PHP - Need Prevent Inserting Duplicate Record In Datbase When Refresh Form

[/code]
Hi //open the connection
$conn = mysql_connect("localhost", "techhom1_test", "pro176");

//pick the database to use
mysql_select_db("testDB", $conn);

//create the sql statement
$sql = "INSERT INTO master_name values ('', '$firstname')";

//execute
if (mysql_query($sql,$conn)){
echo "Record Added!";
} else{
echo "Somethin went wrong";
}


the code above is insert code and working with out any errors but when i am submitting text into data using method below it is sending echo back Somethin went wrong.I think i am confused with the write name of the user table

database is ***_test
my tables are
duty_number
 master_name

in this _test user i am trying to add first name in master_name field so here is the submit code

Code: [Select]
<FORM ACTION=insert.php  METHOD=POST>
<P>text to add:<br>
<input type=text name=master_name size=30>
<p><input type=submit name=submit value=Insert Record></p>
</FORM>


My table below in Data base

[code]Field Type Null Default Comments
name_id smallint(5) Yes NULL
name_dateaddedn datetime Yes NULL
name_datemodified datetime Yes NULL
firstname varchar(75) Yes NULL
lastname varchar(75) Yes NULL

can someone please look at the code and find the error for me Thanks

Am adding text with paragraphs into my database but when i view that data on another page it's just one big block of text. :S Anyone no how i can stop this?

I have 2 files; Newfault.php and thankyou.php

1) Data is entered into a form in Newfault.php
2) The data from this form is retieved in thankyou.php and is then inserted into a table called "Calls"

Problem:

My entered record is being added to my database OK, but it is also adding a blank record for some reason and I can't work out why. Can anyone help? This is for my uni assignment.

Thanks,

Ladykudos

Here is my add driver button

Hello,

I have looked on the forum, but haven't found the answer to this.

I have a basic [name, email, phone, event] web form and want to make sure there person is not submitting twice by doing duplicate check of the mail.  I looked at other solutions that suggested setting up unique IDs for each user, but it would be possible for a user to sign up for more then one event.

Any info would be helpful. Not a newbie, but not an expert.

First of all excuse me if this topic is inappropriate in this forum. But I think it's rather a PHP problem.

I can't figure out multiple duplicate database records on submitting a form.

The database table have two columns:  the first one 'Id' with AUTO_INCREMENT and the second one 'Name'.

Here's the php code for database insertion and the form:
------------------------------------------------------------------

<?php

if($_GET['add_name']){
    $host = *******;
    $user = *******';
    $pass = *******;
    $db = *******;
    $con = mysql_connect($host,$user,$pass) or die;
    mysql_select_db($db,$con);
   
    $name = $_GET['add_name'];
    $sql = "INSERT INTO names (Name) VALUES ('$name')";
    mysql_query($sql);
}

?>

<form action="<?php echo $_SERVER['PHP_SELF'] ?>" method="GET">
Your Name:
<input name="add_name" type="text"  />
<input type="submit" value="Submit" />
</form>

------------------------------------------------------------------

After submitting the form to itself once I have multiple Name entries with different Ids. The curious thing is that with Chrome browser
I get two duplicate records, with Mozilla - three of them.
Seems like mysql_query runs multiple times.

It works fine when submitting the form to a separate script and not to itself.

Do I miss something? It must be very basic.

I have a table called "colors".  It has 2 columns, id and color.

All I'm trying to do is pull the data into the form, then edit the colors (for example - misspelled, etc.)
Then submit it back to the table.

Here is the code that pulls the data into a form so I can edit it:
Code: [Select]
<form action="adminupdatecolors.php" method="post">

<table width="500" border="1" cellpadding="10">

<tr>
  <td>Color Options:</td>
    <td>
    <?
$result = mysql_query("SELECT * FROM colors");
while ($row = mysql_fetch_assoc($result)) {

echo '<input type="text" name="color[]" value="' . $row['color'] . '"/>' . $row['id'] . '<br />';
}

?>
   </td>
</tr>

   <tr>
  <td>&nbsp;</td>
  <td><input type="Submit" value="Update Colors"></td>
  </tr>
</form>
</table>

Here is my update script:
Code: [Select]
<?php

include("config.php");
include("db.php");

$id=$_POST['id'];
$color=$_POST['color'];

