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.
Related
I have created two files on remote server. One is html form which asks to enter some fields and another is a php file which will get all the data and insert into the database.
For this from html file on click of submit button I am calling php file, but the file is not getting execute I think because when I click on submit it again reloads the same html page.
html :
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>MCQ Questions</title>
</head>
<body>
<form method="post" >
<p> Enter the question :</p> <input name="question" type="input"> <br><br>
<p> Enter options :</p>
Enter option 1 : <input name="opt1" type="input"> <br><br>
Enter option 2 : <input name="opt2" type="input"> <br><br>
Enter option 3 : <input name="opt3" type="input"> <br><br>
Enter option 4 : <input name="opt4" type="input"> <br><br>
<p> Enter correct answer :</p>
<input name="ans" type="input"> <br><br>
<input type="submit" value = "Submit" onClick = "uploadQuestion.php">
</form>
</body>
</html>
php file:
<?php
$question=$_POST['question'];
$option1=$_POST['opt1'];
$option2=$_POST['opt2'];
$option3=$_POST['opt3'];
$option4=$_POST['opt4'];
$ans=$_POST['ans'];
$db_server = mysql_connect("address","username","pass");
if(!$db_server) {
die("Database connection failed: " . mysql_error());
}else{
$db_select = mysql_select_db("mlm",$db_server);
if (!$db_select) {
die("Database selection failed:: " . mysql_error());
}
}
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question','$option1',$option2,$option3,$option4,$ans)";
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
?>
I also tried this way :
<input type="submit" value = "Submit" onClick = "http://address/uploadQuestion.php">
But nothing is working. Whats going wrong here? I am a beginner in web development,, can anyone help please? Thank you..
EDIT :
$database = new Database('addredd','username','pass','handbook');
$dbConnection = $database->getDB();
$stmt = $dbConnection->prepare("insert into questions(question,answer_a,answer_b,answer_c,answer_d,answer) values(?,?,?,?,?,?)");
$stmt->execute(array($question,$option1,$option2,$option3,$option4,$ans));
I tried to use pdo statement but getting this error :
Fatal error: Class 'Database' not found in /var/www/html/uploadQuestion.php on line 12
EDIT2 :
I am trying to upload one file on server and want to save it in database also, so for this I have created 2 files one is index.php and another is uploadFile.php.
As you shown now I used pdo for this but when I click on upload image again same page is getting load.
index.php
<form action="index.php" method="post" enctype="multipart/form-data">
<p> Select image to upload:</p>
<input name = "file" type="file" id="fileToUpload"><br><br>
Enter chapter name :
<input name = "chapterName" type = "text"><br><br>
<input type="submit" value = "Upload Image">
</form>
<?php
if (isset($_FILES['file']['tmp_name']))
{
$ch = curl_init();
$cfile = new CURLFile($_FILES['file']['tmp_name'],$_FILES['file']['type'],$_FILES['file']['name']);
$data = array("myfile" => $cfile);
curl_setopt($ch, CURLOPT_URL, "http://host/NewProject/uploadFile.php");
curl_setopt($ch, CURLOPT_FOLLOWLOCATION, true);
curl_setopt($ch, CURLOPT_POST, true);
curl_setopt($ch, CURLOTP_POSTFIELDS, $data);
$response = curl_exec($ch);
if($response == true)
{
echo "File posted";
}
else{
echo "Error: " . curl_error($ch);
}
}
?>
uploadFile.php
<?php
ini_set('display_errors', 1);
if(isset($_FILES['myfile']['tmp_name']))
{
$path = "files/" . $_FILES['myfile']['name'];
move_uploaded_file($_FILES['myfile']['tmp_name'], $path);
$chapterName=$_POST['chapterName'];
$dbh = new PDO('mysql:host=host;dbname=database_name','username', 'password');
$stmt = $dbh->prepare("INSERT INTO chapters (title,file) VALUES (?, ?)");
$stmt->execute(array($chapterName,$path));
if ($dbh->lastInsertId())
{
echo 'File upploaded.';
}
else
{
echo 'File could not upload.';
}
}
?>
Please help.. Thank you..
