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The PHP form information is not being inserted into the database – Phpmyadmin

Im trying to insert Form information into the database.A very simple code.

I’ve verified to all the other answers in the site but it does not work


        <?php 
        session_start();

    // variable declaration
    $school_name= "";
    $dob = "";
    $stu_name = "";
    $stu_id = "";
    $class = "";
    $section = "";
    $project = "";
    $_SESSION['success'] = "";

    // connect to database
  $db = mysql_connect('localhost', 'root', '');
  global $db;
  global $dbcreate;
  //creating database
  $dbcreate="CREATE DATABASE student";
  $retval=mysql_query($dbcreate,$db);
  //Creating table
  mysql_select_db("student",$db);


    if (isset($_POST['add'])){
        // receive all input values from the form
        $school_name = mysql_real_escape_string($_POST['school_name']);
        $dob = mysql_real_escape_string($_POST['dob']);
        $stu_name = mysql_real_escape_string( $_POST['stu_name']);
        $stu_id = mysql_real_escape_string( $_POST['stu_id']);
        $class = mysql_real_escape_string( $_POST['class']);
        $section = mysql_real_escape_string( $_POST['section']);
    $project = mysql_real_escape_string( $_POST['project']);
  }
      //Insert the data into the tables
      if(isset($_POST['add'])){
    $sql1="INSERT INTO  student(school_name,dob,stu_name,stu_id,class,section,project)
    VALUES ({$_POST['school_name']},{$_POST['dob']},{$_POST['stu_name']},{$_POST['stu_id']},{$_POST['class']},{$_POST['section']},{$_POST['project']})";
    $insertquery=mysql_query($sql1,$db);
      }


  ?>
<html>
  <head>
    <meta charset="utf-8">
    <title>Login</title>
    <link rel="stylesheet" href="style.css">
  </head>
  <body>

<div class="login-box">
  <h1>Students Details</h1>
  <form method="post" action="login.php">


  <div class="textbox">
    <i class="fas fa-user"></i>
    <input type="text" placeholder="School Name" name="school_name"value="<?php echo $school_name;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="Date" name="dob"value="<?php echo $dob;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="Student Name" name="stu_name"value="<?php echo $stu_name;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="Student Id" name="stu_id"value="<?php echo $stu_id;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="class" name="class"value="<?php echo $class;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="Section" name="section"value="<?php echo $section;?>">
  </div>

  <div class="textbox">
    <i class="fas fa-lock"></i>
    <input type="text" placeholder="Project" name="project"value="<?php echo $project;?>">
  </div>

  <button type="submit" class="btn" name="add" href="login.php">Add details </button>
</form>
</div>
</body>
</html>

I expected the information to be saved in the database.
I went on to check the database in localhost/phpmyadmin.

But nothing is saved.
don’t worry about the deprecated old mysql function I’m using a old version of XAMPP

Tags:

2

Answers


  1. 0

    try:

        if (isset($_POST['add'])) {
        // receive all input values from the form
        $school_name = mysql_real_escape_string($_POST['school_name']);
        $dob = mysql_real_escape_string($_POST['dob']);
        $stu_name = mysql_real_escape_string( $_POST['stu_name']);
        $stu_id = mysql_real_escape_string( $_POST['stu_id']);
        $class = mysql_real_escape_string( $_POST['class']);
        $section = mysql_real_escape_string( $_POST['section']);
        $project = mysql_real_escape_string( $_POST['project']);

        $sql1="INSERT INTO student(school_name,dob,stu_name,stu_id,class,section,project)
               VALUES ('{$school_name}', '{$dob}', '{$stu_name}', '{$stu_id}', '{$class}', '{$section}', '{$project}')";
        $insertquery = mysql_query($sql1, $db);
    }

    you’re also creating a database every time, only need to do this ONCE, but also need to create the table ONCE.

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  2. 0

    You are using string operations to populate a query. Never do this! Most languages and DB libraries will provide parameterized query functionality; PHP makes you jump through a slightly different hoop for this. Use prepared statements to ensure your string values, which are not quoted here, are both properly escaped and enquoted.

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