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PHP - Select One Uinique Field From Mysql
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. Similar TutorialsHi: I'm going crazy trying to do the following: I'm making a job registration process where the user registers on one php page to the website, must acknowlege and email receipt using an activate php page, then is directed to upload their C.V. (resume) based on the email address they enter in the active page output. I then run an upload page to store the resume in teh MySQL db based on the users email address in the same record. If I isolate the process of the user registering to the db, it works perfectly. If I isolate the file upload process into the db, it works perfect. I simply cannot upload teh file to the existing record based on teh email form field matching the user_email field in the db. With the processes together, teh user is activated, but teh file is not uploaded. Maybe I've simply been at this too long today, but am compeled to get through it by end day. If anyone can help sugest a better way or help me fix this, I will soo greatly appreciate it. My code is as follows for the 2 pages. ---------activate.php------- <?php session_start(); include ('reg_dbc.php'); if (!isset($_GET['usr']) && !isset($_GET['code']) ) { $msg = "ERROR: The code does not match.."; exit(); } $rsCode = mysql_query("SELECT activation_code from subscribers where user_email='$_GET[usr]'") or die(mysql_error()); list($acode) = mysql_fetch_array($rsCode); if ($_GET['code'] == $acode) { mysql_query("update subscribers set user_activated=1 where user_email='$_GET[usr]'") or die(mysql_error()); echo "<h3><center>Thank You! This is step 2 of 3. </h3>Your email is confirmed. Please upload your C.V. now to complete step 3.</center>"; } else { echo "ERROR: Incorrect activation code... not valid"; } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta content="text/html; charset=utf-8" http-equiv="Content-Type" /> <title>Job application activation</title> </head> <body> <center> <br/><br/><br/> <p align="center"> <form name="form1" method="post" action="upload.php" style="padding:5px;"> <p>Re-enter you Email : <input name="email" type="text" id="email"/></p></form> <form enctype="multipart/form-data" action="upload.php" method="POST"> <input type="hidden" name="MAX_FILE_SIZE" value="4000000"> Upload your C.V.: <input name="userfile" type="file" id="userfile"> <input name="upload" type="submit" id="upload" value="Upload your C.V."/></form> </p> </center> </body> </html> --------upload.php---------- <?php session_start(); if (!isset($_GET['usr']) && !isset($_GET['code']) ) { $msg = "ERROR: The code does not match.."; exit(); } if(isset($_POST['upload']) && $_FILES['userfile']['size'] > 0) { $fileName = $_FILES['userfile']['name']; $tmpName = $_FILES['userfile']['tmp_name']; $fileSize = $_FILES['userfile']['size']; $fileType = $_FILES['userfile']['type']; $email = $_POST['email']['user_email']; $fp = fopen($tmpName, 'r'); $content = fread($fp, filesize($tmpName)); $content = addslashes($content); fclose($fp); if(!get_magic_quotes_gpc()) { $fileName = addslashes($fileName); } include 'reg_dbc.php'; $query = "UPDATE subscribers WHERE $email = user_email (name, size, type, content ) ". "VALUES ('$fileName', '$fileSize', '$fileType', '$content')"; mysql_query($query) or die('Error, query failed'); mysql_close($dbname); } ?> <center> <br/> <br/> <br/> <br/> Thank you for uploading your <?php echo "$fileName"; ?> file, completing your registration, and providing us your C.V. for this position. <br/> <br/> <br/> We will contact you if your canditature qualifies. </center> 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. 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 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 Can 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? 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"; } ?> 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 $r1 = mysql_query("SELECT * FROM pushup where paddler_id =1 ORDER BY practice_id ASC "); I want to put something which makes it paddler_id = X where x = 2,3,4,5... and x exists in database and do not print twice the x (because that database may have rows where x = same id How do i do that??
Hi guys,
<form id='contact-form".$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>");
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... Hi i have the following code Code: [Select] $orderFetch->order="3"; mysql_query("UPDATE `categories` SET order=order-1 WHERE `order` > $orderFetch->order"); i wish to minus -1 from all the "order" fields in categories that are over a certain value i.e 3. But however this code is not working can anyone tell me where i'm going wrong? thank you I have this script running: <?php $query = mysql_query("SELECT * FROM jobs WHERE event = 'Yes' ORDER BY title"); while ($row = mysql_fetch_array($query)) { ?> How do I check if a field is empty and NOT display it...? For instance it has a 'applied' field, if that is empty I dont want it to display, however I still need the event = 'Yes' part. In my mysql database i have a field that records peoples details one part of the details is their county/region they live in. It records the county in a normal varchar field however i need a simple php script that searching the database and finds the most common county so i can then return to the screen where the most popular region for my members. Thank you for reading. I have made a simple form where users who have been subscribed and unsubscribe by inserting their email address. In my database using PHPMyAdmin, my database to store the emails is 'Links', the table is 'email' and the fields are the 'id' and 'emailaddress'. What I have tried is making a text input field, where the user ill insert his or her email address, to unsubscribe on the website. As a result the user's field for his or her email address will be delete in the database which is saving the emails for all users who have subscribed. My HTML codes a Code: [Select] <p>Subscribe for newsletters:</p> <img src="images/k-newsletter-icon.png" width="96" height="96" alt="subscri"/> <form action="index.php" method="post"> <input type="text" size="25" placeholder="Your email address..." name="enter"/> <input class="submit" type="submit" value="Subscribe" name="subscribe"/> </form> My PHP codes a Code: [Select] <?php if ($_SERVER['REQUEST_METHOD'] == 'POST') { $email = $_POST['enter']; @mysql_connect ('localhost', 'root', '') or die ('Error'); @mysql_select_db ('links') or die ('Error'); if (!filter_var($email, FILTER_VALIDATE_EMAIL)) { echo "Not an email"; return false; } else { mysql_query("DELETE FROM email WHERE emailaddress ='$email'"); echo "deleted"; } } ?> When I test it,it is not working, as I see the email which was saved, is still in the database! Help! I have a mysql field that I want to store a php variable in and then retrieve it.
