displaying duplicate entry from database with php - php

I am trying to display data from search function in database but unfortunately i get duplicate entries, can anyone check this and help me a little bit:
I am using this query and this php Code where startdate and enddate are from the members table, all the others data are from person table:
$query = "select * FROM person, members where fname= '$fname' AND lname = '$lname'";
for ($i=0; $i <$num_results; $i++)
{
$row = $result->fetch_assoc();
echo '<p><strong>'.($i+1).'. Name: ';
echo htmlspecialchars(stripslashes($row['fname']));
echo '</strong><br />Surname: ';
echo stripslashes($row['lname']);
echo '<br />Board: ';
echo stripslashes($row['board']);
echo '<br />Department: ';
echo stripslashes($row['departmentname']);
echo '<br />Start Date: ';
echo stripslashes($row['startdate']);
echo '<br />End Date: ';
echo stripslashes($row['enddate']);
echo '</p>';
This is what i get as output: So i want to be displayed only on person with name aa nd surname bb (not five times but only once).
1. Name: aa
Surname: bb
Board: Secur
Department: Telec
Start Date: 2016-07-01
Edn Date: 2016-07-31
2. Name: aa
Surname: bb
Board: Secur
Department: Telec
Start Date: 2016-07-19
Edn Date: 2016-07-21
after searching during the edit process the ID is not changing at all
<?php
function renderForm($personid, $personname, $personsurname, $error)
{
?>
<html>
<head>
<title>Edit Record</title>
</head>
<body>
<?php
// if there are any errors, display them
if ($error != '')
{
echo '<div style="padding:4px; border:1px solid red; color:red;">'.$error.'</div>';
}
?>
<form action="" method="post">
<input type="hidden" name="personid" value="<?php echo $personid; ?>"/>
<div>
<p><strong>ID:</strong> <?php echo $personid; ?></p>
<strong>First Name: *</strong> <input type="text" name="personname" value="<?php echo $personname; ?>"/><br/>
<strong>Last Name: *</strong> <input type="text" name="personsurname" value="<?php echo $personsurname; ?>"/><br/>
<p>* Required</p>
<input type="submit" name="submit" value="Submit">
</div>
</form>
</body>
</html>
<?php
}
// connect to the database
$host = "localhost";
$user = "kkoikm_kum";
$pass = "datgbhnkum";
$db = "koikm_kum";
// open connection
$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");
// select database
mysql_select_db($db) or die ("Unable to select database!");
// check if the form has been submitted. If it has, process the form and save it to the database
if (isset($_POST['submit']))
{
// confirm that the 'id' value is a valid integer before getting the form data
if (is_numeric($_POST['personid']))
{
// get form data, making sure it is valid
$personid = $_POST['personid'];
$personname = mysql_real_escape_string(htmlspecialchars($_POST['personname']));
$personsurname = mysql_real_escape_string(htmlspecialchars($_POST['personsurname']));
// check that firstname/lastname fields are both filled in
if ($personname == '' || $personsurname == '')
// generate error message
$error = 'ERROR: Please fill in all required fields!';
//error, display form
renderForm($personid, $personname, $personsurname, $error);
}
else
{
// save the data to the database
mysql_query("UPDATE tblperson SET personname='$personname', personsurname='$personsurname' WHERE personid='$personid'")
or die(mysql_error());
// once saved, redirect back to the view page
header("Location: home.php");
}
}
else
{
// if the 'id' isn't valid, display an error
echo 'Error!';
}
}
else
// if the form hasn't been submitted, get the data from the db and display the form
{
// get the 'id' value from the URL (if it exists), making sure that it is valid (checing that it is numeric/larger than 0)
if (isset($_GET['personid']) && is_numeric($_GET['personid']) && $_GET['personid'] > 0)
{
// query db
$personid = $_GET['personid'];
$result = mysql_query("SELECT * FROM tblperson WHERE personid=$personid")
or die(mysql_error());
$row = mysql_fetch_array($result);
// check that the 'id' matches up with a row in the databse
if($row)
{
// get data from db
$personname = $row['personname'];
$personsurname = $row['personsurname'];
// show form
renderForm($personid, $personname, $personsurname, '');
}
else
// if no match, display result
{
echo "No results!";
}
}
else
// if the 'id' in the URL isn't valid, or if there is no 'id' value, display an error
{
echo 'Error!';
}
}
?>
I have modified this code but seems that i have made mistakes and i cant find the. after being able to edit the name and surname i will add other fields also.