First repair your form, type="" can't be named input u can check here https://www.w3schools.com/tags/att_input_type.asp
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>MCQ Questions</title>
</head>
<body>
<form action="uploadQuestion.php" method="post" enctype="multipart/form-data">
<p> Enter the question :</p> <input name="question" type="text"> <br><br>
<p> Enter options :</p>
Enter option 1 : <input name="opt1" type="text"> <br><br>
Enter option 2 : <input name="opt2" type="text"> <br><br>
Enter option 3 : <input name="opt3" type="text"> <br><br>
Enter option 4 : <input name="opt4" type="text"> <br><br>
<p> Enter correct answer :</p>
<input name="ans" type="text"> <br><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
Then yours php code
<?php
// mysql connection
$db_server = mysql_connect("address","username","pass");
// check for mysql connection
if(!$db_server)
{
die("Database connection failed: " . mysql_error());
}
else
{
// check if database exists
$db_select = mysql_select_db("mlm",$db_server);
if (!$db_select)
{
die("Database selection failed:: " . mysql_error());
}
}
// escape post variables
$question = mysql_real_escape_string($_POST['question']);
$option1 = mysql_real_escape_string($_POST['opt1']);
$option2 = mysql_real_escape_string($_POST['opt2']);
$option3 = mysql_real_escape_string($_POST['opt3']);
$option4 = mysql_real_escape_string($_POST['opt4']);
$ans = mysql_real_escape_string($_POST['ans']);
// make query
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question', '$option1', '$option2', '$option3', '$option4', '$ans')";
// check if query runs
if (!mysql_query($sql)) {
die('Error: ' . mysql_error());
}
?>
Or php with mysqli
<?php
// host, username, password, database name
$db_server = mysqli_connect("address", "username", "pass", "mlm");
// check for connection
if(!$db_server)
{
die("Database connection failed: " . mysqli_error($db_server));
}
// escape post variables
$question = mysqli_real_escape_string($db_server, $_POST['question']);
$option1 = mysqli_real_escape_string($db_server, $_POST['opt1']);
$option2 = mysqli_real_escape_string($db_server, $_POST['opt2']);
$option3 = mysqli_real_escape_string($db_server, $_POST['opt3']);
$option4 = mysqli_real_escape_string($db_server, $_POST['opt4']);
$ans = mysqli_real_escape_string($db_server, $_POST['ans']);
// make query
$sql = "INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ('$question', '$option1', '$option2', '$option3', '$option4', '$ans')";
// check if query runs
if (!mysqli_query($db_server, $sql))
{
die('Error: ' . mysqli_error($db_server));
}
?>
Or php with prepared statements
<?php
// mysql connection
$dbh = new PDO('mysql:host=adress;dbname=database_name', 'username', 'password');
// escape post variables
$question = $_POST['question'];
$option1 = $_POST['opt1'];
$option2 = $_POST['opt2'];
$option3 = $_POST['opt3'];
$option4 = $_POST['opt4'];
$ans = $_POST['ans'];
$stmt = $dbh->prepare("INSERT INTO questions (question,answer_a,answer_b,answer_c,answer_d,answer) VALUES ( ?, ?, ?, ?, ?, ?)");
$stmt->execute(array($question, $option1, $option2, $option3, $option4, $ans));
if ($dbh->lastInsertId())
{
echo 'Sucess.';
}
else
{
echo 'Fail.';
}
?>
Change your from code to this
<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>MCQ Questions</title>
</head>
<body>
<form action="uploadQuestion.php" method="post" enctype="multipart/form-data">
<p> Enter the question :</p> <input name="question" type="input"> <br><br>
<p> Enter options :</p>
Enter option 1 : <input name="opt1" type="input"> <br><br> Enter option 2 : <input name="opt2" type="input"> <br><br> Enter option 3 : <input name="opt3" type="input"> <br><br> Enter option 4 : <input name="opt4" type="input"> <br><br>
<p> Enter correct answer :</p>
<input name="ans" type="input"> <br><br>
<input type="submit" value="Submit">
</form>
</body>
</html>
The below code is a simple form that is sending the data that is inputted to my local database.
