Question

我正在尝试在表格中插入表单值。我不知道未发布值的原因是错误值不能为null我确定我在表单中提到了正确的表值，任何人都可以检查我犯了什么错误

这是我的按钮：

<td><form action="<?php echo base_url();?>Dashboard/ReviewManagement" method="POST">
    <input type="hidden" name="topicid" value="<?php echo $proview->id;?>">
    <input type="hidden" name="courseid" value="<?php echo $proview->courseId;?>">
    <button type="submit" class="btn btn-xs btn-warning" >
        <span class="icon-ok"></span>
    </button>
</form></td>

这是我的控制人：

public function ReviewManagement() {

    $username = $this->session->userdata('user_name');
    $data['userslist'] = $this->Dashboard_model->getUsers();

        $insert_data = array(
            'topicid' =>$this->input->post('topicid'),
            'stage' => $this->input->post('stage'),
            'allocatedresource' => $this->input->post('allocatedresource'),
            'status' => 1,
            'createdby' => $username,
            'createdon' => date('d/m/Y'),
            'expec' =>$this->input->post('expec')

        );

        $this->Dashboard_model->reviewmanagerInsert($insert_data);
        $this->load->view('template/header');
            $this->load->view('dashboard/reviewmanagement',$data);
            $this->load->view('template/footer'); 
        }

Here is my model:

    public function reviewmanagerInsert($data)
    {
   $this->db->insert('reviewmanager',$data); 
}

这是我的观点：

  <form action="<?php echo base_url();?>Dashboard/ReviewManagement" method="post">
        <div class="form-row">
            <div class="col-md-3">Reviewer:</div>
            <div class="col-md-9"> 
                <select name="allocatedresource" class="form-control">
                   <option value="0">Select Reviewer</option>
                <?php foreach ($userslist as $users) { ?>
                    <option value="<?php echo $users->id; ?>"><?php echo $users->username; ?></option>
                <?php } ?>
            </select>
            </div>
        </div>
        <div class="form-row">
            <div class="col-md-3">Stage:</div>
            <div class="col-md-9"> 
                <select class="form-control col-md-8" name="stage" required="">
                    <option value="" selected="">Select Stage</option>
                    <option value="1">PB Review</option>
                    <option value="2">PB2 Review</option>
                    <option value="3">Alpha Review</option>
                    <option value="4">Beta Review</option>
                    <option value="5">Gamma Review</option>
                </select>
            </div>
        </div>

        <div class="form-row">
            <div class="col-md-3">Due Date:</div>
            <div class="col-md-9">
                <div class="input-group">
                    <div class="input-group-addon"><span class="icon-calendar-empty"></span></div>
                    <input type="text" name="expec" autocomplete="off" required="" class="datepicker form-control" value=""/>
                </div>
            </div>
        </div>
        <div class="form-row">
            <div class="col-md-8 col-md-offset-2">
                <button type="submit" class="btn btn-primary btn-block">Assign</button>             
            </div>
        </div>
    </form>

收到此错误：

列“ topicid”不能为空

INSERT INTO `reviewmanager` 
            (`topicid`, `stage`, `allocatedresource`, 
            `status`, `createdby`, `createdon`, `expec`) 
    VALUES (NULL, NULL, NULL, 1, 'admin', '16/08/2018', NULL)

有人可以帮我犯什么错误

谢谢。

Answer 1

仅使用一种具有隐藏字段的表单，并使用类似jquery的值给这些隐藏字段

您的按钮视图

<td>
    <div id="topicid" value="<?php echo $proview->id;?>"></div>
    <div id="courseid" value="<?php echo $proview->courseId;?>"></div>
    <button type="button" class="btn btn-xs btn-warning" >
        <span class="icon-ok" id="submit"></span>
    </button>
</td>

然后在您的主表单中的隐藏字段下方添加

<input type="hidden" name="topicid" id="topic_id">
<input type="hidden" name="courseid" id="course_id">

然后在js下面添加以分配值

$(document).ready(function(){
    $("#submit").click(function(){
        var topicid = $("#topicid").val();
        var courseid = $("#courseid").val();
        $("#topic_id").val(topicid);
        $("#course_id").val(courseid);
    });
});

此答案未经测试。

Answer 2

只需在表单的两个输入下方添加...

<input type="hidden" name="topicid" id="topic_id">
<input type="hidden" name="courseid" id="course_id">

Answer 3

像下面一样解决了

<td><form action="<?php echo base_url();?>Dashboard/ReviewManagement" method="POST">
                                                    <input type="hidden" name="topicid" value="<?php echo $proview->id;?>">
                                                    <input type="hidden" name="courseid" value="<?php echo $proview->courseId;?>">
                                                    <button type="submit" class="btn btn-xs btn-warning" >
                                                        <span class="icon-ok"></span>
                                                    </button>
                                                </form></td>

从下面更改为秒

<form action="<?php echo base_url();?>Dashboard/ReviewManagementInsert" method="post">
                <div class="form-row">
                    <div class="col-md-3">Reviewer:</div>
                    <div class="col-md-9"> 
                        <select name="allocatedresource" class="form-control">
                           <option value="0">Select Reviewer</option>
                        <?php foreach ($userslist as $users) { ?>
                            <option value="<?php echo $users->username; ?>"><?php echo $users->username; ?></option>
                        <?php } ?>
                    </select>
                    </div>
                </div>

控制器代码1：

public function ReviewManagement() {

    if ($this->session->userdata('is_logged')) {
        if (!empty($this->input->post('topicid'))) {
            $id = $this->input->post('topicid');
            $data['userslist'] = $this->Dashboard_model->getUsers();
            $data['id'] = $this->input->post('topicid');
            $this->session->set_userdata('topicid', $data['id']);
            $this->load->view('template/header');
            $this->load->view('dashboard/reviewmanagement', $data);
            $this->load->view('template/footer');
        } elseif (!empty($this->session->userdata('topicid'))) {
            $id = $this->session->userdata('topicid');
            $data['userslist'] = $this->Dashboard_model->getUsers();
            $data['id'] = $this->session->userdata('gettopicdetails');
            $this->load->view('template/header');
            $this->load->view('dashboard/reviewmanagement', $data);
            $this->load->view('template/footer');
        } else {
            echo 'im else';
            die;
        }
    } else {
        $this->session->set_flashdata('error', 'Topic Id not captured.');
        redirect('Dashboard');
    }
}

控制器代码2

public function ReviewManagementInsert() {
    $topicid = $this->session->userdata('topicid');
    $username = $this->session->userdata('user_name');


    $insert_data = array(
        'topicid' => $topicid,
        'stage' => $this->input->post('stage'),
        'allocatedresource' => $this->input->post('allocatedresource'),
        'status' => 1,
        'createdby' => $username,
        'createdon' => date('d/m/Y'),
        'expec' => $this->input->post('expec')
    );

    $this->Dashboard_model->reviewmanagerInsert($insert_data);
    $this->load->view('template/header');
    $this->load->view('dashboard/reviewmanagement');
    $this->load->view('template/footer');
}

在表中插入表单值在Codeigniter中不起作用

3 个答案: