Question

我需要有关codeigniter中视图的帮助。通常我只是以同样的方式行事，这是工作。现在我感到困惑，或者我错了。

这是控制器：

        ...

        $curl_exec = curl_exec($ch);
        $result = json_decode($curl_exec, TRUE);
        $httpcode = curl_getinfo($ch, CURLINFO_HTTP_CODE);

        if($httpcode == 200)
        {
            $msg['success'] = 'Succcessfully completed step 1!';
            $this->load->view('back/vproductadd', $msg);
        } elseif ($httpcode == 500) {
            $msg['double'] = 'Succcessfully completed step 1!';
            $this->load->view('back/vproductadd', $msg);
        } {
            $msg['others'] = 'Succcessfully completed step 1!';
            $this->load->view('back/vproductadd', $msg);
        }

这就是观点：

                <?php
                    if (isset($success)) {
                        echo "<div class='alert alert-danger alert-dismissible'>";
                        echo "<button type='button' class='close' data-dismiss='alert' aria-hidden='true'>&times;</button>";
                        echo "<h4><i class='icon fa fa-ban'></i> Alert!</h4>";
                        echo "$success";
                        echo "</div>" ;
                    }
                  ?>
                  <?php
                    if (isset($double)) {
                        echo "<div class='alert alert-danger alert-dismissible'>";
                        echo "<button type='button' class='close' data-dismiss='alert' aria-hidden='true'>&times;</button>";
                        echo "<h4><i class='icon fa fa-ban'></i> Alert!</h4>";
                        echo "$double";
                        echo "</div>" ;
                    }
                  ?>
                  <?php
                    if (isset($others)) {
                        echo "<div class='alert alert-danger alert-dismissible'>";
                        echo "<button type='button' class='close' data-dismiss='alert' aria-hidden='true'>&times;</button>";
                        echo "<h4><i class='icon fa fa-ban'></i> Alert!</h4>";
                        echo "$others";
                        echo "</div>" ;
                    }
                  ?>

                  // Form
                  <div class="form-group">
                    <label for="exampleInputEmail1">Product Code</label>
                    <input type="text" name="codeproduct" id="codeproduct" class="form-control"  placeholder="Product Code">
                  </div>
                 // ...
                 // End Form

该消息应显示在警告框中，但不会显示。如果我在条件消息显示结束时添加exit();。我的代码有什么问题吗？

Answer 1

你需要把＆＃39;否则＆＃39;在最后一个名言上。

if($httpcode == 200)
    {
        $msg['success'] = 'Succcessfully completed step 1!';
        $this->load->view('back/vproductadd', $msg);
    } elseif ($httpcode == 500) {
        $msg['double'] = 'Succcessfully completed step 1!';
        $this->load->view('back/vproductadd', $msg);
    } else {
        $msg['others'] = 'Succcessfully completed step 1!';
        $this->load->view('back/vproductadd', $msg);
    }

Answer 2

问题是 - 如果你让控制器中的脚本死掉，CI还没有完成。因为它会加载@first所有视图，以便最终将它们放出来。

下面的应用程序流程图应该显示它是如何工作的。 Sed Global substitution

有关详细信息，请查看

将值返回到codeigniter视图

2 个答案: