PHP - Select Field Question
Hi. I have two select fields one called category and the other sub category. Both are in the format.
<select name="category" value=" <?php $qGetCat = "SELECT * FROM club_category" ; $rGetCat = mysql_query($qGetCat); while ($Cat = mysql_fetch_assoc($rGetCat)) { ?>" /> <option value="<?php echo $Cat['categorys'];?>"><?php echo $Cat['categorys']; ?></option> <?php } ?> </select> <select name="sub_category" value=" <?php $qGetSubCat = "SELECT * FROM sub_categorys" ; $rGetSubCat = mysql_query($qGetSubCat); while ($SubCat = mysql_fetch_assoc($rGetSubCat)) { ?>" /> <option value="<?php echo $SubCat['sub_categorys'];?>"><?php echo $SubCat['sub_categorys']; ?></option> <?php } ?> </select> Now I want to make it so that when a particular category is chosen for example self defence. The sub category when clicked shows all the self defence stuff. If watersports was chosen sub category would show things like swimming. What do I need to look up to do this? At the minute there is no link the way I have coded these to fields. In the database its self there is a; category table with categoryID and category_name sub category table with sub_cat_name and categoryID I think this will work but I have never done this so its making my head hurt just thinking about it. Does anyone have any advice, tuts or scripts to do this? Thank you Similar TutorialsCan anybody clarify the usefulness of "AS" and elaborate on how it operates. If I select a 'bananas' AS 'yeelow fruit' will it change the TITLE in the table? Or the table that is being viewed? If I dump a list of 'FRUITS' into and HTML table, will using AS rename each column for me, or is that handled by MY coing of the HTML table? Hello. I am trying to display only one instance of records that have the same memberid in my db. I am using the following statement but it continues to show all of the records that have the same memberid. Any ideas what I may be doing wrong? Code: [Select] $sql = "select DISTINCT memberid, event, category, date, enddate, locality, location, address, city, state, zip, contact, phone, notes, doc1, doc2, doc3, doc4, doc5 from event where date >= '$datenow' ORDER by date ASC"; Thanks for any help! Hi Guys, Can someone please help me before I kill myself! I have a table holding details of images to be displayed in a gallery. The table has columns as 'id', 'image_caption', 'file_path', and 'userid'. The 'id' column is the unique key, and the 'userid' column is obviously id of the user who uploaded the image. What I would like to do is select and display one image from each unique userid and then display each image along with the details in my gallery. It doesn't really matter which users image is selected aslong as there is only one for each unique user. I have been trying a number of ways to do this (DISTINCT / subqueries) but I just cant get anything to work. Could someone please advise me on how I could get this to work. Cheers in advance. I want to select the user from super_administrators, administrators, teachers and students, and give the user permission based from what table he "came". But is giving the "Query failed" error... Code: [Select] <?php //Start session session_start(); //Include database connection details require_once('../config/config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $email = clean($_POST['email']); $password = clean($_POST['password']); //Input Validations if($email == '') { $errmsg_arr[] = 'O campo Email nao foi preenchido.'; $errflag = true; } if($password == '') { $errmsg_arr[] = 'O campo Senha nao foi preenchido.'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: ../index.php"); exit(); } //Create query $qry = "SELECT * FROM super_administrators,administrators,teachers,students WHERE email = '$email' AND passwd = '".md5($_POST['password'])."'"; $result = mysql_query($qry); $member = mysql_fetch_assoc($result); $table = mysql_field_table('$result', '0'); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { if($table == 'super_administrators') { session_regenerate_id(); $_SESSION['SESS_ID'] = $member['id']; $_SESSION['SESS_NAME'] = $member['name']; $_SESSION['SESS_EMAIL'] = $member['email']; session_write_close(); header("location: ../users/sadmin/index.php"); exit(); } if($table == 'administrators') { session_regenerate_id(); $_SESSION['SESS_ID'] = $member['id']; $_SESSION['SESS_SCHOOL_ID'] = $member['school_id']; $_SESSION['SESS_NAME'] = $member['name']; $_SESSION['SESS_EMAIL'] = $member['email']; session_write_close(); header("location: ../users/admin/index.php"); exit(); } if($table == 'teachers') { session_regenerate_id(); $_SESSION['SESS_ID'] = $member['id']; $_SESSION['SESS_SCHOOL_ID'] = $member['school_id']; $_SESSION['SESS_NAME'] = $member['name']; $_SESSION['SESS_EMAIL'] = $member['email']; session_write_close(); header("location: ../users/prof/index.php"); exit(); } if($table == 'students') { session_regenerate_id(); $_SESSION['SESS_ID'] = $member['id']; $_SESSION['SESS_SCHOOL_ID'] = $member['school_id']; $_SESSION['SESS_CLASS_ID'] = $member['class_id']; $_SESSION['SESS_NAME'] = $member['name']; $_SESSION['SESS_REGISTRATION'] = $member['registration']; $_SESSION['SESS_EMAIL'] = $member['email']; session_write_close(); header("location: ../users/aluno/index.php"); exit(); } } else { $errmsg_arr[] = 'Suas informacoes de login estao incorreta. Por favor, tente novamente.'; $errflag = true; $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: ../index.php"); exit(); } } else { die("Query failed"); } ?> Hello, i am trying to get my form to list usernames ascending as an option, but the select field just returns blank. Here is my code: Code: [Select] /** * editUserForm - Displays the users database table in * a nicely formatted field table. */ function editUserForm(){ $q = "SELECT * " ."FROM ".TBL_USERS." ORDER BY username ASC"; $result = $database->query($q); while ($row = mysql_fetch_assoc($result)) { echo '<option value="'.$row["username"].'">'.$row["username"].'</option>'; } } Code: [Select] <form action="/" method="post"> <div class="column"> <p> <select name="user" id="user" placeholder="Select A User To Edit" class="{validate:{required:true}}"> <option>Select A User To Edit</option> <? editUserForm(); ?> </select> </p> <div class="action_bar"> <input type="submit" class="button blue" value="Submit Post" /> <a href="#modal" class="modal button">Cancel</a> </div> </form> All help is appreciated I created a drop-down menu using a MySQL statement in php for a form. My drop down menu works fine, but I want to assign a default value to it (normal text) the value will never change, thus I do not need to extract the default value from the table, I already know the value. Here is the basic snippet from the script: <?php include("../includes/xxx.php"); $cxn = mysqli_connect($host,$user,$password,$dbname); $query = "SELECT DISTINCT `plant_id` FROM `plant` ORDER BY `plant_id`"; $result = mysqli_query($cxn,$query); while($row = mysqli_fetch_assoc($result)) { extract($row); echo "<option value='$plant_id'>$plant_id</option>\n"; } ?>
Hi guys, <form id='contact-form".$KitchConfigID."' name='contact-form".$KitchConfigID."' action='_pages/_kitchenconfig/adm_saveKitchConfig.php?Nav=".$NavHash."&kcid=".$KitchConfigID."' method='POST'> <div class='row'> <div class='col-sm-12'> <div class='md-form mb-0 form-sm'> <input value='".$KitchConfig."' type='text' id='configname".$KitchConfigID."' name='configname".$KitchConfigID ."' class='form-control'> <label for='configname".$KitchConfigID ."' class=''>Config Name:</label> </div> </div> </div> <div class='row'> <div class='col-sm-12'> <div class='md-form mb-0 form-sm'> <select id='designer_".$KitchConfigID."' name='designer_".$KitchConfigID."' class='mdb-select md-form' searchable='Search here..' onchange='changeInput(this.value, ".$SelField.");'>"); $designersql = "SELECT * FROM `tbl_contacts` ORDER BY `FirstName`"; $resultdesigner = $conn->query($designersql); if ($resultdesigner->num_rows > 0) { $resultdesigner->MoveFirst; echo("<option value='designer_".$KitchConfigID."'></option>"); while($rowdesigner = $resultdesigner->fetch_assoc()) { $DesignerName = $rowdesigner["FirstName"]." ".$rowdesigner["LastName"]; $DesignerID = $rowdesigner["ContactID"]; if($rowdesigner["ContactID"]==$rowkitch["Designer"]){ echo("<option value='".$DesignerID."' selected='selected'>".$DesignerName."</option>"); } else { echo("<option value='".$DesignerID."'>".$DesignerName."</option>"); } } } echo("</select> <input type='text' name='".$SelField."' value='' /> <label for='designer_".$KitchConfigID."' class=''>Designer:</label> </div> </div> </div> <div class='md-form mb-0 float-right'> <button type='submit' class='btn btn-outline-danger waves-effect btn-sm float-right'>Save <i class='fas fa-magic ml-1'></i></button> </div>
$upd_KitchConfigID = $_GET["kcid"]; if(isset($_POST["configname$upd_KitchConfigID"])) { $upd_KitchConfig = $_POST["configname$upd_KitchConfigID"]; } else { $upd_KitchConfig = 'NotSet'; $ErrMsg = $ErrMsg . 'No KitchConfig, '; } if(isset($_POST["designer$upd_KitchConfigID"])) { $upd_KitchConfigDesigner = $_POST["designer_$upd_KitchConfigID"]; } else { $upd_KitchConfigDesigner = 1; $ErrMsg = $ErrMsg . 'No Designer, '; } echo("designer_$upd_KitchConfigID: ".$_POST["designer_$upd_KitchConfigID"]."<br><br>");