I'm assuming there is primary/unique key in your person table
$query = "select * FROM person, members where fname= '$fname' AND lname = '$lname' GROUP BY person.person_id";

Related

How to input a number into a database based of a drop down menu consisting of data from another table in PHP?

I wish to input a number into a database based of a drop down menu consisting of data from another table.
Links table:
Category table:
So basically my drop down will consist of the category.cat written information. But when I submit the form it will input category.id into the links.catID column in the database.
The code I have so far is:
<?php
// since this form is used multiple times in this file, I have made it a
function that is easily reusable
function renderForm($links, $url, $catID, $type, $error){
?>
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01//EN" "http://www.w3.org/TR/html4/strict.dtd">
<html>
<head>
<title>New Record</title>
</head>
<body>
<?php
// if there are any errors, display them
if ($error != ''){
echo '<div style="padding:4px; border:1px solid red; color:red;">'.$error.'</div>';
}
?>
<form action="" method="post">
<div>
<strong>Link Title: *<br></strong> <input type="text" name="links" size="40" value="<?php echo $links; ?>" /><br><br/>
<strong>URL: *<br></strong> <input type="text" name="url" size="40" value="<?php echo $url; ?>" /><br><br/>
<?php
require 'db/connect.php';
echo" <strong>Category: *<br></strong>";
echo "<select name='catID' id='catID'>";
$sql = "SELECT * FROM links";
$results = $db->query($sql);
if($results->num_rows){
while($row = $results->fetch_object()){
echo "<option>";
echo "{$row->catID}";
echo "</option>";
}
} echo "</select><br>";
?>
<br>
<strong>Type: *<br></strong> <input type="text" name="type" size="40" value="<?php echo $type; ?>" /><br><br/>
<p>* Required</p><br>
<input type="submit" name="submit" value="Submit">
</div>
</form>
</body>
</html>
<?php
}
// connect to the database
include('connect-db.php');
// check if the form has been submitted. If it has, start to process the form and save it to the database
if (isset($_POST['submit'])){
// get form data, making sure it is valid
$links = mysql_real_escape_string(htmlspecialchars($_POST['links']));
$url = mysql_real_escape_string(htmlspecialchars($_POST['url']));
$catID = mysql_real_escape_string(htmlspecialchars($_POST['catID']));
$type = mysql_real_escape_string(htmlspecialchars($_POST['type']));
// check to make sure all fields are entered
if ($links == '' || $url == '' || $catID == '' || $type == ''){
// generate error message
$error = 'ERROR: Please fill in all required fields!';
// if either field is blank, display the form again
renderForm($links, $url, $catID, $type, $error);
} else {
// save the data to the database
mysql_query("INSERT links SET links='$links', url='$url', catID='$catID', type='$type'")
or die(mysql_error());
// once saved, redirect back to the view page
header("Location: view.php");
}
} else {
// if the form hasn't been submitted, display the form
renderForm('','','','','');
}
?>
Which gives me the following:
May be try this? (Considering the links.links columns and category.cat columns are common)
Store the value of dropdown in a variable say $dropdown_selected_option
Getting the id from the category table using sql:
$sql = "Select id from category where cat = '$dropdown_selected_option'";
$result = mysqli_query($conn, $sql);
Later run a query again to update the given fields in second table;
Update links set
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result))
{
// Run update query here where $row['id'] has the ID from the category table required.
}
}

MYSQL and PHP Loop Failure

What problem i am having right now, is that there is a while loop to post my topics retrieved from MySQL. Works perfectly fine, until I get into inputting data. I have recently created a comment system, where for each topic there will be a comment box to submit. The problem is the while loop runs it over and over again, so when i type a comment for one topic, it posts to all of them.
Here is my code:
//MYSQLI LOGIN DETAILS
$servername = "***";
$username = "***";
$password = "***";
$dbname = "***";
//MYSQLI CREATE CONNECTION
$constatus = new mysqli($servername, $username, $password, $dbname);
//MYSQLI CHECK CONNECTION
if ($constatus->connect_error) {
die("Connection failed: " . $constatus->connect_error);
}
//MYSQLI COUNT COLUMNS
$sql = "SELECT NEWSID, AUTHOR, ADMINSTS, DATE, HEADING, ARTICLE FROM news ORDER BY NEWSID DESC";
$result = $constatus->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo
"<div class=newsboard_topic>" .
"<div class=newsboard_authordate>" . $row["AUTHOR"];
if ($row["ADMINSTS"] == admin) {
echo
"<div class=newsboard_adminfx>
Admin
</div>";
} else if ($row["ADMINSTS"] == sadmin) {
echo
"<div class=newsboard_sadminfx>
Super Admin
</div>";
}
if ($_SESSION['adminsts'] == 'admin' || $_SESSION['adminsts'] == 'sadmin') {
echo "<span class=newsboard_adminactions> <img src='/image/remove.png' style='width:20px; height:20px;'> </span>";
}
echo
"<span class=date>" . $row["DATE"] .
"</span></div>
<h1>" . $row["HEADING"].
"</h1><p class=newsboard_topic_article>" .
$row["ARTICLE"] .
"</p>";
$sqlcomments = "SELECT newscomments.USERID, newscomments.COMMENT, userdata.FIRSTNAME, userdata.LASTNAME, userdata.ADMINSTATUS FROM newscomments JOIN userdata ON newscomments.USERID=userdata.ID WHERE NEWSID=$row[NEWSID] ORDER BY COMMENTID DESC";
$resultcomments = $constatus->query($sqlcomments);
echo "<div class=newsboard_comments>
Comments
<br>";
while($rowcomments = $resultcomments->fetch_assoc()) {
echo $rowcomments["FIRSTNAME"] . " " . $rowcomments["LASTNAME"] . " " . $rowcomments["COMMENT"] . "<br>";
}
if (isset($_SESSION['loggedon']) && $_SESSION['loggedon'] == true) {
echo '
<form method="post">
<input class=postheadline type="text" name="comment" />
<input class=submit type="submit" id="submit" name="submit" value="Comment"/>
</form>';
if (isset($_POST[submit])) {
if (!empty($_POST[comment])) {
$sqlcommentpost = "INSERT INTO newscomments (NEWSID, USERID, COMMENT) VALUES ('$row[NEWSID]', '$_SESSION[profileid]', '$_POST[comment]')";
if ($constatus->query($sqlcommentpost) === TRUE) {
echo "Posted Successfully!";
break;
} else {
echo "Fatal Error. Please try again";
break;
}
}
}
}
echo "</div></div>"; /*Ends newsboard_topic Div & newsboard_comments Div*/
}
} else {
echo "0 results";
}
Live example is online at www.geovillageva.com however you cannot see comments as it is for registered members only because it will have a name for posters.
Lets include the news id also in the form. You can have a hidden input field for this. Then use this news id $_POST[nid] while inserting.
if (isset($_SESSION['loggedon']) && $_SESSION['loggedon'] == true) {
echo '
<form method="post">
<input class=postheadline type="text" name="comment" />
<input type="hidden" name="nid" value="'.$row[NEWSID].'" />
<input class=submit type="submit" id="submit" name="submit" value="Comment"/>
</form>';
if (isset($_POST[submit])) {
if (!empty($_POST[comment])) {
$sqlcommentpost = "INSERT INTO newscomments (NEWSID, USERID, COMMENT) VALUES ('$_POST[nid]', '$_SESSION[profileid]', '$_POST[comment]')";
if ($constatus->query($sqlcommentpost) === TRUE) {
echo "Posted Successfully!";
break;
} else {
echo "Fatal Error. Please try again";
break;
}
}
}
}
So, your input block
if (isset($_SESSION['loggedon']) && $_SESSION['loggedon'] == true) {
echo '
<form method="post">
<input class=postheadline type="text" name="comment" />
<input class=submit type="submit" id="submit" name="submit" value="Comment"/>
</form>';
if (isset($_POST[submit])) {
if (!empty($_POST[comment])) {
$sqlcommentpost = "INSERT INTO newscomments (NEWSID, USERID, COMMENT) VALUES ('$row[NEWSID]', '$_SESSION[profileid]', '$_POST[comment]')";
if ($constatus->query($sqlcommentpost) === TRUE) {
echo "Posted Successfully!";
break;
} else {
echo "Fatal Error. Please try again";
break;
}
}
}
}
needs to go by itself before the display loop. On the query inserting using $row[NEWSID], you need to use $_POST['newsid'] instead (and you may need to add that to your form to be posted along with the comments).
Please note that you need to beef up the security on this considerably or you will be hacked.
You can try a .reset() on your form after the submission was successful(before the break).

Database not updating correctly in PHP / MySQL [duplicate]

This question already has answers here:
Can I mix MySQL APIs in PHP?
(4 answers)
Closed 7 years ago.
I've got a database with a Users table which I'm trying to update.
Currently I have customers.php, which displays form fields with the user information so it can be updated.
This form points to edit_customer_processor.php , which takes the new values, puts them into a MYSQL query... and then despite the query working correctly when I query the DB via the PHPMyAdmin command line, the record doesn't update.
customers.php
<?php
session_start();
if(!$_SESSION["logged_in"]){
header("location:home.php");
die;
}
?>
<?php include 'header.html'; ?>
<div id='maincontent'>
<?php
if (isset($_GET["id"])){
$customer_id = $_GET["id"];
require_once('config.php');
$customer_query = "SELECT * FROM customer WHERE customer_id = $customer_id";
$customer_results = mysql_query($customer_query, $conn);
if (!$customer_results) {
die ("Error selecting car data: " .mysql_error());
}
else {
while ($row = mysql_fetch_array($customer_results)) {
echo "<h3>Edit Customer</h3>";
echo "<FORM method='post' action='edit_customer_processor.php'>";
echo '<p> Name: <input type="text" name="name" size = "40" value=' . $row[name] . '></p>';
echo '<p> Address: <input type="text" name ="address" size="40" value=' . $row[address] . '></p>';
echo '<p> Email: <input type="text" name="email" value=' . $row[email] . '></p>';
echo '<p> Phone: <input type ="text" name="phone" size="20" value=' . $row[phone] . '></p>';
echo '<input type ="hidden" name="customer_id" value="' . $row[customer_id] . '">';
echo '<input type ="hidden" name="formtype" value="edit_customer">';
echo '<input type="submit" name="submit" value= "Update">';
echo '</form>';
}
}
} else {
// If there isn't an ID, display the New Customer form and all customers below, with links
// to their edit pages.
echo "<h3>Enter new customer information and submit.</h3>";
echo "<FORM method='post' action='new_customer_processor.php'>";
echo '<p> Name: <input type="text" name="name" size = "40"></p>';
echo '<p> Address: <input type="text" name ="address" size="40"></p>';
echo '<p> Email: <input type="text" name="email"></p>';
echo '<p> Phone: <input type ="text" name="phone" size="20"></p>';
echo '<input type ="hidden" name="formtype" value="new_customer">';
echo '<input type="submit" name="submit" value= "Submit">';
echo '<input type ="reset" name="reset" value ="Reset">';
echo '</form>';
require_once('config.php');
echo "<h3>Current Customers</h3>";
$query = "SELECT * FROM customer";
$results = mysql_query($query, $conn);
if (!$results) {
die ("Error selecting customer data: " .mysql_error());
}
else {
// In the absence of an ID, all customers will be displayed down
// the bottom of the page
while ($row = mysql_fetch_array($results)) {
echo "<a href=customers.php?id=";
echo $row[customer_id];
echo "><p> $row[name] </p></a>";
echo "<p> $row[address] </p>";
echo "<p> $row[phone] </p>";
echo "<p> $row[email] </p>";
}
}
}
?>
Back to Customers Page
</div>
<?php include 'footer.html' ?>
edit_customer_processor.php
<?php include 'header.html' ?>
<div id="maincontent">
<?php
// Pulling in hidden customer ID from post value
$mysqli = new mysqli( 'localhost', 'root', 'root', 'w_c_a' );
// Check our connection
if ( $mysqli->connect_error ) {
die( 'Connect Error: ' . $mysqli->connect_errno . ': ' . $mysqli->connect_error );
}
// Insert our data
$sql = mysql_query("UPDATE customer
SET name = '".mysql_real_escape_string($_POST['name'])."',
address = '".mysql_real_escape_string($_POST['address'])."',
phone = '".mysql_real_escape_string($_POST['phone'])."',
email = '".mysql_real_escape_string($_POST['email'])."'
WHERE customer_id='".mysql_real_escape_string($_POST['customer_id'])."'");
$update = $mysqli->query($sql);
echo "Customer updated: ";
echo "<a href=customers.php?id=" . $_POST['customer_id'] . ">";
echo "Back to Edit Customer</a>";
?>
</div>
<?php include 'footer.html' ?>
And when I echo the MYSQL query, I get:
UPDATE customer SET name = 'Kellyassdsa', address = 'ads', phone = '0260123123', email = 'asdasd' WHERE customer_id='1'
Which works when I put it in PHPMyAdmin.
I know it'll be some boneheaded little mistake, but I've been trying to get this work for ages now. Any ideas?
Maybe your program just can't connect to your MySQL database.
$customer_results = mysql_query($customer_query, $conn);
I can't see where you gave a value to the var $conn.
If the problem is connection problem then we might need your database info like the name of your table in PhpMyAdmin.
your problem is...
$sql = mysql_query(..);
$update = $mysqli->query($sql);
it should be
$sql = 'UPDATE ...';
$update = $mysqli->query($sql);
i think problem occurs due to line break. pleas make a query in single line without line break.
$sql = mysql_query("UPDATE customer SET name = '".mysql_real_escape_string($_POST['name'])."',address = '".mysql_real_escape_string($_POST['address'])."', phone = '".mysql_real_escape_string($_POST['phone'])."', email = '".mysql_real_escape_string($_POST['email'])."' WHERE customer_id='".mysql_real_escape_string($_POST['customer_id'])."'");
Hope this helps..

