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PHP - (beginner) Writing To Mysql From Form Input
Hello, all!
I am trying to learn PHP and MySQL on my own, and need some debugging help. What exactly is going wrong here? I am following a tutorial and trying to write the code as it says, but am still having trouble with syntax. Running a WAMP, PHP5.3, and MySQL5.5. This is my code: Code: [Select] <html> <body> <form name = "newVenue" method = "post"> Establishment name: <input type = "text" name = "name"> <br> Street Address: <input type = "text" name = "streetAddress"> <br> City: <input type = "text" name = "city"> <br> State: <select name="state"> <option value="AL">AL</option> <option value="AK">AK</option> <option value="AZ">AZ</option> <option value="AR">AR</option> <option value="CA">CA</option> <option value="CO">CO</option> <option value="CT">CT</option> <option value="DE">DE</option> <option value="DC">DC</option> <option value="FL">FL</option> <option value="GA">GA</option> <option value="HI">HI</option> <option value="ID">ID</option> <option value="IL">IL</option> <option value="IN">IN</option> <option value="IA">IA</option> <option value="KS">KS</option> <option value="KY">KY</option> <option value="LA">LA</option> <option value="ME">ME</option> <option value="MD">MD</option> <option value="MA">MA</option> <option value="MI">MI</option> <option value="MN">MN</option> <option value="MS">MS</option> <option value="MO">MO</option> <option value="MT">MT</option> <option value="NE">NE</option> <option value="NV">NV</option> <option value="NH">NH</option> <option value="NJ">NJ</option> <option value="NM">NM</option> <option value="NY">NY</option> <option value="NC">NC</option> <option value="ND">ND</option> <option value="OH">OH</option> <option value="OK">OK</option> <option value="OR">OR</option> <option value="PA">PA</option> <option value="RI">RI</option> <option value="SC">SC</option> <option value="SD">SD</option> <option value="TN">TN</option> <option value="TX">TX</option> <option value="UT">UT</option> <option value="VT">VT</option> <option value="VA">VA</option> <option value="WA">WA</option> <option value="WV">WV</option> <option value="WI">WI</option> <option value="WY">WY</option> </select> <br> Zip: <input type = "text" name = "zip"> <br> email: <input type = "text" name = "email"> <br> password: <input type = "text" name = "password"> <br> <input type="submit" name="Submit" value="Submit"> </form> <?php //If the form isn't empty, assign the value to a variable if (!empty($_POST['name'])) { $name = $_POST['name']; $address = $_POST['streetAddress']; $city = $_POST['city']; $state = $_POST['state']; $zip = $_POST['zip']; $email = $_POST['email']; $password = $_POST['password']; //Connect to the 'Users' database and store the new bar into the 'Venue' table... mysql_connect ("localhost", "newbar", "Jpr5HJ2K5fWvPLXq") or die ('Oh, fuck: '.mysql_error()); mysql_select_db ("users"); $query = "INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', '".$name."', '".$address."', '".$city."', '".$state."', '".$zip."', '".$email."', '".$password."', 0)"; mysql_query($query) or die ('Oh, fuck: '.mysql_error()); echo "Damn, Nathan. This shit actually worked..."; } ?> </body> </html>This is the error I receive: Code: [Select] Oh, fuck: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'INSTERT INTO venues VALUES ('NULL', 'testPic.jpg', 'Nathan's house', '666', 'DAY' at line 1This is my table: Code: [Select] id INT PRIMARY KEY pic_location VARCHAR name VARCHAR address VARCHAR city VARCHAR state VARCHAR zip VARCHAR email VARCHAR password VARCHAR event_name INT Any thoughts as to what is causing this error? Thanks in advance... Similar Tutorialsso what i have going on is that i need help writing a basic script to upload original and copy with resize for thumbnail. then also need to rename both image files with content in the form. My server supports GD Library and that imagemagik or whatever it is lol. upload dirs orig: ../media/photos/ thumb: ../media/photos/thumb/ mySQL DB: name: m_photos fields: id(INT) m_cat(varchar) m_sub_cat(varchar) pic(varchar) description(varchar) p_group(varchar) I have the form written up for how it should look like: Code: [Select] <form action="upload.php" method="post" enctype="multipart/form-data"> Upload an image for processing<br /> <input type="file" name="Image"><br /> <select name="sub_group"> <option value="Photo Shoot">Photo Shoot</option> <option value="Live Performances">Live Performances</option> <option value="Randoms">Randoms</option> <option value="Fan Photos">Fan Photos</option> </select><br /> Location: <input type="text" name="p_group" /><br /> Date of Pic: <input type="text" name="date" /><br /> Description: <input type="text" name="description" /><br /> <input type="submit" value="Upload"> </form> so what needs to happen is for the file upload name. it needs to grab a count from "p_group" database to give it a starting number in a "00" pattern and then the date of pic needs to go in there next then the p_group name. so when the files gets uploaded it would look something like this after upload. "03 - Oct 21, 2011 - The Mex.jpg" NOTE: all pics must be converted to ".jpg" extension. both the orig and thumb will use that same file name cuz they just go into different dirs re-sizing for the thumbnail should be done by aspect ratio and needs to either be 300px width or 400px height. so if anyone would like to help me out please do. im not the greatest at writing in php yet We recently upgraded from PHP4 to PHP5 and the below script that was working perfectly in 4 has completely stopped working and I can't figure out why for the life of me. I'm not an experienced PHP programmer--I've done some forms, but this is the first time I've used a database. What needs to (and was) happen: A user enters their username in the form and gets a readout of their participation so far that month. The problem(s): I know that it's storing the variable 'user' because it echoes it back properly, but the database is no longer allowing me to select the row based on that variable. I know it's not that I can't connect to the database because if instead of '$user' I change the code to a username I know is in there, I get the proper readout. This all started as soon as I transferred over to PHP5--before that, no problems at all. The database information is all correct, I just took it out for privacy's sake. <form id="feedback" method="post" action="index.php"> <input name="user" type="text" value="Enter user name" size="20" maxlength="50" /><br /> <input name="send" id="send" type="submit" value="Submit" /> </form> <?php if (isset($_POST['user'])) { $_session['user'] = $_POST['user']; } ?> <p>You entered your username as: <strong><? echo $_session['user'];?></strong>. If this is not correct or you do not see your information below, please re-enter your username and click Submit again.</p> <?php $con = mysql_connect("database","username","password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("database", $con); $result = mysql_query("SELECT * FROM March WHERE Username='$user'") or die ('Error: '.mysql_error ()); while($row = mysql_fetch_array($result)) { echo "<table border='0'>"; echo "<tr>"; echo "<td><strong>Username:</strong> </td><td>" . $row['Username'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Discussion:</strong></td><td>" . $row['Mar2Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 2 Poll:</strong></td><td>" . $row['Mar2P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Discussion:</strong></td><td>" . $row['Mar9Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 9 Poll:</strong></td><td>" . $row['Mar9P'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Discussion:</strong></td><td>" . $row['Mar16Q'] . "</td></tr>"; echo "<tr><td><strong>Mar. 16 Poll:</strong></td><td>" . $row['Mar16P'] . "</td></tr>"; echo "<tr><td><strong>March Participation To-Date:</strong></td><td>" . $row['Participation'] . "</td></tr>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> ANY help would be greatly appreciated! I've got a couple hundred people who use this on a regular basis and are starting to ask why it's not working. Hei everyone. I am new on php development and i do not have budget on offering a course so i am trying to learn my self. i need some help on the following : i want to create infopanel.php where the script will get info from. For example if user logs in the script will have to call : <?php include '/incl/infopanel.php' ; ?>The info panel to load all details from database: <?php $name='$myname'; $sur_name='$sname'; $age='$myage'; $country='$mycountry'; ?>So the info with my has to be from database . so can some one help me how to call all information from database to a specific file that i will not have to call always information on every page I hope i have been so clean with my request I am getting this error : You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 I have absolutely not a clue what that means or where I went wrong, but ANY help would be appreciated. I've checked my DB username, pass and db name and they are all correct. I am getting the error when I type in $subject_set = mysql_query("SELECT * FROM pages WHERE subject_id = ${subject["id"]}", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } this is my code: <?php require_once("includes/connection.php"); ?> <?php require_once("includes/functions.php"); ?> <?php include("includes/header.php"); ?> <div id="sidebar"> <ul> <?php $subject_set = mysql_query("SELECT * FROM subjects", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } $subject_set = mysql_query("SELECT * FROM pages WHERE subject_id = ${subject["id"]}", $connection); if (!$subject_set) { die("database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } ?> </ul> </div> <div id="content"> <h1 class="sidebar">Content Area</h1> </div> </body> </html> <?php mysql_close($connection); ?> Hi, I am a newbie to php and am trying to copy mysql data to a text file. The format I need to create is I can get files to upload to folder and add a new row for each in to a mysql db using the code below, how do I get all the filles in one row with multiple uploads Code: [Select] //Check if the form was submitted. if(isset($_POST['uploadButton'])) { //Specify folder path where the file is going. $path = 'uploads/'; //Upload file one by one. foreach($_FILES['file']['name'] as $key => $val) { if($val != '') { $file_name = $_FILES['file']['name'][$key]; //Get file name. $file_tmp = $_FILES['file']['tmp_name'][$key]; //Get temporary file name. $file = $path . $file_name; mysql_query("INSERT INTO images (img1) VALUES('$file');") or trigger_error('Unable to INsert: ' . mysql_error()); //Move uploaded file if(move_uploaded_file($file_tmp, $file)) { echo 'File was succesfully uploaded to: ' . $file . '<br />'; } else { //Display error message if there was a problem uploading file. echo 'Error uploading file "' . $key . '."<br />'; } echo $target_path; } } } ?> Hi all, hope some can help because this is giving me more grey hairs than I allready have.
I am trying to solve the following: I'm pulling in JSON data, decoding it to a array which I write to a database. However in the array are social security numbers that I don't want to store (privacyregulations). So I need to check if a row has a social security number and if it has one I want to replace it with 'anonymized'. I have got it to work successfully on a local Xampp test environment using the following script (extract off course): //first anonymize the social security number (ssn) if exists $data = json_decode($jsondata, true);
// then fill the tables foreach($data as $row) {
$insertstatement = mysqli_query($link,"INSERT INTO $tablename (personid, type, ssn) VALUES ('".$row["personid"]."', '".$row["type"]."', '".$row["ssn"]."') ON DUPLICATE KEY UPDATE type = VALUES(type), bsn = VALUES(ssn)");
This leads to a filled table where the ssn's are filled with 'anonymized' if they exist in the array and empty if it does not exist for that row. This is exactly what I want ;-).
Personid type ssn
However, when I run the same script on a Lamp production environment (lamp stack on an Azure server) it does not work. Then the scripts results in totally ignoring the whole row that contains a ssn??
Personid type ssn Hope someone can help to get this working on my production environment as well.
Richard
I have a search box with an input box. When the user enters a string in this box and then when it clicks submit, the form needs to go to the http://ww.mywebsite.com/wp-content/uploads/<?php echo $_POST["name"]; ?>.jpg where $_POST["name"] is what the user entered. how to make it work?? <form action="http://ww.mywebsite.com/wp-content/uploads/<?php echo $_POST["name"]; ?>.jpg" method="post" target="_blank"> <!-- form fields here --> <INPUT TYPE = "Text" NAME = "name"> <input type="submit" /> </form> Hi
I'm a beginner in php and are facing a problem i cant seem to find the correct solution on.