I want to make a table for each paddler_id (Field) which exists into pushup Database My code for paddler_id = 1 is require "../sql.php"; $result = mysql_query("EXPLAIN pushup"); $r = mysql_query("SELECT * FROM paddlerinfo t1 LEFT JOIN pushup t2 on t2.paddler_id=t1.id LEFT JOIN practicedate t3 on t3.practice=t2.practice_id ORDER BY t1.firstname ASC "); $r1 = mysql_query("SELECT * FROM pushup where paddler_id =1 ORDER BY practice_id ASC "); echo "<table width='30%' border='1' cellpadding='0' cellspacing='0' style='font-family: monospace'>"; echo "<tr>"; echo "<td>DATE</td>"; while ($row = mysql_fetch_array($result)) { if (in_array($row["Field"], array("paddler_id", "practice_id", "p_id"))) continue; echo "<td>",($row["Field"]), "</td>"; } echo "</tr>"; while (($data = mysql_fetch_array($r1, MYSQL_ASSOC)) !== FALSE) { unset($data["p_id"],$data["paddler_id"]); echo "<tr>"; foreach ($data as $k => $v) { echo "<td>"; echo"$v"; echo "</td>"; } echo "</tr>"; } echo "</table>"; mysql_free_result($result); mysql_free_result($r); mysql_free_result($r1); mysql_close(); The problem is on line $r1 = mysql_query("SELECT * FROM pushup where paddler_id =1 ORDER BY practice_id ASC "); I want to put something which makes it paddler_id = X where x = 2,3,4,5... and x exists in database and do not print twice the x (because that database may have rows where x = same id How do i do that?? Or something like that... I am not sure how to put this.. Anyway, I'll just get started with explaining my problem. I have an admin-page in which you can delete the comments given on blogs, using checkboxes and clicking on a button with the value 'verwijderenSubmit'. The deletion part works just fine, nothing wrong. However, I also want to be able to EDIT the comments with an other button called 'bewerkenSubmit', using the same checkboxes that I use for deletion. Selecting the right CID (CommentID) is no problem, because that works the same as the deletion-part, but selecting the right textarea to update into the database is the problem... I uploaded a file here with the whole code: http://dhost.info/ddfs/myproblem.html I escaped the textarea within with square brackets, because otherwise the whole textarea would screw up.. I also added <!-- RELEVANT CODE --> to select the parts that I need to change. Well, I hope you understand my problem and can help. I have a search function in php where the text characters are matched to characters in a tables field--- works perfectly.... I need to make the input box have a droplist of words from database, this is also easy for me to do. the problem here is there is no definitive value! the options list always outputs a blank in the url--- its supposed to search a matching value and then output the matching value to url... Here is the droplist code: $output['RATESTITLE']='<input class="inputbox" type="text" size="24px" name="ratestitle" value="'.$sch->filter['ratestitle'].'" onfocus="if (this.value ==\''.$output['LANGUAGE_SEARCH_RATESTITLE'].'\') {this.value = \'\'}" />'; this outputs a input text box--- i want to have a droplist of options to populate this text box... If you must know this is the third day im at it... i have this code SELECT COUNT(*) as num FROM videos where title=$title and if the title has punctuation marks for example "psssh's psh" it would interpret it like 'SELECT COUNT(*) as num FROM videos where title=psh's psh ' right? and then it shows 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 's psh ORDER by (eno+0) DESC' at line 2 i dont know what to do, i tried changing ' and "... and adding ' and " also... i also need something that with work with other symbols like & / * etc. Why won't this increment the database field by 1? $sql2="update 'users' set 'TimesLoggedOn' = 'TimesLoggedOn' + 1 where 'BarcodeID' = $barcode"; mysql_query($sql2); Thanks! Hi Guys I have built a simple form, which has text fields Name, Telephone Number, Best Time to Call and E-mail. For security purposes, I am testing each against the function shown below which looks for dangerous code snippets, in an effort to protect against email header injection attacks. When it comes to the E-mail field, I am not actually testing whether a valid e-mail address has been entered, as it is the telephone number which is essential, not the e-mail. My question is, do you think this is a security weakness? Many thanks Code: [Select] //http://www.tonyspencer.com/2005/12/15/email-injection-exploit-through-a-php-contact-form/ //preg_match string to match goes within forward slashes, i.e. /str/, and i at the end makes it case insensitive function containsInjectionAttempt($input) { if (preg_match("/\r/i", $input) || preg_match("/\n/i", $input) || preg_match("/%0a/i", $input) || preg_match("/%0d/i", $input) || preg_match("/Content-Type:/i", $input) || preg_match("/<script>/i", $input) || preg_match("/bcc:/i", $input) || preg_match("/to:/i", $input) || preg_match("/cc:/i", $input)) { return true; } else { return false; } } Hello, guys! Just a small first ever thread, where