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr/>
<form action="" method="post">
<label>Student Name :</label>
<input type="text" name="stu_name" id="name" required="required" placeholder="Please Enter Name"/><br /><br />
<label>Student Email :</label>
<input type="email" name="stu_email" id="email" required="required" placeholder="john123#gmail.com"/><br/><br />
<label>Student City :</label>
<input type="text" name="stu_city" id="school" required="required" placeholder="Please Enter Your City"/><br/><br />
<input type="submit" value=" Submit " name="submit"/><br />
</form>
</div>
<!-- Right side div -->
</div>
<?php
if(isset($_POST["submit"])){
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('".$_POST["stu_name"]."','".$_POST["stu_email"]."','".$_POST["stu_city"]."')";
if ($conn->query($sql) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error."');</script>";
}
$conn->close();
}
?>
</body>
</html>
The issue that I am finding is that I am trying to change the Student City input field into a drop down where the values is retrieve from the database and put into a drop down list for a new user to select.
Could someone advise on what needs to be done please.
i am trying to use the below code and send the below list to my database.
<option value="US">United States</option>
<option value="UK">United Kingdom</option>
<option value="France">France</option>
<option value="Mexico">Mexico</option>
but i am finding it hard to send the these values to the database with my above code as well as where to place this code.
As a rough example of how you could build the dropdown menu using data from your db this should give you the general idea perhaps.
/* store formatted menu options in temp array */
$html=array();
/* query db to find schools/cities */
$sql='select distinct `student_school` from `students` order by `student_school`';
$res=$mysqli_query( $conn, $sql );
/* process recordset and store options */
if( $res ){
while( $rs=mysqli_fetch_object( $res ) ){
$html[]="<option value='{$rs->student_school}'>{$rs->student_school}";
}
}
/* render menu */
echo "<select name='stu_city'>", implode( PHP_EOL, $html ), "</select>";
You need to refactor your code by moving the if (isset($_POST)) above the html:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli ( $servername, $username, $password, $dbname );
// Check connection
if ($conn->connect_error) {
die ( "Connection failed: " . $conn->connect_error );
}
$sql = "SELECT city_name FROM cities" ;
if ($conn->query ( $sql ) === TRUE) {
$cities = ... // build the cities from the query result
} else {
$cities = '<option value="none">No cities found</option>' ;
}
if (isset ( $_POST ["submit"] )) {
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('" . $_POST ["stu_name"] . "','" . $_POST ["stu_email"] . "','" . $_POST ["stu_city"] . "')";
if ($conn->query ( $sql ) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error . "');</script>";
}
$conn->close ();
}
?>
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr />
<form action="" method="post">
<label>Student Name :</label> <input type="text" name="stu_name"
id="name" required="required" placeholder="Please Enter Name" /><br />
<br /> <label>Student Email :</label> <input type="email"
name="stu_email" id="email" required="required"
placeholder="john123#gmail.com" /><br />
<br /> <label>Student City :</label> <select name="stu_city" multiple><?php echo $cities; ?>
</select>><br />
<br /> <input type="submit" value=" Submit " name="submit" /><br />
</form>
</div>
<!-- Right side div -->
</div>
</body>
</html>
Use the Select tag: Lets say you hav a column in your database with Student City, like this, lets say the database field is called city
City 1
City 2
City 3
Step 1: Query the database and fetch all the Cities
$sql = "SELECT city FROM table_name";
$result = $conn->query($sql);
Then you come to your dropdown:
<select name="stu_city" id="..." required>
<?php
while($cities = $conn->fetch_array($result){
extract($cities);
echo "<option value='...'>$city</option>";
}
?>
</select>
