为什么php表单不提交？

我正在为一个网站的在线招生开发一个脚本。下面是页面的

php

代码。问题是它没有提交

    <?php
include ("include/header.php"), include ("include/config.php");

 if(isset($_POST['applyAdmission'])) {

  $admission_no = $_POST['admission_no'];
  $f_name = $_POST['f_name'];
  $l_name = $_POST['l_name'];
  $p_add = $_POST['p_add'];
  $c_add = $_POST['c_add'];
  $dob = $_POST['dob'];
  $education = $_POST['education'];
  $mobile = $_POST['mobile_no'];
  $course = $_POST['course'];
  $subjects = $_POST['subjects'];
  $timing = $_POST['timing'];

  $filepath_pic = $_FILES['picture']['name'];
  $res_move_pic = move_uploaded_file($_FILES['picture']['tmp_name'], "/admission/".$filepath_pic);

  $filepath_sign = $_FILES['sign']['name'];
  $res_move_sign = move_uploaded_file($_FILES['sign']['tmp_name'], "/admission/".$filepath_sign);

  $agree_terms = $_POST['agree_terms'];
  $agree_cond = $_POST['agree_cond'];


  if ($res_move_pic == 1 && $res_move_sign == 1 ) {

    $query = "INSERT into online_admission (f_name, l_name, p_add, c_add, dob, degree, mobile_no, course, subjects, timing, pic, sign, agree_terms, agree_cond, applied_on) 
             values ('$f_name','$l_name','$p_add','$c_add','$dob','$education','$mobile','$course','$subjects','$timing','$filepath_pic','$filepath_sign','$agree_terms','$agree_cond','now()')";
    $res = mysql_query($query) or die("ERROR: Unable to insert into database.");

    if ($res == 1) {
      header('Location:http://adarshclasses.in/admission_success.php/');
      exit();
     } else {
      header('Location:http://adarshclasses.in/admission_failed.php/');
      exit();
        }
     } else {
       echo "Error in updateing profile pic and sign";
   }
 } else {
   //echo "Please submit the form, thanks!";
}

;?>

将

更改为

。按钮可以使用javascript，但仅使用php按钮无法使用post数据。您无法通过

获取

POST

数据。为此，您必须使用

改变这个

<button type="text" name="submit" class="btn btn-success btn-lg btn-block" name="applyAdmission"> Submit my application form </button>

您可以查看更多详细信息

已更改

<button type="text">

到

更改

 button type="text" to type="button" Or input type ="submit/button"

include ("include/header.php"), include ("include/config.php");

<button type="text"...

---

您需要更改此代码：

<button type="text" name="submit" class="btn btn-success btn-lg btn-block" name="applyAdmission"> Submit my application form </button>

---

我看到了一个小语法错误，我认为解决这个问题将解决您的问题

改变

include ("include/header.php"), include ("include/config.php");

<button type="text"...

到

为了向您显示语法错误，下面是一个示例：

<?php

error_reporting(E_ALL);
ini_set('display_errors', 'On');

include("test.php"), include("someother.php");

输入类型不正确 您还应该更改按钮类型

改变

include ("include/header.php"), include ("include/config.php");

<button type="text"...

---

您正在使用的提交的HTML页面的输入类型背后的原因也请发布HTML。因此，我们可以找出问题所在。添加表单以获得完整性您的意思是单击“提交”按钮时不会发生任何事情？或者输入没有被保存？？请你解释清楚。并显示您的HTML。是否包含数据库文件？//这是什么意思？？您好，我用这种方式编辑了
提交我的申请表
，现在它看起来可以工作了，但是在上传签名和图片时，他们的问题更多了，您能不能也弄清楚。您面临的问题是什么？它说“无法插入数据”意味着SQL查询注释有问题。这不是错误。使用此
mysql\u错误（$connectionvariable）你所说的
mysql\u错误（$connectionvariable）是什么意思我是一名前端开发人员，不太了解，请简要告诉我。考虑到OPs第一行代码中存在语法错误，我认为这清楚地表明他们还没有进行任何Php调试。在旧版本的php中，你可能可以做任何事情。它不像新版本那样敏感。不过我不确定
include ("include/header.php"), include ("include/config.php");

include ("include/header.php");
include ("include/config.php");

<?php

error_reporting(E_ALL);
ini_set('display_errors', 'On');

include("test.php"), include("someother.php");

 Parse error: syntax error, unexpected ',' in ...\tests\includeTest.php on line 6

<button type="text"...

<button type="submit"...

<input type="submit" name="" value="Submit">