PHP MySQL Delete record only if it exist

Having a bit of trouble, I have created a form to deleted a record from a linked MySQL database using PHP that works, but I am having an problem with how to make an error display if a uadnumber value for example already exists.
<form name="deleterecord" action="indexdelete.php" method="post">
UAD Username: <br/><input type="text" name="uadnumber" /><br/>
<input type="submit" onclick="return deletedatabase();" value="Delete" />
</form>
<?php
// find the values from the form
$uadnumber = $_POST['uadnumber'] ;
?>
<?php
$con=mysqli_connect($db_hostname,$db_username,$db_password,$db_database);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$retval = mysqli_query($con,"DELETE FROM users WHERE uadnumber='$uadnumber'");
if(!$retval<0)
{
die('Could not delete data: ' . mysql_error());
}
echo "Deleted all linked data from user $uadnumber successfully"."<br><br>";
echo "<hr>";
echo "Below is the remaining users within the database";
mysqli_close($con);
?>
if (isset($_POST["uadnumber"])) {
$uadnumber = (int)$_POST["uadnumber"];
$db = new mysqli($db_hostname,$db_username,$db_password,$db_database);
if ($db->query("SELECT uadnumber FROM users WHERE uadnumber='".$uadnumber."'")->num_rows > 0) {
if ($db->query("DELETE FROM users WHERE uadnumber='".$uadnumber."'")->affected_rows > 0) {
echo 'Desired rows have been removed.';
} else {
echo 'No rows have been removed.';
}
} else {
echo 'There are no rows identified by given value.';
}
}