$item5 = $_POST['PIGM1'];
if (empty($item5)) {
$email_body .= '';
}
if (isset($item5))
{
$email_body = '<html><head><style type="text/css">td { width: 450; }</style></head><body>';
$email_body .= '<table border="0" width="900">';
$email_body .= '<tr><td><strong>Header in table</strong></td><td></td></tr>';
$email_body .= '<tr><td><strong>Ordered</strong></td><td>' . $item5 . ' pcs. of some item</td></tr>';
$email_body .= '</table><br />';
}
Can anyone help me in the right direction? Hi all, I am trying to replicate/modify this checkbox for a different script and I don't understand what the php code is doing in this form? Could someone explain to me whats happening by including that php code in the form? Code: [Select] <input type="checkbox" onclick="showArchive()" NAME="showarchive" id="showarchive" <?php echo ($showarchive ? 'checked="checked"' : '');?> /> Thanks! Hey guys i've been trying to find out how to create and and write in a file, so that you can for example add a nav button to your website by form (you know like a admin panel). It tried some things but it doesnt work cause it wont write to the file... here is the code i've used: $ourFileName = $_POST['sitename']; $ourFileHandle = fopen($ourFileName, 'w') or die("can't open file"); fclose($ourFileHandle); $myFile = $_POST['sitename']; $fh = fopen($myFile, 'w') or die("can't open file"); $stringData = "test\n"; fwrite($fh, $stringData); $stringData = "test\n"; fwrite($fh, $stringData); fclose($fh); the "sitename" is the id from the form where you choose what the menu tab should be called. I would really glad if you could help me out with this one thanks! MinG Hi folks, Hi Everyone, I'm new to PHP freaks, and I'm hoping someone might be able to help me. I have written some code for a html page and used php to retrieve confirm whether or not data is in a text file. I also tried to write some code to insert the data supplied to my html page to the text file but it's not working. Can someone help me figure out what my issue is. I have attached my text file, and my php code as well. Below you'll find the code I used for my html page. Thank you for all your help, Phee <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title>Telephone Directory</title> </head> <body> <form action='SignGuestBook.php' method='post'> <h1>Sign Guest Book</h1> <hr> <br> <table align='Left'> <tr> <td>Name: </td> <td><input name='name' /></td> </tr> <tr> <td>E-mail: </td> <td><input name='email' /></td> </tr> <tr> <td><input type="submit" value='Sign' /></td> <td><input type="reset" value='Reset Form' /></td> </tr> </table> <h3></h3> <br> <h4></h4> <br> <h5></h5> <br> <h6></h6> <br> <h7></h7> <br> <hr> <a href="http://helios.ite.gmu.edu/~smohamu2/IT207/Lab%20Assignment%208/AddNew.html">View Guest Book</a> </form> </body> </html> [attachment deleted by admin] Hi,
I need to create a landing page with a form. That form needs to be recorded somewhere instead of sent to email. I know I can write it to a SQL database, and then to an excel file. But I only need a temporary solution so I figured I'd just go straight to CSV.
Is this bad practice? What potential problems might I encounter other than security issues?
How to do it? I have read some methods but they seem to not work. This is what I have got so far:
<?php I have a calendar select date function for my form that returns the date in the calendar format for USA: 02/16/2012. I need to have this appear as is for the form and in the db for the 'record_date' column, but I need to format this date in mysql DATE format (2012-02-16) and submit it at the same time with another column name 'new_date' in the database in a hidden input field. Is there a way to do this possibly with a temporary table or something? Any ideas would be welcome. Doug I have read around and can't seem to find the right coding for what I need on this forum and some other other forums. I have a contact form (as listed below) and I need 2 locations (Print Name and Title) fields to auto-populate on a separate form (can be a doc, pdf, etc. any form of document which is easiest) and this form can be totally back end and the individual using the form never is going to see the form. It's going on a contract form, that we would like to auto-populate. Also is there a simple attachment code so individuals can attach documents to the code? <p style: align="center"><form action="mailtest.php" method="POST"> <?php $ipi = getenv("REMOTE_ADDR"); $httprefi = getenv ("HTTP_REFERER"); $httpagenti = getenv ("HTTP_USER_AGENT"); ?> <input type="hidden" name="ip" value="<?php echo $ipi ?>" /> <input type="hidden" name="httpref" value="<?php echo $httprefi ?>" /> <input type="hidden" name="httpagent" value="<?php echo $httpagenti ?