Example:
Color table has the field named colorBackground with the value of #FF0000
CSS table has the field named cssBackground with the value of .background { background-color: #444444; }
What I want to do is make Color/colorBackground a variable {{$backgroundcolor = rsColor['colorBackground']
Then I want to change the CSS/cssBackground value to .background { background-color:$backgroundcolor; }
Creating the $backgroundcolor works fine. I'm just not sure how to put the $backgroundcolor in my CSS/cssBackground field.
I have a script that uploads and images, creates 2 versions of it, re-sizes the images and inserts all the info into a DB. The odd thing is that if I don't have an image to upload, it will still run the script but it will completely skip over the description. If I upload an image, it works fine. Not too sure what the issue is. Is it because my variable of $desc causing a conflict? Here's my code: Code: [Select] <?php ini_set("display_errors",1); include 'dbconn.php'; $change=""; $abc=""; define ("MAX_SIZE"," 10000"); function getExtension($str) { $i = strrpos($str,"."); if (!$i) { return ""; } $l = strlen($str) - $i; $ext = substr($str,$i+1,$l); return $ext; } $errors=0; if($_SERVER["REQUEST_METHOD"] == "POST") { $image =$_FILES["file"]["name"]; $uploadedfile = $_FILES['file']['tmp_name']; if ($image) { $filename = stripslashes($_FILES['file']['name']); $extension = getExtension($filename); $extension = strtolower($extension); if (($extension != "jpg") && ($extension != "jpeg") && ($extension != "png") && ($extension != "gif")) { $change='<div class="msgdiv">Unknown Image extension </div> '; $errors=1; } else { $size=filesize($_FILES['file']['tmp_name']); if ($size > MAX_SIZE*1024) { $change='<div class="msgdiv">You have exceeded the size limit!</div> '; $errors=1; } if($extension=="jpg" || $extension=="jpeg" ) { $uploadedfile = $_FILES['file']['tmp_name']; $src = imagecreatefromjpeg($uploadedfile); } else if($extension=="png") { $uploadedfile = $_FILES['file']['tmp_name']; $src = imagecreatefrompng($uploadedfile); } else { $src = imagecreatefromgif($uploadedfile); } echo $scr; list($width,$height)=getimagesize($uploadedfile); $newwidth=150; $newheight=($height/$width)*$newwidth; $tmp=imagecreatetruecolor($newwidth,$newheight); $newwidth1=50; $newheight1=($height/$width)*$newwidth1; $tmp1=imagecreatetruecolor($newwidth1,$newheight1); imagecopyresampled($tmp,$src,0,0,0,0,$newwidth,$newheight,$width,$height); imagecopyresampled($tmp1,$src,0,0,0,0,$newwidth1,$newheight1,$width,$height); $famname = $_POST['famname']; $desc = $_POST['description']; $petname = $_POST['petname']; $letter = $_POST['letter']; $filename = "images/". $_FILES['file']['name']; $filename1 = "images/small". $_FILES['file']['name']; imagejpeg($tmp,$filename,100); imagejpeg($tmp1,$filename1,100); imagedestroy($src); imagedestroy($tmp); imagedestroy($tmp1); }} } //If no errors registred, print the success message if(isset($_POST['Submit']) && !$errors) { $insert="insert ignore into tbl_tribue (petname,familyname,letter,description,imgpath,thumbpath) values ('$petname','$famname','$letter','$desc','$filename','$filename1')"; mysql_query($insert); //mysql_query("update gallery set imgpath='$filename'"); $change=' <div class="msgdiv">Pet Added Successfully!</div>'; print $insert; } ?> I had turned on the error and the print to see the output. Anyone have any ideas? Thanks! Can anyone tell me the best way to handle empty search field on a mysql Query? The query below if $name is left blank but the other fields are filled it shows no results, currently i use if.... else.... but that means i have a lot of code replicated, is there a better way? $sql = "SELECT private.username, se.age, se.ro, se.suc, se.bu, se.fu, com.in, com.ol, se.di, se.bo, se.uin FROM Reg_Profile_public AS se INNER JOIN Reg_Profile_Private AS private USING (uin) INNER JOIN Reg_Profile_public_Com AS com USING (uin) WHERE se.age BETWEEN '$low' AND '$high' AND se.ro=''$name"; I need to update a particular table and there be a whole lot of field names. I'd like to do it without having to use them the same way you can access then like $row[12] after a "select" query. The standard for updating (at least how I write it) goes like this ... Code: [Select] $query = "UPDATE linguistics SET welcome1='$var3', welcome2='$var4' WHERE language = '$lengua'"; Well, I'd like to do it where I don't have to name welcome1 and welcome2, but instead refer to their numerical position within the database. On a "select" query we can do this. Can this be done on an update? Something perhaps like ... Code: [Select] $query = "UPDATE linguistics SET 3='$var3', 4='$var4' WHERE language = '$lengua'"; ... where 3 and 4 represent the 4th and 5th fields respectively. I know that this particular phrasing doesn't work because I've tried it, but surely there has to be something. This will save me LOTS of time. Thanks!! |
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