$result = mysql_query("SELECT id FROM colors ORDER BY id DESC LIMIT 0,1");

if ($row = mysql_fetch_assoc($result)) {
$id = $row['id']; 
}
$sql = "DELETE FROM colors WHERE id='$id'";
mysql_query($sql) or die("Error: ".mysql_error());

foreach ($color as $colorvalue) {
$sql2 = "INSERT INTO colors (id,color) VALUES ($id,'$color')";
mysql_query($sql2) or die("Error: ".mysql_error());

}

header("Location: " . $config_basedir . "adminhome.php");

?>

Hi, I cannot get my form to find the number "555" entered for a user in the database. There is a table field called "dob" with value 555 for a particular user. Below is the code I am trying to search the table and output a match. Please if anyone can help that would be great. thank you.

Code: [Select]
<?php 
$host = "";
$database  = "";
$username  = "";
$password  = "*";
$tbl_name   = "users";
$conn = mysql_connect($host, $username, $password) or die("Could not connect: " . mysql_error());
mysql_select_db($database);
 

$dob = $_POST['dob'];
 
//THE SEARCH FUNCTION
$result = mysql_query ( "SELECT * FROM users WHERE dob  LIKE '%$dob%' ");

if(isset($_POST['submit'])) 

{
while ($row = mysql_fetch_assoc($result)) {
    $message = $row["dob"];
    
 
}

 }
?>
 <html>
 <body>

<form action="login_successBACKUP02.php" method="post">
 
 
    <input name="dob" type="text" size="20" id="dob" />
    Date of Birth<br />
   
    <input type="submit" value="Store in database and search" />
    <input type="reset" value="Reset fields" />
 
 
</form> 
 
 
</body>
</html>

Hello PHP experts.
Please help me on this.
I have created a form at http://lakshwebdesign.com/clients/onev8llc/qualitynew-site/instantfreequote.html
How do I add these 3 PHP functionalities to it:-

1) On submit, all the data be sent to an email address.
2) Also, record it in database.
3) Record date on which the form is submitted.

I have knowledge of PHP and I can understand the codes somewhat so I am not a complete beginner of PHP.

Any help will be highly appreciated.

Thanks in advance.

Hello all,

As the title says: For sake of example, let's say I have a form with a couple text boxes and a submit button. When you hit submit, the data from the text boxes is translated into a database record.

Howver, refershing the page will this record again and again -- which I do not want.

How can I prevent this?

Keeping in mind the business logic of my applicatoin allows the same record to be entered twice -- however, it should only happen if the user intentionally visits the form agian, and enters the same data. It should not happen on a page refresh.

I assume this is a common problem...? Any thoughts?

Thanks!

Well the problem is simple i want to prevent submitting empty textfield or textarea with php code

Here is my code ... but it doesn't work as i wish to 
if((empty($text) == "") || strlen($text) < 1)
{
        {
        echo "L&#363;dzu Aizpildat Aili";
        }
}

Also somebody advised me to use this code but i found it far more useless then mine ...
function check_empty($text)
{
        if(empty($text) || !empty($text) || $text || !$text || $text ^ $text || 1/0)
                check_empty(!!!!!$text);
        
        return never_ever;
} // check_empty

I have two scripts: script1.php and script2.php.   Script1 creates if it doesn't already exist and adds to a session named "SESSION1" and displays it:   Script2 similarly adds to a session named "SESSION2", but then needs to display the session used by the first script (i.e. SESSION1), and then goes back to its original session (SESSION2).   Script1 works perfect.  But when Script2 is executed, it changes the session ID in the SESSION1 cookie to the same value as used in its SESSION2 cookie.  If Script1 is later executed, it obviously lost its previous session values as it is now using a new session ID.   If I comment out the two session_name() lines, it will not overwrite the other session, however, this doesn't provide the functionality I need.   What is causing this and how do I prevent it????   script1.php

<?php
// script 1.  Will be accessed as http://one.example.com

$t=time();

//Access the primary session for script 1
session_name('SESSION1');
session_start();
$_SESSION['s1_'.$t]=$t;
echo("SESSION1<pre>".print_r($_SESSION,1)."</pre>");
?>

script2.php

<?php
// script 2.  Will be accessed as http://two.one.example.com

$t=time();