You need to refactor your code by moving the if (isset($_POST)) above the html:
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Student";
// Create connection
$conn = new mysqli ( $servername, $username, $password, $dbname );
// Check connection
if ($conn->connect_error) {
die ( "Connection failed: " . $conn->connect_error );
}
$sql = "SELECT city_name FROM cities" ;
$result = $conn->query ( $sql );
if (isset ( $_POST ["submit"] )) {
$sql = "INSERT INTO students (student_name, student_email, student_school)
VALUES ('" . $_POST ["stu_name"] . "','" . $_POST ["stu_email"] . "','" . $_POST ["stu_city"] . "')";
if ($conn->query ( $sql ) === TRUE) {
echo "<script type= 'text/javascript'>alert('New Record Inserted Successfully');</script>";
} else {
echo "<script type= 'text/javascript'>alert('Error: " . $sql . "<br>" . $conn->error . "');</script>";
}
$conn->close ();
}
?>
<html>
<head>
<title>!!!!!!!!!!!!!!</title>
<link rel="stylesheet" type="text/css" href="style.css">
</head>
<body>
<div id="main">
<h1>Insert data into database using mysqli</h1>
<div id="login">
<h2>Student's Form</h2>
<hr />
<form action="" method="post">
<label>Student Name :</label> <input type="text" name="stu_name"
id="name" required="required" placeholder="Please Enter Name" /><br />
<br /> <label>Student Email :</label> <input type="email"
name="stu_email" id="email" required="required"
placeholder="john123#gmail.com" /><br />
<br /> <label>Student City :</label> <select name="stu_city" multiple>
<?php
if ($result == TRUE) {
while($cities = $conn->fetch_array($result)){
extract($cities);
echo "<option value=''>$city_name</option>";
}
}
else {
echo "<option value='none'>No cities found</option>";
}
?>
</select>><br />
<br /> <input type="submit" value=" Submit " name="submit" /><br />
</form>
</div>
<!-- Right side div -->
</div>
</body>
</html>
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){
$("select[name='sel']").append(e);
});
}
</script>
I'm creating a form to save the 'id' and the 'eventname' to save in the database. i'm trying to save the form data into the table but not able to..here's the database code and the form code. please help. i tried but not able to try out the problem. I'm not getting any error but the data's what i enter in the form is not getting saved into the database..
//table structure
CREATE TABLE `event` (
`Memberid` int(7) NOT NULL,
`events` varchar(50) COLLATE latin1_general_ci NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;
//form code
<!DOCTYPE html>
<html lang="en">
<head></head>
<body>
<?php
if(isset($_POST['add']))
{
// database connection
$host_name = 'localhost';
$host_user = 'root';
$host_pass = '';
$conn = mysql_connect($host_name, $host_user, $host_pass);
if(! $conn ){
die('Could not connect: ' . mysql_error());
}
//code for SQL injection
if(! get_magic_quotes_gpc() ) {
$login = addslashes ($_POST['login']);
$eventList = addslashes ($_POST['eventList']);
} else {
$login = $_POST['login'];
$eventList = $_POST['eventList'];
}
//inserting data into the database table
$sql = "INSERT INTO EVENT " . "(login, eventList) " . "VALUES " .('$login','$eventList')";
mysql_select_db('database');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
} else
{
?>
<div class="container">
<!-- Codrops top bar -->
<section class="main">
<form class="form-3" method="POST" action="<?php $_PHP_SELF ?>">
<p class="clearfix">
<label for="login">Username</label>
<input type="text" name="login" id="login" placeholder="Username">
</p>
<p class = "clearfix">
<label for="select">Event Name</label>
<select style="width:200px;" name="eventList" tabindex="5">
<option value="select">Select Event</option>
<optgroup label="Do Re Me">
<option value="Battle of Bands">Battle of Bands</option>
<option value="Eastern">Eastern</option>
<option value="Solo">Solo</option>
</optgroup>
</select> <br>
</p>
<p class="clearfix">
<input type="submit" name="submit" value="Sign in">
</p>
</form>
</section>
</div>
<?php
}
?>
</body>
</html>
This question already has answers here:
How can I get a PHP value from an HTML form?
(2 answers)
Closed 1 year ago.
