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PHP - Php: Input Mysql Query
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
Similar TutorialsHere is my code: // Start MySQL Query for Records $query = "SELECT codes_update_no_join_1b" . "SET orig_code_1 = new_code_1, orig_code_2 = new_code_2" . "WHERE concat(orig_code_1, orig_code_2) = concat(old_code_1, old_code_2)"; $results = mysql_query($query) or die(mysql_error()); // End MySQL Query for Records This query runs perfectly fine when run direct as SQL in phpMyAdmin, but throws this error when running in my script??? Why is this??? Code: [Select] 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 '= new_code_1, orig_code_2 = new_code_2WHERE concat(orig_code_1, orig_c' at line 1 If you also have any feedback on my code, please do tell me. I wish to improve my coding base. Basically when you fill out the register form, it will check for data, then execute the insert query. But for some reason, the query will NOT insert into the database. In the following code below, I left out the field ID. Doesn't work with it anyways, and I'm not sure it makes a difference. Code: Code: [Select] mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); Full code: Code: [Select] <?php include_once("includes/config.php"); ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <title><? $title; ?></title> <meta http-equiv="Content-Language" content="English" /> <meta http-equiv="Content-Type" content="text/html; charset=UTF-8" /> <link rel="stylesheet" type="text/css" href="style.css" media="screen" /> </head> <body> <div id="wrap"> <div id="header"> <h1><? $title; ?></h1> <h2><? $description; ?></h2> </div> <? include_once("includes/navigation.php"); ?> <div id="content"> <div id="right"> <h2>Create</h2> <div id="artlicles"> <?php if(!$_SESSION['user']) { $username = mysql_real_escape_string($_POST['username']); $password = mysql_real_escape_string($_POST['password']); $name = mysql_real_escape_string($_POST['name']); $server_type = mysql_real_escape_string($_POST['type']); $description = mysql_real_escape_string($_POST['description']); if(!$username || !$password || !$server_type || !$description || !$name) { echo "Note: Descriptions allow HTML. Any abuse of this will result in an IP and account ban. No warnings!<br/>All forms are required to be filled out.<br><form action='create.php' method='POST'><table><tr><td>Username</td><td><input type='text' name='username'></td></tr><tr><td>Password</td><td><input type='password' name='password'></td></tr>"; echo "<tr><td>Sever Name</td><td><input type='text' name='name' maxlength='35'></td></tr><tr><td>Type of Server</td><td><select name='type'> <option value='Any'>Any</option> <option value='PvP'>PvP</option> <option value='Creative'>Creative</option> <option value='Survival'>Survival</option> <option value='Roleplay'>RolePlay</option> </select></td></tr> <tr><td>Description</td><td><textarea maxlength='1500' rows='18' cols='40' name='description'></textarea></td></tr>"; echo "<tr><td>Submit</td><td><input type='submit'></td></tr></table></form>"; } elseif(strlen($password) < 8) { echo "Password needs to be higher than 8 characters!"; } elseif(strlen($username) > 13) { echo "Username can't be greater than 13 characters!"; } else { $check1 = mysql_query("SELECT username,name FROM servers WHERE username = '$username' OR name = '$name' LIMIT 1"); if(mysql_num_rows($check1) < 0) { echo "Sorry, there is already an account with this username and/or server name!"; } else { $ip = $_SERVER['REMOTE_ADDR']; mysql_query("INSERT INTO servers (username, password, name, type, description, ip, votes, beta) VALUES ($username, $password, $name, $server_type, $description, $ip, 0, 1)"); echo "Server has been succesfully created!"; } } } else { echo "You are currently logged in!"; } ?> </div> </div> <div style="clear: both;"> </div> </div> <div id="footer"> <a href="http://www.templatesold.com/" target="_blank">Website Templates</a> by <a href="http://www.free-css-templates.com/" target="_blank">Free CSS Templates</a> - Site Copyright MCTop </div> </div> </body> </html> Why exactly is the _get returning just nothing everytime? Code: [Select] $seshid = $_GET['finish']; // Little things in here ... ///Add lesson plan if(isset($_POST['addinc'])){ if(!empty($_POST['inc'])){ $inc = addslashes($_POST['inc']); mysql_query("INSERT INTO `group_sessions_lplan` (`group_sessions_lplan_id`, `group_session_id`, `plan`) VALUES ('', '$seshid', '".strip_tags($inc, "<a><b><strong><u><i><span>")."')"); header("Location: ?id=".$seshid."#I"); exit(); } } Not too sure myself, anyone know how I can get it to show its actual value? The code below is a function that checks to see if an email address exists in a database, if so it alerts the user. The db has one table and one field. It works fine when there is ONE record! However, if there are > 1 it doesn't work. How can I step through each record and compare it to what the user entered? Of course, $_POST is the user's value and the db record is the $myAddy value. <?php function emailLookup() { include ('file:///Library/WebServer/Documents/re_connect_scripts/emailLookup.php'); while ($row = mysqli_fetch_array($result)) { extract ($row); $myAddy = $addy; } if ($_POST["add_email"] == $myAddy) { global $lookupError; $lookupError = 'This email address is already on the list.'; global $counter; $counter++; } else { return; } } ?> I'm restarting this under a new subject b/c I learned some things after I initially posted and the subject heading is no longer accurate. What would cause this behavior - when I populate session vars from a MYSQL query, they stick, if I populate them from an MSSQL query, they drop. It doesn't matter if I get to the next page using a header redirect or a form submit. I have two session vars I'm loading from a MYSQL query and they remain, the two loaded from MSSQL disappear. I have confirmed that all four session vars are loading ok initially and I can echo them out to the page, but when the application moves to next page via redirect or form submit, the two vars loaded from MSSQL are empty. Any ideas? Hello all,
I am completely sideways right now trying to get this to work. Basically I am attempting to make a form that fetches results based on a drop down and then populates a text box that the user can then edit. I have several issues, first of all, I would have to use check boxes for if a user has a certain file on record. Which would be editable, however, I have had issues displaying tinyint values of -1 as true and 0 as false.
