I am facing problems connecting my HTML form to database. I am very new at this. Please do help me.
This is the HTML Login form code
logout.html
<form name="form" onsubmit="submit1()" action="connectivity-sign-up.php" method="POST" >
<div id="errorBox"></div>
<input type="text" name="Name" value="" placeholder="First Name" class="input_name" >
<input type="text" name="LastName" value="" placeholder="Last Name" class="input_name" >
</div>
<div id="email_form">
<input type="text" name="Email" value="" placeholder="Your Email" class="input_email">
</div>
<div id="Re_email_form">
<input type="text" name="enterEmail" value="" placeholder="Re-enter Email" class="input_Re_email">
</div>
<div id="password_form">
<input type="password" name="Password" value="" placeholder="New Password" class="input_password">
</div>
<!--birthday details start-->
<div>
<h3 class="birthday_title">Birthday</h3>
</div>
<div>
<select name="birthday_month" >
<option value="" selected >Month</option>
<option value="1">Jan</option>
<option value="2">Feb</option>
<option value="3">Mar</option>
<option value="4">Apr</option>
<option value="5">May</option>
</select>
<select name="birthday_day" >
<option value="" selected>Day</option>
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
<option value="5">5</option>
</select>
<select name="birthday_year">
<option value="" selected>Year</option>
<option value="2013">2013</option>
<option value="2012">2012</option>
<option value="2011">2011</option>
<option value="2010">2010</option>
<option value="2009">2009</option>
</select>
</div>
<!--birthday details ends-->
<div id="radio_button">
<input type="radio" name="radiobutton" value="Female">
<label >Female</label>
<input type="radio" name="radiobutton" value="Male">
<label >Male</label>
</div>
<div>
<p id="sign_user" onClick="Submit()" value= "Submit" >Sign Up </p>
<input type="submit" value="Submit">
</div>
</form>
The Submit1() function is the validate function. When I click Submit it should first validate and then send the data to the form. Now the validate function works properly but how do I call it such that it will send the data once validated.And here is the PHP connectivity part
File name : connectivity-sign-up.php
<?php define('DB_HOST', 'localhost');
define('DB_NAME', 'customerdb');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error());
function Signin()
{
$fname = $_POST['Name'];
$lname = $_POST['LastName'];
$email = $_POST['Email'];
$password = $_POST['Password'];
$query = "INSERT INTO custtable (fname,lname,email,password) VALUES ('$fname','$lname','$email','$password')";
$data = mysql_query ($query)or die(mysql_error());
if($data)
{
echo "YOUR REGISTRATION IS COMPLETED...";
}
}
?>
When user click submit button call Signin() function.
if (isset($_POST['submit'])) {
Signin();
}
-
<?php define('DB_HOST', 'localhost');
define('DB_NAME', 'customerdb');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
$con = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db = mysql_select_db(DB_NAME, $con) or die("Failed to connect to MySQL: " . mysql_error());
if (isset($_POST['submit'])) {
Signin();
}
function Signin()
{
$fname = $_POST['Name'];
$lname = $_POST['LastName'];
$email = $_POST['Email'];
$password = $_POST['Password'];
$query = "INSERT INTO custtable (fname,lname,email,password) VALUES ('$fname','$lname','$email','$password')";
$data = mysql_query($query) or die(mysql_error());
if ($data) {
echo "YOUR REGISTRATION IS COMPLETED...";
}
}
?>
Update html : <input type="submit" name="submit" value="Submit">
<!--birthday details ends-->
<div id="radio_button">
<input type="radio" name="radiobutton" value="Female">
<label>Female</label>
<input type="radio" name="radiobutton" value="Male">
<label>Male</label>
</div>
<div>
<p id="sign_user" onClick="Submit()" value="Submit">Sign Up </p>
<input type="submit" name="submit" value="Submit">
</div>
</form>
When click Sign Up, the information will be send to connectivity-sign-up.php, and execute the code in connectivity-sign-up.php. And Signin() is not invoked, there is just a declare.