I am currently trying to complete a project where the specifications are to use a search form to search through a packaging database. The database has lots of variables ranging from Sizes, names, types and meats. I need to create a search form where users can search using a number of different searches (such as searching for a lid tray that is 50 cm long).
I have spent all day trying to create some PHP code that can search for info within a test database I created. I have had numerous amounts of errors ranging from mysql_fetch_array errors, boolean errors and now currently my latest error is that my table doesn't seem to exist. Although i can enter data into it (html and php pages where I can enter data), I don't know what is causing this and I have started again a few times now.
Can anyone give me some idea or tips of what I am going to have to do currently? Here is just my small tests at the moment before I move onto the actual sites SQL database.
Creation of database:
<body>
<?php
$con = mysql_connect("localhost", "root", "");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
if (mysql_query("CREATE DATABASE db_test", $con))
{
echo "Database created";
}
else
{
echo "Error creating database: " . mysql_error();
}
mysql_select_db("db_test", $con);
$sql = "CREATE TABLE Liam
(
Code varchar (30),
Description varchar (30),
Category varchar (30),
CutSize varchar (30),
)";
mysql_query($sql, $con);
mysql_close($con);
?>
</body>
HTML search form page:
<body>
<form action="form.php" method="post">
Search: <input type="text" name="term" /><br />
<input type="submit" name="submit" value="Submit" />
</form>
</body>
The PHP code I am using to attempt to gather info from the database
(I have rewritten this a few times, this code also displays the "table.liam doesn't exist")
<body>
<?php
$con = mysql_connect ("localhost", "root", "");
mysql_select_db ("db_test", $con);
if (!$con)
{
die ("Could not connect: " . mysql_error());
}
$sql = mysql_query("SELECT * FROM Liam WHERE Description LIKE '%term%'") or die
(mysql_error());
while ($row = mysql_fetch_array($sql)){
echo 'Primary key: ' .$row['PRIMARYKEY'];
echo '<br /> Code: ' .$row['Code'];
echo '<br /> Description: '.$row['Description'];
echo '<br /> Category: '.$row['Category'];
echo '<br /> Cut Size: '.$row['CutSize'];
}
mysql_close($con)
?>
</body>
try this out let me know what happens.
Form:
<form action="form.php" method="post">
Search: <input type="text" name="term" /><br />
<input type="submit" value="Submit" />
</form>
Form.php:
$term = mysql_real_escape_string($_REQUEST['term']);
$sql = "SELECT * FROM liam WHERE Description LIKE '%".$term."%'";
$r_query = mysql_query($sql);
while ($row = mysql_fetch_array($r_query)){
echo 'Primary key: ' .$row['PRIMARYKEY'];
echo '<br /> Code: ' .$row['Code'];
echo '<br /> Description: '.$row['Description'];
echo '<br /> Category: '.$row['Category'];
echo '<br /> Cut Size: '.$row['CutSize'];
}
Edit: Cleaned it up a little more.
Final Cut (my test file):
<?php
$db_hostname = 'localhost';
$db_username = 'demo';
$db_password = 'demo';
$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, $con);
?>
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="utf-8" />
<title></title>
</head>
<body>
<form action="" method="post">
Search: <input type="text" name="term" /><br />
<input type="submit" value="Submit" />
</form>
<?php
if (!empty($_REQUEST['term'])) {
$term = mysql_real_escape_string($_REQUEST['term']);
$sql = "SELECT * FROM liam WHERE Description LIKE '%".$term."%'";
$r_query = mysql_query($sql);
while ($row = mysql_fetch_array($r_query)){
echo 'Primary key: ' .$row['PRIMARYKEY'];
echo '<br /> Code: ' .$row['Code'];
echo '<br /> Description: '.$row['Description'];
echo '<br /> Category: '.$row['Category'];
echo '<br /> Cut Size: '.$row['CutSize'];
}
}
?>
</body>
</html>
You're getting errors 'table liam does not exist' because the table's name is Liam which is not the same as liam. MySQL table names are case sensitive.