//Access the primary session for script 2
$default_name=session_name('SESSION2');
session_start();
$_SESSION['s2_'.(2*$t)]=2*$t;
echo("SESSION2<pre>".print_r($_SESSION,1)."</pre>");

//Use session created by script 1
$old_id_script2 = session_id();
session_write_close();
$old_name_script2 = session_name('SESSION1');
session_start();
echo("SESSION1<pre>".print_r($_SESSION,1)."</pre>");

//Go back to primary session
session_write_close();
$old_id_script1 = session_id($old_id_script2);
$old_name_script1 = session_name($old_name_script2);
session_start();
echo("SESSION2<pre>".print_r($_SESSION,1)."</pre>");

echo("default_name: $default_name<br>");
echo("old_id_script2: $old_id_script2<br>");
echo("old_name_script2: $old_name_script2<br>");
echo("old_id_script1: $old_id_script1<br>");
echo("old_name_script1: $old_name_script1<br>");
?>

Edited by NotionCommotion, 30 November 2014 - 11:45 AM.

Hi all,   Its been a long time since last help request from real professional from here but I'm again in trouble with a much more spectacular plan I'm working on.

For those who are interested in the plan then here it is:
My Idea was to make a new build starting from scratch and make it as dynamical as possible. So my goal is not to make almost anything fixed in the code. I have made a decision to make a one supper large table for multiple different entries so no more joining and no more views for me.! In this help request I'm having trouble with Posting values to a page processing page lets call it record_changer.php The sole purpose of this file is to get form posts and decide what to do. Either update, delete, or insert.

record_changer.php
 

<?php

include '../../config/config.inc.php';

if(is_ajax()){
# Checks if action value exists
  if(isset($_POST["action"]) && !empty($_POST["action"])){
    $action = $_POST["action"];
# Switch case for value of action
    switch($action){
			case "insert": datatable_insert_function();
			break;
			case "update": datatable_update_function();
			break;
			case "delete": datatable_delete_function();
			break;
    }
  }
}

# Function to check if the request is an AJAX request
function is_ajax(){ return isset($_SERVER['HTTP_X_REQUESTED_WITH']) && strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest'; }

function datatable_insert_function(){
}

function datatable_update_function(){
}

function datatable_delete_function(){
}

?>

The problem.

The problem is that the $insert places two entries to the DB. I cant seem to understand why.?

	# Test _POST values
	$_POST['UserID'] = '2';
	$_POST['WorkID'] = '22';
	$_POST['Status'] = '1';
	$_POST['Code'] = '1';
	$_POST['Title'] = '1';


	$columns = array();
	foreach(array_keys($_POST) as $name){
		# Exclude Action and ID
		if($name == 'Action' || $name == 'ID' || $name == 'submit' ){ continue; }
		$columns[] = $name;
	}

	print_r($columns);

	echo "<br>";

	$data = array_fill_keys($columns, 'NULL');

	print_r($data);

	foreach($data as $key => $value){
		$data[$key] = empty($_POST[$key]) ? 'NULL' : "'".mysql_real_escape_string($_POST[$key])."'";
	}

	echo "<br>";

	print_r($data);

	$insert = mysql_query('INSERT INTO datatable (ID, '.implode(', ',$columns).')VALUES (null, '.implode(',',$data).')') or die(mysql_error());

No errors no nothing. Just two entries of correct data. 
PS. Sorry for a lot of prints in the code it is work and idea in the progress. The posts at the moment are fixed in the code so it is easier to refresh and debug.

Please help if you spot the problem. Im really out of ideas. Some fresh eyes might make a difference.

And Please for those who want to say it is a bad idea and why and why and so on.. Move a long.!!! Im not interested in whinging i have a great use for this and just having trouble with the two entries.

Thanks.
Edited by ztimer, 14 January 2015 - 03:27 PM.