<?php define('DB_HOST', 'localhost');
define('DB_NAME', 'customerdb');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error());
function Signin()
{
$fname = $_POST['Name'];
$lname = $_POST['LastName'];
$email = $_POST['Email'];
$password = $_POST['Password'];
$query = "INSERT INTO custtable (fname,lname,email,password) VALUES ('$fname','$lname','$email','$password')";
$data = mysql_query ($query)or die(mysql_error());
if($data)
{
echo "YOUR REGISTRATION IS COMPLETED...";
}
}
//invoke Signin
Signin();
?>
If you remove the method declaration for Signin() and just have the entire page as a script, then the logic you have in the function will execute. This would result in the following file:
<?php define('DB_HOST', 'localhost');
define('DB_NAME', 'customerdb');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " . mysql_error());
$fname = $_POST['Name'];
$lname = $_POST['LastName'];
$email = $_POST['Email'];
$password = $_POST['Password'];
$query = "INSERT INTO custtable (fname,lname,email,password) VALUES ('$fname','$lname','$email','$password')";
$data = mysql_query ($query)or die(mysql_error());
if($data)
{
echo "YOUR REGISTRATION IS COMPLETED...";
}
?>
Another alternative is to actually call the function somewhere in the page like so:
Signin()
The better option would be executing the Signin() function somewhere in the page.And you would need to add a connetion between the connectivity-sign-up.php and logout.html file
Related
I am trying to create a simple form that will insert the given data received by my HTML form, into my SQL table named 'Vendors', however I am struggling to work with its functionality.
There are 7 text fields that I am wanting to add to my Vendors table, and these are so named:
vendorName
addressL1 (Line 1)
addressL2
postcode
email
telephone
description
The HTML for this form can be found below:
<!DOCTYPE HTML>
<html>
<head>
</head>
<body>
<form action="" method="post">
<ul class="form-style-1">
<li>
<label style="color:#4D4D4D;" >Vendor Name <span class="required">*
</span></label>
<center> <input type="text" name="vendorName" class="field-long"
required="required" placeholder="Vendor Name" /> </center>
</li>
<li>
<label style="color:#4D4D4D;">Vendor Address <span class="required">*
</span></label>
<center> <input type="text" name="addressL1" required="required"
class="field-long" placeholder="Address Line 1" /> </center>
</br>
<center> <input type="text" name="addressL2" required="required"
class="field-long" placeholder="Address Line 2" /> </center>
</br>
<center> <input type="text" name="postcode" required="required"
class="field-short" placeholder="Postcode" /> </center>
</li>
<li>
<label style="color:#4D4D4D;">Vendor Contact Details <span
class="required">*</span></label>
<center> <input type="text" name="email" required="required"
class="field-long" placeholder="Email Address" /> </center>
</br>
<center> <input type="text" name="telephone" required="required"
class="field-long" placeholder="Phone Number" /> </center>
</select>
</li>
<li>
<label style="color:#4D4D4D;">Vendor Description </label>
<center> <textarea name="description" id="field5" class="field-long
field-textarea" placeholder="Description"></textarea> </center>
</li>
<li>
<center> <input type="submit" class="AddButton" value="POST"></input>
</center>
</li>
</ul>
</form>
</body>
</html>
And the PHP I have used is:
<?php
date_default_timezone_set('Europe/London');
$server = "";
$connectionInfo = array( "Database"=>"");
$conn = sqlsrv_connect($server,$connectionInfo);
if (!$conn)
{
die("Connection failed");
}
$_SERVER['REQUEST_METHOD'];
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
$VendorName = $_POST['vendorName'];
$AddressLine1 = $_POST['addressL1'];
$AddressLine2 = $_POST['addressL2'];
$Postcode = $_POST['postcode'];
$VendorEmail = $_POST['email'];
$VendorNumber = $_POST['telephone'];
$VendorDes = $_POST['description'];
$time = time();
$timestamp = date("Y-m-d H:i:s", $time);
$describeQuery = ("INSERT INTO Vendors (VendorName, VendorAL1,
VendorAL2, VendorPost, VendorEmail, VendorNumber, VendorDes,
Added)
VALUES ('".$VendorName."', '".$AddressLine1."',
'".$AddressLine2."', '".$Postcode."',
'".$VendorEmail."', '".$VendorNumber."',
'".$VendorDes."', '".$timestamp."')");
$results = sqlsrv_query($conn, $describeQuery);
if(sqlsrv_query($conn, $describeQuery))
{
$alert = "Vendor Successfully Added";
echo "<script type='text/javascript'>alert('$alert');
</script>";
}
else
{
echo 'Information not inserted';
}
}
sqlsrv_close($conn);
?>
Each time I submit the form, it goes straight to the 'Information not inserted' ELSE statement and doesn't import the data into my database.
I have removed my server name and database name for precautionary reasons, however I can assure you they are correct as I have worked on a previous project and used the same method of connecting.
Any help on this would be greatly appreciated, and if there are any formatting mistakes, apologies in advance, I am not an avid user of stack overflow.
Use Mysqli Please, I have updated the script.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO Vendors (VendorName, VendorAL1,
VendorAL2, VendorPost, VendorEmail, VendorNumber, VendorDes,
Added)
VALUES ($VendorName, $AddressLine1, $AddressLine2,$Postcode,$VendorEmail,$VendorNumber,$VendorDes,$timestamp)";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
?>
Trying to insert data into my table but I keep getting an undefined index because there are "no value set when I submit my form". The if (isset($_POST['submit'])) removes my error even when I run the .php alone but no data is inserted when I submit my form. Any help is appreciated. Thank you
My form.html
<form name="supportForm" class="form" action="database.php" method="POST" onsubmit="return validateForm()">
<label>Name:</label>
<input type="text" name="name"/>
<br/>
<label>Client ID:</label>
<input type="text" name="clientID"/>
<br/>
<label>E-mail address:</label>
<input type="email" name="email"/>
<br/>
<label>Phone number:</label>
<input type="tel" name="tel"/>
<br/>
<br/>
Support Type:<br>
<input type="radio" name="suppType" value="Question/Inquiry">Question/Inquiry<br>
<input type="radio" name="suppType" value="Software">Software Issue<br>
<input type="radio" name="suppType" value="Hardware">Hardware Issue<br>
<input type="radio" name="suppType" value="Connectivity">Connectivity<br>
</br>
Operating System:
<select id="select">
<option disabled selected value="">Choose a product</option>
<option value="w7" name="OS">Windows 7</option>
<option value="w8" name="OS">Windows 8/8.1</option>
<option value="w10" name="OS">Windows 10</option>
</select>
<br> </br>
Problem Description:
<br><textarea id="ta" rows="10" cols="80" name="pDesc"></textarea></br>
<input type="checkbox" name="terms" value="agree">
I agree to the terms and conditions.
<br> </br>
<input type="hidden" name="submitted" value="true">
<input type="submit" name="submit" onClick="validateSubmit()">
</form>
My PHP file
<?php
//Creates static credentials
define('DB_NAME', 'data');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
//Creates connection to the database
$con = new mysqli(DB_HOST, DB_USER, DB_PASSWORD);
//Checks for connection
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}
//If there are no connection, error
if (!$con) {
die ('Could not connect' . mysqli_error());
}
//Select the 'data' database
$con->select_db(DB_NAME);
//Checks if database 'data' has been selected
if (mysqli_select_db($con, DB_NAME)) {
echo "Database exists <br>";
} else {
echo "Database does not exist";
}
//Successful connection message
echo "Connected successfully <br>";
if (isset($_POST['submit'])) {
//Retrieving values from support form
$name = $_POST['name'];
$clientID = $_POST['clientID'];
$email = $_POST['email'];
$tel = $_POST['tel'];
$suppType = $_POST['suppType'];
$OS = $_POST['OS'];
$pDesc = $_POST['pDesc'];
//Inserting values into a table
$sql = "INSERT INTO info (fullname, clientID, email, tel,
suppType, OS, pDesc)
VALUES ($name, $clientID, $email, $tel,
$suppType, $OS, $pDesc)";
if (!mysqli_query($con, $sql)) {
echo "No data";
} else {
echo "Data recorded successfully";
}
}
//Closes connection
mysqli_close($con);
You must write name="OS" in <select> not in <option>
<select id="select" name="OS">
<option disabled selected value="">Choose a product</option>
<option value="w7">Windows 7</option>
<option value="w8">Windows 8/8.1</option>
<option value="w10">Windows 10</option>
</select>
And Sql must be like this you need apostrophes ('') around variables
$sql = "INSERT INTO `info` (fullname, clientID, email, tel, suppType, OS, pDesc)
VALUES ('$name', '$clientID', '$email', '$tel', '$suppType', '$OS', '$pDesc')";
You not showing us the validateForm() function, therefore we won't really know whats happening there, nonetheless I have edited your form and did a validation using php,
what you need to do first is to check if all values are set before jumping to insert into db, and make sure email is a proper email, also the select option the name attribute needs to be on the select tag not on the option tag, the option must only have values.
Then Validate,Filter and sanitize user input before storing to the
database. Treat every userinput on your form as if its from a very dangerous hacker.
There's something called prepared statements, in mysqli and PDO you should try to learn that and use it :) you will enjoy it, I will leave it to you to research as to why you need to use prepared statements.
This is how your code should look
<form name="supportForm" class="form" action="database.php" method="POST">
<label>Name:</label>
<input type="text" name="name"/>
<br/>
<label>Client ID:</label>
<input type="text" name="clientID"/>
<br/>
<label>E-mail address:</label>
<input type="email" name="email"/>
<br/>
<label>Phone number:</label>
<input type="tel" name="tel"/>
<br/>
<br/>
Support Type:<br>
<input type="radio" name="suppType" value="Question/Inquiry">Question/Inquiry<br>
<input type="radio" name="suppType" value="Software">Software Issue<br>
<input type="radio" name="suppType" value="Hardware">Hardware Issue<br>
<input type="radio" name="suppType" value="Connectivity">Connectivity<br>
</br>
Operating System:
<select id="select" name="OS">
<option value="0">Choose a product</option>
<option value="w7">Windows 7</option>
<option value="w8">Windows 8/8.1</option>
<option value="w10">Windows 10</option>
</select>
<br> </br>
Problem Description:
<br>
<textarea id="ta" rows="10" cols="80" name="pDesc"></textarea>
</br>
<input type="checkbox" name="terms" value="agree">
I agree to the terms and conditions.
<br> </br>
<input type="hidden" name="submitted" value="true">
<input type="submit" name="submit">
</form>
Then database.php
<?php
//Creates static credentials
define('DB_NAME', 'data');
define('DB_USER', 'root');
define('DB_PASSWORD', '');
define('DB_HOST', 'localhost');
$errors = ""; //checking for errors
//Creates connection to the database
$con = new mysqli(DB_HOST, DB_USER, DB_PASSWORD);
//Checks for connection
if ($con->connect_error) {
die("Connection failed: " . $con->connect_error);
}
//If there are no connection, error
if (!$con) {
die('Could not connect' . mysqli_error());
}
//Select the 'data' database
$con->select_db(DB_NAME);
//Checks if database 'data' has been selected
if (mysqli_select_db($con, DB_NAME)) {
echo "Database exists <br>";
} else {
echo "Database does not exist";
}
//Successful connection message
echo "Connected successfully <br>";
if (isset($_POST['submit'])) {
//check values are set
if (empty($_POST['name'])) {
echo "enter name";
$errors++;
} else {
$name = userIput($_POST['name']);
}
if (empty($_POST['clientID'])) {
echo "enter id";
$errors++;
} else {
$clientID = userIput($_POST['clientID']);
}
if (empty($_POST['email'])) {
echo "enter email";
$errors++;
} else {
$email = userIput($_POST['email']);
if (!preg_match("/([\w\-]+\#[\w\-]+\.[\w\-]+)/", $email)) { //validate email,
echo "enter valid email";
$errors++;
}
}
if (empty($_POST['tel'])) {
echo "enter tel";
$errors++;
} else {
$tel = userIput($_POST['tel']);
}
if (!isset($_POST['suppType'])) {
echo "select one option";
$errors++;
} else {
$suppType = userIput($_POST['suppType']);
}
if (isset($_REQUEST['OS']) && $_REQUEST['OS'] === "0") {
echo "please select product";
$errors++;
} else {
$OS = userIput($_POST['OS']);
}
if (empty($_POST['pDesc'])) {
echo "enter Description";
$errors++;
} else {
$pDesc = userIput($_POST['pDesc']);
}
if ($errors <= 0) { // No errors
//prepare and insert query
$sql = $con->prepare("INSERT INTO info (fullname, clientID, email, tel,suppType, OS, pDesc) VALUES (?, ?, ?, ?, ?, ?, ?)");
$sql->bind_param("sssssss", $name, $clientID, $email, $tel, $suppType, $OS, $pDesc);
if ($sql->execute()) {
echo "records inserted successfully";
} else {
echo "Could not insert " . mysqli_error();
}
$sql->close();
$con->close();
}
}
function userIput($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
Hope this will help a little, and you will learn a thing or two, and I'm always available for suggessions, just incase I missed something. Thanks
My issue is that I am making a dating site form with following categories
firstname,lastname,username,password,email,mysex,yoursex,relationship,date of birth, and country
I did the whole php code so when submitted would send info to sql server. But there is some error on php half, date of birth(DOB) day,month,year. And mysex and yoursex values are not submitting. Sql gives undefined index. Here is my code.(The falf that is not submitting)
<form action="process.php" method="post" id="register" class="col-xs-12">
<input class="register-switch-input" type="radio" name="mysex" value="hombre" id="me-male" > <label class="register-switch-label" for="me-male"> Hombre </label>
<input class="register-switch-input" type="radio" name="mysex" value="mujer" id="me-female"> <label class="register-switch-label" for="me-female"> Mujer </label> <br>
<input type="radio" name="yoursex" value="hombre" id="your-male" checked> <label for="your-male"> Hombre </label>
<input type="radio" name="yoursex" value="mujer" id="your-female"> <label for="your-female"> Mujer </label>
<input type="radio" name="yoursex" value="cualquiera" id="cualquiera"> <label for="cualquiera">Cualquiera </label> <br>
<label class="form-label">Nacimiento:</label>
<select name="DOBMonth" >
<option> - Month - </option>
<option value="January">January</option>
<option value="Febuary">Febuary</option>
""
<select name="DOBDay" >
<option> - Day - </option>
""
<select name="DOBYear">
<option> - Year - </option>
<option value="2003">2003</option>
The "" is because the selects have a million options and are unnessecary.
Here is my php code
<?php
$servername = "localhost";
$username = "diego966";
$password = "ddddd966";
$dbname= "signup";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$firstname = $_POST["firstname"];
$lastname = $_POST["lastname"];
$username = $_POST['username'];
$password = $_POST['password'];
$email = $_POST['email'];
$mysex = $_POST["mysex"];
$yoursex = $_POST["yoursex"];
$relationship = $_POST['relationship'];
$DOBday = $_POST['DOBday'];
$DOBmonth = $_POST['DOBmonth'];
$DOByear = $_POST['DOByear'];
$country = $_POST['country'];
$sql="INSERT INTO accounts(firstname, lastname, username, password, email, mysex, yoursex, relationship, DOBday, DOBmonth, DOByear, country)
VALUES ('$firstname', '$lastname','$username', '$password', '$email','$mysex', '$yoursex', '$relationship', 'DOBday', 'DOBmonth ','DOByear', '$country')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
As partly mentioned in the comments:
DOBMonth is DOBmonth in PHP
DOBYear is DOByear in PHP
DOBDay is DOBday in PHP
your radio-field needs to be defined by name="yoursex[]" in HTML and then you check in PHP if isset($_POST["yoursex"]) and then you can use $_POST["yoursex"][0]
Sanitizing has been mentioned but can not be mentioned enough
I have creacted an SQL Database. I succsess fully insert data into database, but i want to search from database.
HTML
<body>
INSERT AREA
<br>
<form action="demo.php" method="post"/>
<p>imei: <input type="text" name="input1"/> </p>
<select name="input2">
<option value="1">1</option>
<option value="2">2</option>
<option value="3">3</option>
<option value="4">4</option>
</select>
<br>
<br>
<input type="submit" src="submit.png" alt="Submit Form" />
</form>
Search AREA
<br>
<form action="form.php" method="post">
Search: <input type="text" name="term" /><br />
<input type="submit" value="Submit" />
</form>
</body>
demo.php
<?php
define('DB_NAME', '#');
define('DB_USER', '#');
define('DB_PASSWORD', 'mypass');
define('DB_HOST', 'localhost');
$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db(DB_NAME, $link);
if (!$db_selected) {
die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}
$value = $_POST['input1'];
$value2 = $_POST['input2'];
$sql = "INSERT INTO demo (input1, input2) VALUES ('$value', '$value2')";
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
mysql_close();
?>
So all these work perfect , i insert data to SQL Database fine, but check the next php code for search, isn't return any info it just open with-out return value. *i have wipe the database info.
Database name : demo
Host name :localhost
form.php
<?php
$db_hostname = 'localhost';
$db_username = 'my username';
$db_password = 'my pass';
$db_database = 'demo';
// Database Connection String
$con = mysql_connect($db_hostname,$db_username,$db_password);
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($db_database, $con);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<form action="demo.php" method="post"/>
<p>imei: <input type="text" name="input1"/> </p>
<select name="input2">
<option value="1">111111111111111111111</option>
<option value="2">222222222222222222</option>
<option value="3">33333333333333333</option>
<option value="4">4444444444444444</option>
</select>
<br>
<br>
<input type="image" src="submit.png" alt="Submit Form" />
</form>
<br>
<form action="form.php" method="post">
<input type="text" name="term" /><br />
<input type="submit" value="Submit" />
</form>
<?php
if (!empty($_REQUEST['term'])) {
$term = mysql_real_escape_string($_REQUEST['term']);
$sql = "SELECT * FROM demeo WHERE Description LIKE '%".$term."%'";
$r_query = mysql_query($sql);
while ($row = mysql_fetch_array($r_query)){
echo 'Primary key: ' .$row['PRIMARYKEY'];
echo '<br /> Code: ' .$row['input1'];
echo '<br /> Description: '.$row['input2'];
}
}
?>
</body>
</html>
By looking at your first example, it seems that in the second one, your table and column names are incorrect.
Replace your current query:
SELECT * FROM demeo WHERE Description LIKE '%".$term."%'
With this query:
SELECT * FROM demo WHERE input2 LIKE '%".$term."%'}
Also consider the suggestions of the other users to avoid sql injections.
Here's a pic of what i'm trying to achieve
http://oi60.tinypic.com/b9gf20.jpg
Heres all my code...
PHP FILE: contact_form.php
<?php
define('DB_NAME', 'xxx');
define('DB_USER', 'xxx');
define('DB_PASSWORD', 'xxx');
define('DB_HOST', 'xxx');
$connection = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD);
if(!$connection){
die('Database connection failed: ' . mysqli_connect_error());
}
$db_selected = mysqli_select_db($connection, DB_NAME);
if(!$db_selected){
die('Can\'t use ' .DB_NAME . ' : ' . mysqli_connect_error());
}
echo 'Connected successfully';
if (isset($_POST['itemname'])){
$itm = $_POST['itemname'];
}else{
$itm = '';
}
if($_POST['mile']){
$mi = $_POST['mile'];
}else{
echo "Miles not received";
exit;
}
if($_POST['email']){
$email = $_POST['email'];
}else{
echo "email not received";
exit;
}
$sql = "INSERT INTO seguin_orders (itemname, mile, email) VALUES ('$itm', '$mi', '$email')";
if (!mysqli_query($connection, $sql)){
die('Error: ' . mysqli_connect_error($connection));
}
CONACT FORM: formz.php
<html>
<header>
</header>
<body>
<form action="/demoform/contact_form.php" class="well" id="contactForm" method="post" name="sendMsg" novalidate="">
<big>LOAD PAST ORDERS:</big>
<select id="extrafield1" name="extrafield1">
<option value="">Please select...</option>
</select>
</br>
<input type="text" required id="mile" name="mile" placeholder="Miles"/>
</br>
<input id="email" name="email" placeholder="Email" required="" type="text" value="demo#gmail.com" readonly="readonly"/>
</br>
<input id="name" name="itemname" placeholder="ITEM NAME 1" required="" type="text" />
</br>
<input type="reset" value="Reset" />
<button type="submit" value="Submit">Submit</button>
</form>
</body>
</html>
Much thanks and appreciation for any time and help with this thanks.
<form>
<?php
while ($row = mysqli_fetch_assoc($result))
{
echo '<option value =" ' . $row['column'] . ' ">' . $row['column'] . '</option>';
};
?>
</form>
When you click select ,then send a ajax request to another php script to select database about the columns you want;then change it;like this:
<input name="email" />
<select onclick="ajaxFunc();" name="sel">
</select>
<script>
function ajaxFunc(){
var mail = $("input[name='email']").val();
$.post('anotherFile.php',{mail:mail},function(e){
$("select[name='sel']").append(e);
});
}
</script>