Secondly, writing the data back to the table after clicking a submit button, I have still issues with checkboxes here and am thoroughly confused to if I need to use an array or what here.
Thirdly, should I be using PDO or mysql for this? We have an older NAS device that I will be running this on and the firmware is older.
All the info is stored in the same table, and would be selecting a single record at a time for a user to edit. I was thinking I would just write back ALL values to the table when submit is clicked? Is that a bad idea?
As you can tell I just need someone to kick my butt in the right direction, I feel like I have read all the different ways to skin a cat but confused myself in the process.
Thanks
Hi. I have an input field where a date is entered, format dd-mm-yy. I need to query the database to see if this date exists. How can I convert the date to yyyymmdd before the query? Thanks I have a family site with a member list and a forum that both run on MySql. One of the items in the member list is the birthday. What I like to achieve is that a day or 2 before a member's birthday, a post wil be automaticly inserted in the forum, (a post that contains that this persons birthday is comming up in a few days) without any user input. I like to have this done automaticly because I don't feel like setting up a cron job for every member seperatly. There are just too many members. What I need is a script that will create a cron job or something simular the moment a new member registers and updates his profile and sets his birthday. Hi people. I have a form which inputs into a database. Here is the code that inserts a yes no option... Code: [Select] <select name = "consent"> <option value = "Yes" <?php if ($_POST['consent'] == 'Yes') { echo 'selected="selected"'; } ?>>Yes</option> <option value = "No" <?php if ($_POST['consent'] == 'No') { echo 'selected="selected"'; } ?>>No</option> </select> However, I have been asked if I can make it a yes or no checkbox instead. Please can you tell me how I need to code it so that the "yes" or "no" is recorded in the DB. At the moment I just have this Code: [Select] <input name="consent" type="checkbox" value="Yes" />Yes<br /> <input name="consent" type="checkbox" value="No" />No<br /> Thanks in advance VinceG Guys, quick one. Im writing a script for a form to post info into a MySql table. Now rather than just having a single row for input I'd like to have lets say 10 rows, so I can add 10 records to the database. What I'm pondering is 2 things: 1: can i just repeat Code: [Select] <input type="text" name="opponent" size="27" /> over and over, or is it going to need its own name each time for example; Code: [Select] <input type="text" name="opponent2" size="27" /> <input type="text" name="opponent3" size="27" /> 2: when it comes to the processing script is it more economical to have the forms input field named the same over and over (if it IS possible) and if not whats the most econimcal way to code my Code: [Select] $opp= $_POST['opponent']; $query="INSERT INTO fixtures (match_date, season, opponent) VALUES ('$date', '$season', '$opp',)"; Your help and comments are appreciated as always guys Tom 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... Hi
I need a sql query help from you guys.
It is a sql query for get all upline referrer details from database for particular person
when new person register his details are storing in wp_members_tbl with uername, password, firstname, lastname, email, phone, address, referrer etc.
Below the query is for user
$wp_aff_members_db = $wpdb->get_row("SELECT * FROM $members_table_name WHERE refid = '".$_SESSION['user_id']."'", OBJECT);
And below the query is for this user's upline referrer
$wp_aff_members_db = $wpdb->get_row("SELECT * FROM $members_table_name WHERE refid = '$referrer'", OBJECT);
Now i need the query for find and get this referrer's upline referrer.