>" /> <div align="center"> <p class="style1">Name</p> <input type="text" name="name"> <p class="style1">Address</p> <input type="text" name="address"> <p class="style1">Email</p> <input type="text" name="email"> <p class="style1">Phone</p> <input type="text" name="phone"> <p class="style1">Debtor</p> <input type="text" name="debtor"> <p class="style1">Debtor Address</p> <input type="text" name="debtora"> <br /> <br /> <a href="authoforms.php" target="_blank" style="color:#ffcb00" vlink="#ffcb00">Click here to view Assignment Agreement and Contract Agreement</a> <p class="style1"><input type='checkbox' name='chk' value='I Have read and Agree to the terms.'> I have read and agree to the Assignment and Contract Agreement <br></p> <p class="style1">Print Name</p> <input type="text" name="pname"> <p class="style1">Title</p> <input type="text" name="title"> <p class="style1">I hear by agree that the information I have provided is true, accurate and the information I am submitting is <br /> not fraudulent. Please click the agree button that you adhere to Commercial Recovery Authority Inc.'s terms:</p> <select name="agree" size="1"> <option value="Agree">Agree</option> <option value="Disagree">Disagree</option> </select> <br /> <br /> <p class="style1">Employee ID:</p> <input type="text" name="employee"> <br /> <input type="submit" value="Send"><input type="reset" value="Clear"> </div> </form> </p> The mailtest php is this ?php $ip = $_POST['ip']; $httpref = $_POST['httpref']; $httpagent = $_POST['httpagent']; $name = $_POST['name']; $address = $_POST['address']; $email = $_POST['email']; $phone = $_POST['phone']; $debtor = $_POST['debtor']; $debtora = $_POST['debtora']; $value = $_POST['chk']; $pname = $_POST['pname']; $title = $_POST['title']; $agree = $_POST['agree']; $employee = $_POST['employee']; $formcontent=" From: $name \n Address: $address \n Email: $email \n Phone: $phone \n Debtor: $debtor \n Debtor's Address: $debtora \n 'Client' has read Assignment and Contract Agreement: $value \n Print Name: $pname \n Title: $title \n I hear by agree that the information I have provided is true, accurate and the information I am submitting is not fraudulent. Please click the agree button that you adhere to Commercial Recovery Authority Inc.'s terms: $agree \n \n Employee ID: $employee \n IP: $ip"; $recipient = "mail@crapower.com"; $subject = "Online Authorization Form 33.3%"; $mailheader = "From: $email \r\n"; mail($recipient, $subject, $formcontent, $mailheader) or die("Error!"); echo "Thank You!" . " -" . "<a href='index.php' style='text-decoration:none;color:#ffcb00;'> Return Home</a>"; $ip = $_POST['visitoraddress'] ?> Hello I am having problems uploading an image through a HTML form. I want the image to be uploaded to the server and the image name to be written to the mysql database. Below is the code I am using: Code: [Select] <?php if (isset($_POST['add'])){ echo "<br /> add value is true"; $name = $_POST['name']; $description = $_POST['description']; $price = $_POST['price']; $category_id = $_POST['category_name']; $image = $_FILES['image']['name']; //file path of the image upload $filepath = "../images/"; //mew name for the image upload $newimagename = $name; //new width for the image $newwidth = 100; //new height for the image $newheight = 100; include('../includes/image-upload.php'); mysql_query("INSERT INTO item (item_name, item_description, item_price, item_image) VALUES ('$name','$description','$price','$image')"); ?> Here is the image-upload.php file code: Code: [Select] <?php //assigns the file to the image $image =$_FILES["image"]["name"]; $uploadedfile =$_FILES["image"]["tmp_name"]; if ($image) { //retrieves the extension type from image upload $extension = getextension($image); //converts extension to lowercase $extension = strtolower($extension); //create image from uploaded file type if($extension=="jpg" || $extension=="jpeg") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); }else if($extension=="png") { $uploadedfile = $_FILES['image']['tmp_name']; $src = imagecreatefrompng($uploadedfile); }else{ $src = imagecreatefromgif($uploadedfile); } //creates a list of the width and height of the image list($width,$height)=getimagesize($uploadedfile); //adds color to the image $tmp = imagecreatetruecolor($newwidth,$newheight); //create image imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); //set file name $filename = $filepath.$newimagename.".".$extension; $imagename = $newimagename.".".$extension; //uploads new file with name to the chosen directory imagejpeg($tmp,$filename,100); //empty variables imagedestroy($src); imagedestroy($tmp); } ?> Any help would be appreciated, fairly new to all this! Thanks!!! Hey guys I am a little confused, I have a listing page that displays all of my rows in my SQL table.
