如何修复 PHP 代码中的未定义变量？

Question

这是我在 view/tampil 中单击更新按钮时的错误。php:

A PHP Error was encountered
Severity: Warning
Message: Missing argument 1 for Home::ubahdata()
Filename: controllers/Home.php
Line Number: 46
Backtrace:

File: A:\Sites\PHP_CI\CRUD\application\controllers\Home.php
Line: 46
Function: _error_handler

File: A:\Sites\PHP_CI\CRUD\index.php
Line: 315
Function: require_once
A PHP Error was encountered

Severity: Notice
Message: Undefined variable: nohp
Filename: controllers/Home.php
Line Number: 52

Backtrace:

File: A:\Sites\PHP_CI\CRUD\application\controllers\Home.php
Line: 52
Function: _error_handler

File: A:\Sites\PHP_CI\CRUD\index.php
Line: 315
Function: require_once

当我点击更新按钮时，它显示错误未定义的变量。 undefined variable 是什么意思？我认为我的错误不是来自未定义的变量。

models/mahasiswa_model.php :

public function getUser($nohp)
    {
        $query = $this->db->get_where('mahasiswa', array('nohp' => $nohp));
        return $query->row_array();
    }
    public function ubah_model_data($user, $nohp)
    {
        $this->db->where('mahasiswa.nohp', $nohp);
        return $this->db->update('mahasiswa', $user);
    }
    public function ubah_model($nohp)
    {
        $query = $this->db->get_where('mahasiswa', array('nohp' => $nohp));
        return $query->row_array();
    }

这是我来自 controller/home 的代码。php :

public function ubahdata($nohp)
    {
        $user['nohp'] = $this->input->post('nohp');
        $user['nama'] = $this->input->post('nama');
        $user['alamat'] = $this->input->post('alamat');

        $query = $this->mahasiswa_model->ubah_model_data($user, $nohp);
    }
    public function ubah($nohp)
    {
        $data['mahasiswa'] = $this->mahasiswa_model->ubah_model($nohp);
        $this->load->view('home/tampil', $data);
    }

这是我来自 view/tampil 的代码。php :

<form method="post" action="<?= base_url('home/ubahdata/'); ?>">
                     <?= $this->session->flashdata('message'); ?>
                     <div class="form-group row">
                         <div class="col-sm mb-3 mb-sm-0">
                             <label for="nohp">Nomer Handphone</label>
                             <input type="text" class="form-control form-control-user" id="nohp" name="nohp" placeholder="Masukkan Nomer Handphone" value="<?php echo $tampil->nohp; ?>">
                         </div>

Answer 1

问题是你的方法需要在调用时给出一个参数，该参数在您的 Home 控制器方法 ubahdata 中是 $nohp。因此，如果您想解决此问题，您要么必须在调用该方法时传递该参数，要么删除 it.I 不明白此变量的用途并查看您的 HTML 代码，您通过 POST 方法传递该变量，因此您不需要将其作为参数传递，您可以将其删除。这只是我的理解。

public function ubahdata()
    {
        $user['nohp'] = $this->input->post('nohp');
        $user['nama'] = $this->input->post('nama');
        $user['alamat'] = $this->input->post('alamat');

        $query = $this->mahasiswa_model->ubah_model_data($user,  $user['nohp']);
    }
    public function ubah($nohp)
    {
        $data['mahasiswa'] = $this->mahasiswa_model->ubah_model($nohp);
        $this->load->view('home/tampil', $data);
    }

如何修复 PHP 代码中的未定义变量？

How to fix undefined variable in PHP code?

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