I ask a question where I hope anyone is able to help me out I am working on a project with a friend of mine for a company. I have a problem with the function on a form. I wonder; How do i change the option value of a form? Let's say if I choose a category like "Cake", I want the second drop down box to display different brands of cake. But if I choose bicycle for example, I want the option values to be different brands of bicycles. I hope you guys understand what I was trying to say here. Anyways, here is a sample image where I want this function to occur: Sample: Cheers and thanks in advance, Marius I have a question concerning outputting information taken out of the database. Query is as follows: Code: [Select] $query = "SELECT * FROM users"; $results = $this->db->query($query); $result = $results->fetch_assoc(); return $result['username']; The value that returns is the first username in the database. But here I am selecting all the usernames, can I select others ones with the fetch_assoc? First off, I am new to PHP, and want to learn it, but it seems very complicated. Where do I begin? I found a script online for a contact form and I am having trouble making it work for my needs. I wanted to include a phone number field in the email message. This is what I have so far for the PHP, but it does not include the phone field in the body of the message. Everything else works fine. Apologize in advance if this message has been posted many times before. <?php $post = (!empty($_POST)) ? true : false; if($post) { include 'functions.php'; $name = stripslashes($_POST['contactname']); $email = trim($_POST['email']); $subject = "New Message from your website"; $message = stripslashes($_POST['message']); $phone = stripslashes($_POST['phone']); $error = ''; // Check name if(!$name) { $error .= 'Please enter your name.<br />'; } // Check email if(!$email) { $error .= 'Please enter an e-mail address.<br />'; } if($email && !ValidateEmail($email)) { $error .= 'Please enter a valid e-mail address.<br />'; } // Check message (length) if(!$message || strlen($message) < 15) { $error .= "Please enter your message. It should have at least 15 characters.<br />"; } if(!$error) { $mail = mail(WEBMASTER_EMAIL, $subject, $message, "From: ".$name." <".$email.">\r\n" ."Reply-To: ".$email."\r\n" ."X-Mailer: PHP/" . phpversion()); if($mail) { echo 'OK'; } } else { echo '<div class="notification_error">'.$error.'</div>'; } } ?> Hi Guys
I'm just getting back into coding after taking a break from it. I want to have a text field that can take up to 1000 characters. What is the best field type for the MySQL field for this? I can't remember if there are any particular special types.
Thanks
This is what i have Code: [Select] $traderss1=mysql_query("SELECT * FROM `poke_owned` WHERE `box`='1' AND `trainer`='" . $trading->username . "' ORDER BY `name` ASC"); How can i do this: Code: [Select] $traderss1=mysql_query("SELECT * FROM `poke_owned` WHERE `box`='1' AND `trainer`='" AND `traded`='0' . $trading->username . "' ORDER BY `name` ASC"); I know its not as simple as just adding AND `traded`='0' cause i tried that and i just got this Code: [Select] Parse error: syntax error, unexpected '=' in /home/tpkrpgn1/public_html/trade5.php on line 57 MySQL table (in case you need to know it?) Code: [Select] (`id`, `trainer`, `gender`, `name`, `party1`, `party2`, `party3`, `party4`, `party5`, `party6`, `level`, `exp`, `hp`, `totalhp`, `att`, `def`, `box`, `days`, `battle_wins`, `battle_losses`, `move1`, `move2`, `move3`, `move4`, `attbon`, `defbon`, `totalexp`, `sold`, `traded`, `item`, `soldtrainer`) I am working on a quiz app image 1 shows the index.php page image 2 shows the first question image 3 shows the second question image 4 shows the third question image 5 shows the result after completing the quiz image 6 shows the database 'quizzer' and its tables image 7 shows the 'questions' table image 8 shows the 'choices' table THIS LINK CONTAIN ALL THE CODE (and images) I HAVE DONE SO FAR https://www.mediafir...o7f5q0fe6y/quiz 1.Now my question is how to select the question RANDOMLY from 'questions' table along with 'choices' (by adding code to the existing file or create a new one). 2.If user refresh/reload the page before starting ('Start Quiz') or click 'Take Again' after finishing the quiz, the question should appear randomly. 3.Basically I want to change the order of question appearing in the browser each time I refresh. 4.My work so far is mentioned above.........Please help me with this "RANDOM" problem !! P.S - Will it be possible, by creating a random function in PHP which will check for repeat questions in a session and check for the 'id' of the question and if it is new display it on the page. If so what should I do and if no then how to do? |