An example of what I am developing can be seen he http://www.drivencar...k/used-cars.php
So as you can see I have developed a listing page that fetches all of the table rows, each row is equivalent to one vehicle.
I now want to let users filter the results using the form to the left, I was going to use AJAX originally however I feel as if it would take way to long to learn how to develop it that way.
Here is the code setup I am using to achieve the example I have shown:
<?php
include('database.php');
try {
$results = $db->query("SELECT Make, Model, Colour, FuelType, Year, Mileage, Bodytype, Doors, Variant, EngineSize, Price, Transmission, PictureRefs, ServiceHistory, PreviousOwners, Options, FourWheelDrive FROM import ORDER BY Make ASC");
} catch (Exception $e) {
echo "Error.";
exit;
}
try {
$filterres = $db->query("SELECT DISTINCT Make FROM import ORDER BY Make ASC");
} catch (Exception $e) {
echo "Error.";
exit;
}
?>
As you can see the first block of code has two SQL selectors, the $results is used to fetch the whole table and list all vehicles.
The second block is used to display the 'Make' column for the form.
This block of code is the actual form:
<form>
<select class="form-control select-box" name="">
<option value="make-any">Make (Any)</option>
<?php while($make = $filterres->fetch(PDO::FETCH_ASSOC))
{
echo '
<option value="">'.$make["Make"].'</option>
';
} ?>
</select>
<button href="#" class="btn btn-block car-search-button btn-lg btn-success"><span class="glyphicon car-search-g glyphicon-search"></span> Search cars
</button>
</form>
As you can see this block is using a while loop to display all of the 'Make's' in the 'Make' column and uses a DISTINCT clause so that it doesn't show identical options.
Here is the block that lists the results to the page:
<?php while($row = $results->fetch(PDO::FETCH_ASSOC))
{
echo '
<div class="listing-container">
<a href="carpage.php"><h3 class="model-listing-title clearfix">'.$row["Make"].' '.$row["Model"].' '.$row["Variant"].'</h3></a>
<h3 class="price-listing">£'.number_format($row['Price']).'</h3>
</div>
<div class="listing-container-spec">
<img src="'.(explode(',', $row["PictureRefs"])[0]).'" class="stock-img-finder"/>
<div class="ul-listing-container">
<ul class="overwrite-btstrp-ul">
<li class="diesel-svg list-svg">'.$row["FuelType"].'</li>
<li class="saloon-svg list-svg">'.$row["Bodytype"].'</li>
<li class="gear-svg list-svg">'.$row["Transmission"].'</li>
<li class="color-svg list-svg">'.$row["Colour"].'</li>
</ul>
</div>
<ul class="overwrite-btstrp-ul other-specs-ul h4-style">
<li>Mileage: '.number_format($row["Mileage"]).'</li>
<li>Engine size: '.$row["EngineSize"].'cc</li>
</ul>
<button href="#" class="btn h4-style checked-btn hover-listing-btn"><span class="glyphicon glyphicon-ok"></span> History checked
</button>
<button href="#" class="btn h4-style more-details-btn hover-listing-btn tst-mre-btn"><span class="glyphicon glyphicon-list"></span> More details
</button>
<button href="#" class="btn h4-style test-drive-btn hover-listing-btn tst-mre-btn"><span class="test-drive-glyph"></span> Test drive
</button>
<h4 class="h4-style listing-photos-count"><span class="glyphicon glyphicon-camera"></span> 5 More photos</h4>
</div>
';
} ?>
So down to my question... How can I filter these results displayed in the listing block using the select element, when a user selects a 'Make' from the select element I want them to be able to submit the form and return all rows in the SQL table containing the same 'Make' string and hide other rows that are false.
Any ideas how I can achieve this or any easier ways?
Thanks
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