Question

<html>

<head>
<script src="check.js"></script>
<title>Sign Up</title>
</head>

<body>
<div id="container">
   <div id="child"

  <form action="insert.php" method="post">
    <br /><br />
    <input type="text" placeholder="Firstname" name="fn" maxlength="12" 
    autofocus/> <br /><br />
   <input type="text" placeholder="Lastname" name="ln" maxlength="12" /> <br 
    /><br />
   <input type="text" placeholder="username/student id" name="sid" 
    maxlength="12" /> <br /><br />
   <label>
    <input placeholder="password" name="password" id="password" 
     type="password" />
   </label>
  <br></br>
  <label>
  <input placeholder="confirm password" type="password"  
     id="confirm_password"  onkeyup='check();'/> 
    <br></br>
       <span id='message' ;"></span>
     </label>
    <br /><br />
   <input class="form-submit-button" type="submit" value="Sign in">
   </form>
    </div>
   </div>
   </body>

     </html>

此html页面应收集用户详细信息并将其链接到php文件，即insert.php

insert.php

<?php
$link = mysqli_connect("localhost", "root", "root", "demo");

if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
 }

 $sql = "INSERT INTO student_details VALUES ($_POST['sid'], 
  $_POST['fn'], $_POST['ln'], $_POST['password'])";

  if(mysqli_query($link, $sql)){
  echo "Records added successfully.";
  } else{
  echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
   }

  mysqli_close($link);
 ?>

此文件从表单中收集数据并将其插入表中。当我在html页面填写表格并提交时，没有任何事情发生。当我在localhost中单独执行代码时，它正在处理静态值。我是php和html的新手。我无法找到错误。

动态地将值插入myysql表

0 个答案: