PHP: Inserting Values from the Form into MySQL

PHP: Inserting Values from the Form into MySQL

I created a users table in mysql from the terminal and I am trying to create simple task: insert values from the form. This is my dbConfig file 


and This is my Index.php . 


 
    
    
    
    

    

    test

    



     

    






After hitting my save button, nothing happens, database is still empty. I tried echo'ing the INSERT query and it takes all values from the form as it is supposed to. After I try to check if this worked from terminal, I login into my sql try to return all data from users table and I get empty set.

Solutions/Answers:

Answer 1:

The following code just declares a string variable that contains a MySQL query:

$sql = "INSERT INTO users (username, password, email)
    VALUES ('".$_POST["username"]."','".$_POST["password"]."','".$_POST["email"]."')";

It does not execute the query. In order to do that you need to use some functions but let me explain something else first.

NEVER TRUST USER INPUT: You should never append user input (such as form input from $_GET or $_POST) directly to your query. Someone can carefully manipulate the input in such a way so that it can cause great damage to your database. That’s called SQL Injection. You can read more about it here

To protect your script from such an attack you must use Prepared Statements. More on prepared statements here

Include prepared statements to your code like this:

$sql = "INSERT INTO users (username, password, email)
    VALUES (?,?,?)";

Notice how the ? are used as placeholders for the values. Next you should prepare the statement using mysqli_prepare:

$stmt = mysqli_prepare($sql);

Then start binding the input variables to the prepared statement:

$stmt->bind_param("sss", $_POST['username'], $_POST['email'], $_POST['password']);

And finally execute the prepared statements. (This is where the actual insertion takes place)

$stmt->execute();

NOTE Although not part of the question, I strongly advice you to never store passwords in clear text. Instead you should use password_hash to store a hash of the password

Answer 2:

There are two problems in your code.

  1. No action found in form.
  2. You have not executed the query mysqli_query()

dbConfig.php

<?php

$conn=mysqli_connect("localhost","root","password","testDB");

if(!$conn)
{
die("Connection failed: " . mysqli_connect_error());
}

?>

index.php

 include('dbConfig.php');

<!Doctype html>
<html>
<head>
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1">
<meta name="description" content="$1">
<meta name="viewport" content="width=device-width, initial-scale=1">

<link rel="stylesheet" type="text/css" href="style.css">

<title>test</title>


</head>
<body>

 <?php

  if(isset($_POST['save']))
{
    $sql = "INSERT INTO users (username, password, email)
    VALUES ('".$_POST["username"]."','".$_POST["password"]."','".$_POST["email"]."')";

    $result = mysqli_query($conn,$sql);
}

?>

<form action="index.php" method="post"> 
<label id="first"> First name:</label><br/>
<input type="text" name="username"><br/>

<label id="first">Password</label><br/>
<input type="password" name="password"><br/>

<label id="first">Email</label><br/>
<input type="text" name="email"><br/>

<button type="submit" name="save">save</button>

</form>

</body>
</html>

Answer 3:

Try this:

dbConfig.php

<?php
$mysqli = new mysqli('localhost', 'root', 'pwd', 'yr db name');
    if($mysqli->connect_error)
        {
        echo $mysqli->connect_error;
        }
    ?>

Index.php

<html>
<head><title>Inserting data in database table </title>
</head>
<body>
<form action="control_table.php" method="post">
<table border="1" background="red" align="center">
<tr>
<td>Login Name</td>
<td><input type="text" name="txtname" /></td>
</tr>
<br>
<tr>
<td>Password</td>
<td><input type="text" name="txtpwd" /></td>
</tr>
<tr>
<td>&nbsp;</td>
<td><input type="submit" name="txtbutton" value="SUBMIT" /></td>
</tr>
</table>
control_table.php
<?php include 'config.php'; ?>
<?php
$name=$pwd="";
    if(isset($_POST['txtbutton']))
        {
            $name = $_POST['txtname'];
            $pwd = $_POST['txtpwd'];
            $mysqli->query("insert into users(name,pwd) values('$name', '$pwd')");
        if(!$mysqli) 
        { echo mysqli_error(); }
    else
    {
        echo "Successfully Inserted <br />";
        echo "<a href='show.php'>View Result</a>";
    }

         }  

    ?>

Answer 4:

<!DOCTYPE html>
<?php
$con = new mysqli("localhost","root","","form");

?>



<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
<script type="text/javascript">
$(document).ready(function(){
 //$("form").submit(function(e){

     $("#btn1").click(function(e){
     e.preventDefault();
    // alert('here');
        $(".apnew").append('<input type="text" placeholder="Enter youy Name" name="e1[]"/><br>');

    });
    //}
});
</script>

</head>

<body>
<h2><b>Register Form<b></h2>
<form method="post" enctype="multipart/form-data">
<table>
<tr><td>Name:</td><td><input type="text" placeholder="Enter youy Name" name="e1[]"/>
<div class="apnew"></div><button id="btn1">Add</button></td></tr>
<tr><td>Image:</td><td><input type="file"  name="e5[]" multiple="" accept="image/jpeg,image/gif,image/png,image/jpg"/></td></tr>

<tr><td>Address:</td><td><textarea  cols="20" rows="4" name="e2"></textarea></td></tr>
<tr><td>Contact:</td><td><div id="textnew"><input  type="number"  maxlength="10" name="e3"/></div></td></tr>
<tr><td>Gender:</td><td><input type="radio"  name="r1" value="Male" checked="checked"/>Male<input type="radio"  name="r1" value="feale"/>Female</td></tr>
<tr><td><input  id="submit" type="submit" name="t1" value="save" /></td></tr>
</table>
<?php
//echo '<pre>';print_r($_FILES);exit();
if(isset($_POST['t1']))
{
$values = implode(", ", $_POST['e1']);
$imgarryimp=array();
foreach($_FILES["e5"]["tmp_name"] as $key=>$val){


move_uploaded_file($_FILES["e5"]["tmp_name"][$key],"images/".$_FILES["e5"]["name"][$key]);

                     $fname = $_FILES['e5']['name'][$key];
                     $imgarryimp[]=$fname;
                     //echo $fname;

                     if(strlen($fname)>0)
                      {
                         $img = $fname;
                      }
                      $d="insert into form(name,address,contact,gender,image)values('$values','$_POST[e2]','$_POST[e3]','$_POST[r1]','$img')";

       if($con->query($d)==TRUE)
         {
         echo "Yoy Data Save Successfully!!!";
         }
}
exit;





                      // echo $values;exit;
                      //foreach($_POST['e1'] as $row) 
    //{ 

    $d="insert into form(name,address,contact,gender,image)values('$values','$_POST[e2]','$_POST[e3]','$_POST[r1]','$img')";

       if($con->query($d)==TRUE)
         {
         echo "Yoy Data Save Successfully!!!";
         }
    //}
    //exit;


}
?>

</form>

<table>
<?php 
$t="select * from form";
$y=$con->query($t);
foreach ($y as $q);
{
?>
<tr>
<td>Name:<?php echo $q['name'];?></td>
<td>Address:<?php echo $q['address'];?></td>
<td>Contact:<?php echo $q['contact'];?></td>
<td>Gender:<?php echo $q['gender'];?></td>
</tr>
<?php }?>
</table>

</body>
</html>

Answer 5:

<?php
    $username="root";
    $password="";
    $database="test";

    #get the data from form fields
    $Id=$_POST['Id'];
    $P_name=$_POST['P_name'];
    $address1=$_POST['address1'];
    $address2=$_POST['address2'];
    $email=$_POST['email'];

    mysql_connect(localhost,$username,$password);
    @mysql_select_db($database) or die("unable to select database");

    if($_POST['insertrecord']=="insert"){
        $query="insert into person values('$Id','$P_name','$address1','$address2','$email')";
        echo "inside";
        mysql_query($query);
        $query1="select * from person";
        $result=mysql_query($query1);
        $num= mysql_numrows($result);

        #echo"<b>output</b>";
        print"<table border size=1 > 
        <tr><th>Id</th>
        <th>P_name</th>
        <th>address1</th>
        <th>address2</th>
        <th>email</th>
        </tr>";
        $i=0;
        while($i<$num)
        {
            $Id=mysql_result($result,$i,"Id");
            $P_name=mysql_result($result,$i,"P_name");
            $address1=mysql_result($result,$i,"address1");
            $address2=mysql_result($result,$i,"address2");
            $email=mysql_result($result,$i,"email");
            echo"<tr><td>$Id</td>
            <td>$P_name</td>
            <td>$address1</td>
            <td>$address2</td>
            <td>$email</td>
            </tr>";
            $i++;
        }
        print"</table>";
    }

    if($_POST['searchdata']=="Search")
    {
        $P_name=$_POST['name'];
        $query="select * from person where P_name='$P_name'";
        $result=mysql_query($query);
        print"<table border size=1><tr><th>Id</th>
        <th>P_name</th>
        <th>address1</th>
        <th>address2</th>
        <th>email</th>
        </tr>";
        while($row=mysql_fetch_array($result))
        {
            $Id=$row[Id];
            $P_name=$row[P_name];
            $address1=$row[address1];
            $address2=$row[address2];
            $email=$row[email];
            echo"<tr><td>$Id</td>
            <td>$P_name</td>
            <td>$address1</td>
            <td>$address2</td>
            <td>$email</td>
            </tr>";
        }
        echo"</table>";
    }
    echo"<a href=lab2.html> Back </a>";
?>