Improve my SQL database form PHP code

Can someone have a look at my code Ive finally got working after 2 days and lots of help from here - thank you!
There are a few tweaks i would like to do on it -
for the transaction ID, if i search for any letter in the transaction id, i am shown records - I only want it to show me a record if the FULL transaction ID has been entered and matches the record in the database. Transaction id example: 87K07228GD157974M
if you want to retrieve your code, you must type in your name, email and transaction date, this works perfect BUT the time is also included with the date but i don't want anyone to have to enter the time as well ONLY the date i.e.....
you currently have to enter: 2013-03-07 01:39:23 - but i want to enter in the format of DD/MM/YY - is this possible?
I also don't know if the code is secure also, any advice would be appreciated.
Thanks,
here is the code:
findme.html
<html>
<head>
<title>Search</title>
</head>
<body bgcolor=#ffffff>
<h2>Search Transaction ID</h2>
<form name="search" method="post" action="findme.php">
Seach for: <input type="text" name="find" />
<input type="submit" name="search" value="Search" />
</form>
OR
<h2>Search Name, E-Mail & Transaction Date</h2>
<form name="search" method="post" action="findme1.php">
Full Name (on paypal account) <input type="text" name="name" /> <br><br>
Paypal E-Mail Address <input type="text" name="email" /> <br><br>
Transaction Date - DD/MM/YY <input type="text" name="date" />
<input type="submit" name="search" value="Search" /><br><br>
If searching via Name, E-Mail & Transaction date, all fields must be completed to obtain your code.
</form>
</body>
</html>
findme.php
<html>
<head><title>Searching for a student...</title>
</head>
<body bgcolor=#ffffff>
<?php
echo "<h2>Search Results:</h2><p>";
//If they did not enter a search term we give them an error
if ($find == "")
{
echo "<p>You forgot to enter a search term!!!";
exit;
}
// Otherwise we connect to our Database
mysql_connect("location.com", "ipn", "password!") or die(mysql_error());
mysql_select_db("ipn") or die(mysql_error());
// We perform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);
//Now we search for our search term, in the field the user specified
$iname = mysql_query("SELECT * FROM ibn_table WHERE itransaction_id LIKE '%$find%'");
//And we display the results
while($result = mysql_fetch_array( $iname ))
{
echo "<b>Name: </b>";
echo $result['iname'];
echo " ";
echo "<br>";
echo "<b>E-mail: </b>";
echo $result['iemail'];
echo "<br>";
echo "<b>Transaction Date: </b>";
echo $result['itransaction_date'];
echo "<br>";
//And we remind them what they searched for
echo "<b>Search Term </b>(Transaction ID): </b> " .$find;
//}
echo "<br>";
echo "<br>";
echo "<b>Login Code: </b>";
echo $result['ipaymentstatus'];
echo "<br>";
}
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($iname);
if ($anymatches == 0)
{
echo "Sorry, but we can not find an entry to match your search, please make sure the correct details have been entered...<br><br>";
}
?>
</body>
</html>
findme1.php
<html>
<head><title>Searching for a student...</title>
</head>
<body bgcolor=#ffffff>
<?php
echo "<h2>Search Results:</h2><p>";
//If they did not enter a search term we give them an error
if ($name == "")
if ($email == "")
{
echo "<p>Please enter Full Name, E-Mail Address & Transaction Date EXACTLY how they appear on your PayPal Account...";
exit;
}
// Otherwise we connect to our Database
mysql_connect("location.com", "ipn", "password") or die(mysql_error());
mysql_select_db("ipn") or die(mysql_error());
// We perform a bit of filtering
$find = strtoupper($find);
$find = strip_tags($find);
$find = trim ($find);
//Now we search for our search term, in the field the user specified
$name = mysql_query("SELECT * FROM ibn_table WHERE iemail = '$email' AND iname = '$name' AND itransaction_date = '$date'");
//And we display the results
while($result = mysql_fetch_array( $name ))
{
echo "<b>Name: </b>";
echo $result['iname'];
echo " ";
echo "<br>";
echo "<b>E-mail: </b>";
echo $result['iemail'];
echo "<br>";
echo "<b>Transaction Date: </b>";
echo $result['itransaction_date'];
echo "<br>";
//And we remind them what they searched for
echo "<b>Search Term </b>(Transaction ID): " .$name;
//}
echo "<br>";
echo "<br>";
echo "<b>Login Code: </b>";
echo $result['ipaymentstatus'];
echo "<br>";
}
//This counts the number or results - and if there wasn't any it gives them a little message explaining that
$anymatches=mysql_num_rows($name);
if ($anymatches == 0)
{
echo "Sorry, but we can not find an entry to match your search, please make sure the correct details have been entered...<br><br>";
}
?>
</body>
</html>
Fields in my database are:
iname
iemail
itransaction_id
ipaymentstatus
itransaction_date
Thanks!
As stated in comment for transaction ID you have :
$iname = mysql_query("SELECT * FROM ibn_table WHERE itransaction_id LIKE '%$find%'");
what LIKE with %$find% does is match any part from transaction ID with $find that is why you get results with single letter. Change that to :
$iname = mysql_query("SELECT * FROM ibn_table WHERE itransaction_id = '$find'");
for date issue you can decide what to take from user like you stated date then for example :
if you take :
$date = "12-11-2012"; //(dd-mm-yyyy)
$split = explode("-", $date);
then you can use this to generate SQL date/time format :
$sql_date = date("Y-m-d h:i:s", mktime(0, 0, 0, (int) $split[1], (int) $split[0], (int) $split[2]))
and in sql query :
transaction_date LIKE '$sql_date%'
And at last don't use mysql_* it is deprecated. Instead use mysqli.

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