Please help me with a solution
Regards
Edited by rajasekaran1965, 23 October 2014 - 10:09 AM. hi dudes how do i write a mysql query with 3 columns, where the first column is 'year', the second is 'month' (integer) and the third is 'day' (integer), ordered by desc, but with an extra quirk, where if any of the three columns is zero (which means there is no data for that date column - assume i have a year and a month, but no day)? my code looks like the following Code: [Select] ORDER BY exhib_date_year DESC, exhib_date_month DESC, exhib_date_day DESC Hi Guys, I have been having problems with a piece of PHP and mysql. I was wondering if you could help me out? The error is = Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in /customers/klueless.net/klueless.net/httpd.www/daisysite/home.php on line 120 Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in /customers/klueless.net/klueless.net/httpd.www/daisysite/home.php on line 122 And this is the code which goes with these errors = Code: [Select] $msgquery = "SELECT * FROM spotty_messages WHERE (id_receiver = '" . $userid . "') AND read = '0'"; $messageres = mysql_query($msgquery); $messrow = mysql_fetch_array($messageres); $messagecount = mysql_num_rows($messageres); Hi All, Wondered if someone could help me out with a sql query that I am having difficulty with? My database consists of 3 tables, clients, video, category. The video table stores the primary key value of the clients table and the category table as a foreign key. What I am trying to achieve is return all the videos that are associated to a particular client and group them under the relevant category. If there are now videos that match the category then I do not want to display the category. Here is my code so far: Code: [Select] <?php $sql = "SELECT category.cat_id, category.name AS catname FROM category"; $result = mysql_query($sql) or die (mysql_error()); while($categoryrow = mysql_fetch_assoc($result)) { ?> </p> <div class="themeheader"><h5><?php echo $categoryrow['catname']; ?></h5></div> <Br /> <?php $vsql = "SELECT video.video_id, video.title, video.description, video.thumbnail FROM video WHERE video.cat_id = '" . $categoryrow['cat_id'] . "' AND video.client_id = $customerid ORDER BY video.video_id DESC"; $vresult = mysql_query($vsql) or die (mysql_error()); ?> <div class="videos"> <ul> <?php while($videorow = mysql_fetch_assoc($vresult)) { ?> <li id="categoryList"><a href="film-details.php?video_id=<?php echo $videorow['video_id']; ?>"><img src="+_1m4g35/<?php echo $videorow['thumbnail']; ?>" alt="<?php echo $videorow['title']; ?>" title="<?php echo $videorow['title']; ?>" width="291" height="142" border="0" /></a> <h2><?php echo $videorow['title']; ?></h2> <p><?php $limit = 100; if (strlen($videorow['description']) > $limit) $description = substr($videorow['description'], 0, strrpos(substr($videorow['description'], 0, $limit), ' ')) . '... <a href="film-details.php?video_id='.$videorow['video_id'].'">read more</a>'; echo $description; ?> </p> <?php } //end video loop?> </ul> <br class="clearfloat" /> </div> <?php } //end category loop ?> </div> The above code is the closest I have got but it still outputs the categories even when there are no videos that match the category id and the client id. Any help in the right direction gratefully received as I am gradually going insane! $loc="SELECT * FROM table WHERE username='a', id='1' AND uin='123'"; $get=myspl_query($loc) or die(mysql_error()); Please somone tell me the right command as how to use multiple AND in mysql query. Thanks I have 2 tables (see code below) (1) "tbl_users" has all users info, including "user_alias" (VARCHAR) and "user_id" (INT and PRIMARY KEY) (2) "tbl_projects" has all project info, including "project_client" (INT) which i linked to "user_id" from "tbl_users" and displays "user_alias" Code: [Select] CREATE TABLE `tbl_users` ( `user_id` int(11) NOT NULL auto_increment, `user_first` varchar(25) NOT NULL, `user_last` varchar(25) NOT NULL, `user_email` varchar(35) NOT NULL, `user_pw` varchar(12) NOT NULL, `user_role` int(1) NOT NULL, `user_alias` varchar(35) NOT NULL, PRIMARY KEY (`user_id`) ) CREATE TABLE `tbl_projects` ( `project_id` int(11) NOT NULL auto_increment, `project_client` int(1) NOT NULL, PRIMARY KEY (`project_id`) ) So... I managed to display all users and all projects but I want to show all users with ONLY their projects assigned. I mean, when I click on a user, I want to see his or hers contact info and all the projects that specific user is linked to. For instance, i might have project 1, project 2, project 3 and project 4. Project 1, 2 and 4 are linked to user 1 so when i click on user 1 i see projects 1, 2 and 3 (not 4). Code: [Select] <?php require_once('config.php'); mysql_select_db($database, $makeconnection); //this displays all projects $sql_get_projects="SELECT * FROM tbl_projects ORDER BY project_id ASC"; $get_projects = mysql_query($sql_get_projects, $makeconnection) or die(mysql_error()); $row_get_roles = mysql_fetch_assoc($get_projects); $totalRows_get_projects = mysql_num_rows($get_projects); //this displays all users $sql_find_users = "SELECT * FROM tbl_users WHERE user_id = $user_id"; $find_users = mysql_query($sql_find_users, $makeconnection) or die(mysql_error()); $row_get_users = mysql_fetch_assoc($find_users); $totalRows = mysql_num_rows($find_users); ?> Any help would be great!! Hello! First post. I am trying to code out a news website (kinda) I can't seem to fetch the news articles out of SQL. What am I doing wrong? (start is fetched by $_REQUEST. Assume it is 1) $end=$start+5; $stories = mysql_query("SELECT * FROM 'stories' DESC ORDER BY id LIMIT $start, $end"); $num = mysql_num_rows($stories); print $num; |
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