Answer 6:

When you click the Button

if(isset($_POST['save'])){
        $sql = "INSERT INTO `members`(`id`, `membership_id`, `email`, `first_name`)
        VALUES ('".$_POST["id"]."','".$_POST["membership_id"]."','".$_POST["email"]."','".$_POST["firstname"]."')";
    **if ($conn->query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }**
}

This will execute the Query in the variable $sql

    if ($conn->query($sql) === TRUE) {
        echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }

Our Awesome Free Tools

References

Bootstrap collapse on radio button

Bootstrap collapse on radio button

I have found this LINK on Stack overflow and what i want to achieve is that when user clicks on a radio button that div stays in that state.

For example if i click on Yes the panel should be visible and if i click again on Yes it MUST not close the panel like it does now.
Anyone knows how to fix it?
here is the source code:

Bootstrap collapse panel with radio buttons


Display panel: Yes No

Header

Content

Solutions/Answers:

Answer 1:

Bootstrap and css only solution:
Remove the panel’s id and add a unique class to the class instead. Change the data-target selector for .collapseOne:not(.in) on “Yes” and .collapseOne:not.in for “No”.

<h2>Bootstrap collapse panel with radio buttons</h2>

<br />
<div class="container">
  <div class="row">
    <div class="col-xs-12">
      <span>Display panel: </span>

        <input name="collapseGroup" type="radio" data-toggle="collapse" data-target=".collapseOne:not(.in)"/> Yes

        <input name="collapseGroup" type="radio" data-toggle="collapse" data-target=".collapseOne.in" checked/> No


      <div class="panel-group" id="accordion">
        <div class="panel panel-default">
          <div class="panel-heading">
            <h4 class="panel-title">
                    Header
                </h4>
          </div>
          <div class="collapseOne panel-collapse collapse">
            <div class="panel-body">
              <p>Content</p>
            </div>
          </div>
        </div>
      </div>

    </div>
  </div>
</div>

Answer 2:

You can remove data-toggle="collapse" and data-target="#collapseOne"

Add values to your inputs value="yes" and value="no"

And hide/show collapse with javascript on input change event

$('[name="collapseGroup"]').on('change', function() {  
  if($(this).val() === "yes") {
    $('#collapseOne').collapse('show');
  } else {
    $('#collapseOne').collapse('hide');
  }
});

Codepen

Answer 3:

You can do it using jQuery (adding IDs to inputs):

var yesRadio = $('#yes'),
    noRadio = $('#no');

yesRadio.click(function () {
  if($('#collapseOne').hasClass('in')) {
    return false;
  }
});

noRadio.click(function () {
  if(!$('#collapseOne').hasClass('in')) {
    return false;
  }
});

CODEPEN

Answer 4:

You can do this by some JQuery

HTML:

 <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.3/jquery.min.js"></script>
<!-- radio button with calss -->
<input name="collapseGroup" type="radio" class="yes" data-toggle="collapse" data-target="#collapseOne"/> Yes
<input name="collapseGroup" type="radio" class="no" data-toggle="collapse" data-target="#collapseOne" checked/> No
<!-- content to show/hide -->
<p>
   content
 </p>

Jquery:

$(document).ready(function(){
    $('.yes').click(function(){
        $('p').slideDown(); //to show
    });
    $('.no').click(function(){
        $('p').slideUp();  //to hide
    });
});

CSS:

p{
   background: white;
   border: 1px solid #c6c6c6;
   border-radius: 5px;
   padding: 15px;
   display: none;
 }

and here is the live example https://jsfiddle.net/54jxaoas/1/

Answer 5:

I think better and easier solution is to use just a checkbox

<input type="checkbox" id="check" />

if ($('#check').is(':checked')) {
$('p').slideDown(); 
 }
else {
$('p').slideUp(); 
}

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References