="<?php echo $key ?>"><?php echo $countryName ?></option>
    <?php } ?>
</select>