I am trying to get this query correct.  I want to insert a record into the database upon form submission but only if the record does not already exist.  If the record exists, then I want it to be updated in the database.    What is happening:  Upon form submit, a new record is entered into the database every time.  Note:  The contact_id column is both primary key and unique in my database.  Here is my code:

if($_POST['submit']){
  $con=mysqli_connect("localhost","username","password","database_name");
  // Check connection
  if (mysqli_connect_errno())
    {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
	
	$org = mysql_real_escape_string($_POST['organization']);
  	$namefirst = mysql_real_escape_string($_POST['firstName']);
        $namelast = mysql_real_escape_string($_POST['lastName']);
	$emailaddy = mysql_real_escape_string($_POST['email']);
	$phonenum = mysql_real_escape_string($_POST['phone']);
	$appquestion = mysql_real_escape_string($_POST['appquestion']);
	$banner = mysql_real_escape_string($_POST['banner']);
	$bulletin = mysql_real_escape_string($_POST['bulletin']);
	$giveaway = mysql_real_escape_string($_POST['giveaway']);
	$app = mysql_real_escape_string($_POST['app']);
	$tshirt = mysql_real_escape_string($_POST['tshirt']);
	$tshirtp = mysql_real_escape_string($_POST['tshirtp']);
	$print = mysql_real_escape_string($_POST['print']);
	$party = mysql_real_escape_string($_POST['party']);
	$orgnotes = mysql_real_escape_string($_POST['notes']);
	
 
$sql="INSERT INTO database_name (contact_id, first_name, last_name, email_address, phone_number, org, appquestion, banner, bulletin, giveaway, app, tshirt, promised_tee, print, party, org_notes)
      VALUES
      ('','$namefirst','$namelast','$emailaddy','$phonenum','$churchorg','$appquestion','$banner','$bulletin','$giveaway','$app','$tshirt','$tshirtp','$print','$party','$orgnotes')	
	    
	  ON DUPLICATE KEY UPDATE first_name = '$namefirst', last_name = '$namelast', email_address = '$emailaddy', phone_number = '$phonenum', org = '$org', appquestion = '$appquestion', banner = '$banner', bulletin = '$bulletin', giveaway = '$giveaway', app = '$app', tshirt = '$tshirt', promised_tee = '$tshirtp', print = '$print', party = '$party', org_notes = '$orgnotes'" ;

						

    if (!mysqli_query($con,$sql))
    {
      die('Error: ' . mysqli_error($con));
    }
      echo "1 record added";



  mysqli_close($con);
}

From everything I have read, I need to use ON DUPLICATE KEY UPDATE to replace the old information with new information in the database upon form submission.  While the insert part of my code is working, the portion with ON DUPLICATE KEY UPDATE is not working.   Why might this portion of the code not be working?  Is there a better way to insert else update the information?   Thank you for any help or guidance you can give me!  I've been working on this concept for three days and have read a ton of information about it, but am still not able to get it to work.

I am trying to set up a item entry page form.png:

Upon submission it shows unsuccessful even though I have checked the fields on mysql table and seem to be good am I missing something?

Code: [Select]
<form action="" method="post" enctype="multipart/form-data" name="Product_Entry">
<TABLE>
<TR>
<TD>Product ID</TD><TD><input name="SKU_ProductID" value="<?php if (isset($_post['SKU_ProductID'])) echo $_POST['SKU_ProductID']; ?>" type="text" size="11" maxlength="11" /></TD>
</TR>
<TR>
<TD>Merchant SKU ID</TD><TD><input name="SKU_MerchSKUID" value="<?php if (isset($_post['SKU_MerchSKUID'])) echo $_POST['SKU_MerchSKUID']; ?>" type="text" size="18" maxlength="30" /></TD>
</TR>
<TR>
<TD>Game Title</TD><TD><input name="Game_Title" value="<?php if (isset($_post['Game_Title'])) echo $_POST['Game_Title']; ?>" type="text" size="40" maxlength="40" /></TD>
</TR>
<TR>
<TD>Platform</TD><TD><input name="Platform" value="<?php if (isset($_post['Platform'])) echo $_POST['Platform']; ?>" type="text" size="20" maxlength="20" /></TD>
</TR>
<TR>
<TD>Genre</TD><TD><input name="Genre" value="<?php if (isset($_post['Genre'])) echo $_POST['Genre']; ?>" type="text" size="11" maxlength="11" /></TD>
</TR>
<TR>
<TD>Weight</TD><TD><input name="Weight" value="<?php if (isset($_post['Weight'])) echo $_POST['Weight']; ?>" type="text" size="7" maxlength="7" /></TD>
</TR>
<TR>
<TD>Supplier</TD><TD><input name="Supplier" Value="<?php if (isset($_post['Supplier'])) echo $_POST['Supplier']; ?>" type="text" size="25" maxlength="25" /></TD>
</TR>
<TR>
<TD>Suppliers Price</TD><TD><input name="Supplier_Price" value="<?php if (isset($_post['Supplier_Price'])) echo $_POST['Supplier_Price']; ?>" type="text" size="10" maxlength="12" /></TD>
</TR>
<TR>
<TD>Cashback</TD><TD><input name="Cashback" value="<?php if (isset($_post['Cashback'])) echo $_POST['Cashback']; ?>" type="text" size="6" maxlength="8" /></TD>
</TR>
<TR>
<TD>Cashback Amount</TD><TD><input name="Cashback_Amount" value="<?php if (isset($_post['Cashback_Amount'])) echo $_POST['Cashback_Amount']; ?>" type="text" size="7" maxlength="11" /></TD>
</TR>
<TR>
<TD><input type="hidden" name="submitted" value="true"/></TD><TD><input name="Submit" type="submit" value="Add_Product" /></TD>
</TR></form></TABLE>

