【发布时间】：2017-01-25 10:28:40
【问题描述】：
我已经创建了 valid() 和 insert() 函数，但我不知道如何检查表单是否有效。我只想在验证检查后调用 insert() 函数。
我删除了我的基本 html 表单代码。
view.php
<?php
include('../controller/control.php');
if(isset($_REQUEST['submit']))
{
$u = $_REQUEST['fname'];
$p = $_REQUEST['lname'];
$i = $_FILES['pic']['name'];
$g = $_REQUEST['gender'];
$c = $_REQUEST['country'];
$s = $_REQUEST['state'];
$ci = $_REQUEST['city'];
$z = $_REQUEST['zipcode'];
$cno = $_REQUEST['cno'];
$h = $_REQUEST['hobby'];
$chk="";
if($h)
{
foreach($h as $chk1)
{
$chk .= $chk1.",";
}
}
$a = $_REQUEST['address'];
$pdf = $_FILES['pdf']['name'];
move_uploaded_file($_FILES['pic']['tmp_name'],"upload/".$_FILES['pic']['name']);
move_uploaded_file($_FILES['pdf']['tmp_name'],"upload/".$_FILES['pdf']['name']);
$obj = new control();
$obj->validate($u,$p,$i,$c,$s,$ci,$z,$cno,$a,$pdf);
$obj->insert($u,$p,$i,$g,$c,$s,$ci,$z,$cno,$chk,$a,$pdf);
header("location:form.php");
}
?>
Controll.php
<?php
include('../model/model.php');
class control
{
public function validate($u,$p,$i,$c,$s,$ci,$z,$cno,$a,$pdf)
{
$obj = new model();
$obj->validate($u,$p,$i,$c,$s,$ci,$z,$cno,$a,$pdf);
}
public function insert($u,$p,$i,$g,$c,$s,$ci,$z,$cno,$chk,$a,$pdf)
{
$obj = new model();
$obj->insert($u,$p,$i,$g,$c,$s,$ci,$z,$cno,$chk,$a,$pdf);
}
}
?>
model.php
<?php
class model
{
public function validate($u,$p,$i,$c,$s,$ci,$z,$cno,$a,$pdf)
{
if(empty($u))
{
echo "<script>alert('Please enter Username')</script>";
}
else if(empty($p))
{
echo "<script>alert('Please enter password')</script>";
}
}
public function model()
{
$mysqli = new mysqli("localhost", "root", "", "php_mvc");
}
public function insert($u,$p,$i,$g,$c,$s,$ci,$z,$cno,$chk,$a,$pdf)
{
$mysqli = new mysqli("localhost", "root", "", "php_mvc");
$mysqli->query("INSERT INTO `php_mvc`.`form` (`fname`, `lname`, `pic`, `gender`, `country`, `state`, `city`, `zipcode`, `cno`, `hobby`, `address`, `pdf`) VALUES ('$u', '$p', '$i', '$g', '$c', '$s', '$ci', '$z', '$cno', '$chk', '$a', '$pdf');");
if($mysqli)
{
echo "<script>alert('Inserted Successfully')</script>";
}
header("location:form.php");
}
}
?>
【问题讨论】：
-
从模型的 validate() 方法返回布尔值（真/假）。如果验证失败，则返回 false。然后检查返回值是否为真，然后只调用插入。
-
我已经尝试过，但我无法从model.php中的validate()获取值到view.php
-
给自己留几行
$chk = implode(',', $h ?? []);或$chk = implode(',', $h ? $h: []);(PHP 5)
标签： php function validation model-view-controller