<?php # Product Entry

$page_title = 'Product Entry';

if (isset($_POST['Submit'])) {

$errors = array();

if (empty($_POST['SKU_ProductID'])) {
$errors[] = 'Please enter Product ID.';
} else {
$sid = trim($_POST['SKU_ProductID']);
}

if (empty($_POST['SKU_MerchSKUID'])) {
$errors[] = 'Please enter Merchant SKU.';
} else {
$mid = trim($_POST['SKU_MerchSKUID']);
}

if (empty($_POST['Game_Title'])) {
$errors[] = 'Please enter Game Title.';
} else {
$gt = trim($_POST['Game_Title']);
}

if (empty($_POST['Platform'])) {
$errors[] = 'Please enter Platform.';
} else {
$pl = trim($_POST['Platform']);
}

if (empty($_POST['Genre'])) {
$errors[] = 'Please enter Genre.';
} else {
$ge = trim($_POST['Genre']);
}

if (empty($_POST['Weight'])) {
$errors[] = 'Please enter Weight.';
} else {
$we = trim($_POST['Weight']);
}

if (empty($_POST['Supplier'])) {
$errors[] = 'Please enter Supplier.';
} else {
$sup = trim($_POST['Supplier']);
}

if (empty($_POST['Supplier_Price'])) {
$errors[] = 'Please enter Supplier Price.';
} else {
$sp = trim($_POST['Supplier_Price']);
}

if (empty($_POST['Cashback'])) {
$errors[] = 'Please enter Cashback %.';
} else {
$cb = trim($_POST['Cashback']);
}

if (empty($_POST['Cashback_Amount'])) {
$errors[] = 'Please enter Cashback Amount.';
} else {
$cba = trim($_POST['Cashback_Amount']);
}

if (empty($errors)) {

require_once ('connect.php');

[b]$q = "INSERT INTO `Products` (`SKU_ProductID`, `SKU_MerchSKUID`, `Game_Title`, `Platform`, `Genre`, `Weight`, `Supplier`, `Supplier_Price`, `Cashback`, `Cashback_Amount`) VALUES ('$sid', '$mid', '$gt', '$pl', '$ge', '$we', '$sup', '$sp', '$cb', '$cba')";
$r = @mysql_query ($dbc, $q);

if ($r) { // If it ran OK.

// Print a message:
echo '<h1>Thank you!</h1>
<p>Product Inserted!</p><p><br /></p>';

} else { // If it did not run OK.

// Public message:
echo '<h1>System Error</h1>
<p class="error">You could not be registered due to a system error. We apologize for any inconvenience.</p>'; 

// Debugging message:
echo '<p>' . mysqli_error($dbc) . '<br /><br />Query: ' . $q . '</p>';

} // End of if ($r) IF.

mysqli_close($dbc); // Close the database connection.[/b]
// Include the footer and quit the script:
include ('includes/footer.html'); 
exit();

} else {

echo '<H1>Error!</H1>
<p class="error">The Following error(s) occurred:<br />';

foreach ($errors as $msg) {

echo " - $msg<br />\n";
}

echo '</p><p>Please try again.</p><p><br /></p>';

}

}

?>
If there is anything else needed let me know