PHP - Why Doesn't It Updated?
Is there something wrong with this query?
elseif($_POST["titlee"] && $_POST["contente"]) { $titlee = $_POST['titlee']; $contente = $_POST['contente']; $to = $_POST['edit']; mysql_query("UPDATE custom_pages SET title='$titlee' AND content='$contente' WHERE id='$to'"); echo '<div class="post"> <div class="postheader"><h1>Updated</h1></div> <div class="postcontent"> <p>Your custom page has been updated.</p> </div> <div class="postfooter"></div> </div> '; } Similar TutorialsI am trying to set the hidden input in the javascript but it doesn't set it in the php. This is the form with the hidden input. It corresponds to a field in the database table with the name of draggables. <form method="post" accept-charset="utf-8" id="myForm2" action="/mealplans/add"> <input type="hidden" name="draggables" id="drag12"/> <button id="myId" type="submit">Submit Form</button> </form> This is where I change the value in the javascript: if(data == "draggable3"){ alert("data " + data); var x = document.getElementById("drag12").value; x = "draggable3"; alert(x); } //end if
I've also tried getElementsByName but that didn't work either. var y = document.getElementsByName("draggables").value; //only changes it on the client This is actually a cakephp site but this problem is with the PHP so I am publishing it here. this is the script for viewing the data in edit mode <tr> <td width="67" height="24"><select name="qualifi"> <option selected="selected">None</option> <?php $qry=mysql_query("select * from qualification"); while($obj=mysql_fetch_array($qry)) { ?> <option value="<?php echo $obj[0]; ?>"<?php if($obj[0]==$qualifi) echo("selected"); ?>><?php echo $obj[1]; ?></option> <?php } ?> </select> </td> <?php echo"<td width='301'><input name='board' type='text' size='50' value='$board'/></td>"; echo"<td width='60'><input name='start' type='text' size='10' value='$start'/></td>"; echo"<td width='60'><input name='end' type='text' size='10' value='$end'/></td>"; echo"<td width='60'><input name='percentage' type='text' size='10' value='$percentage'/></td> </tr>"; } ?> </table> and the update query is like //qualification if($_SESSION[empcode]!="") { if($_SESSION[qualifi]!="None") { mysql_query("update HRMEMPQUAL set EMPCODE='$_SESSION[empcode]',QUALCODE='$_SESSION[qualifi]',BOARD='$_SESSION[board]',FROMYEAR='$_SESSION[start]',TOYEAR='$_SESSION[end]',MARK_PERCENT='$_SESSION[percentage]' where EMPCODE='$_SESSION[empcode]'"); } } how can i save all the data correctly while any change is take place...their are 3 rows of data.only the last fled data is updated all the 3 rows What's the best way to display when a page was last edited/modified? i am a beginerr in php.. and here is my code..which shows error in the querry line.. erro goes like this Parse error: syntax error, unexpected T_STRING in C:\xampp\htdocs\restafinal\process_edit_page.php on line 56 and code goes like this.. echo part works well...that query line is the error line..please tell me whats the problem... Code: [Select] <?php //$id=$_GET["id"]; $id= $_GET['id']; $menu_name= $_POST['menu_name']; $position= $_POST['position']; $visible= $_POST['visible']; $note= $_POST['note']; $content= $_POST['content']; //echo part works well.. echo "menu_name"; echo $menu_name; echo "<br/>"; echo "position"; echo $position; echo "<br/>"; echo "visible"; echo $visible; echo "<br/>"; echo "note"; echo $note; echo "<br/>"; echo "id"; echo $id; echo $content; [b]$result = mysql_query(UPDATE page SET note =$note, nevigation = $menu_name WHERE id = $id,$connection)[/b] // test to see if the update occurred if (mysql_affected_rows() == 1) { // Success! } else { $message = "The page could not be updated."; $message .= "<br />" . mysql_error(); } ?> Hi. I am trying to figure out the best approach to pull THE most recent record for display, and ONLY the most recent one. I am really not sure how to go about doing this. Ive just started into stored procedure programming and php, but havent found anything yet on this one. Thanks. I have a script to update certain rows that contain certain data in them. Here is the code: if ($_POST['newstype'] == "1") { $query = mysql_query("SELECT * from `news` WHERE type='1'"); if (mysql_num_rows($query) == 1) { mysql_query("UPDATE `news` SET type = '2' WHERE type = '1' ORDER BY `news`.`dtime` ASC LIMIT 1") or die(mysql_error()); } else { echo "did not edit main news"; } } elseif ($_POST['newstype'] == "2") { $query = mysql_query("SELECT * from `news` WHERE type='2'"); if (mysql_num_rows($query) == 3) { mysql_query("UPDATE `news` SET type = '3' WHERE type= '2' ORDER BY `news`.`dtime` ASC LIMIT 1") or die(mysql_error()); } else { echo "did not edit recent news"; } } elseif ($_POST['newstype'] == "3") { $query = mysql_query("SELECT * from `news` WHERE type='3'"); if (mysql_num_rows($query) == 3) { mysql_query("UPDATE `news` SET type = '4' WHERE type= '3' ORDER BY `news`.`dtime` ASC LIMIT 1") or die(mysql_error()); } else { echo "did not edit old news"; } } else { echo "didnt update anything!"; } Here is my database structure. Why is it not updating it and always saying Did not edit main/recent/old news. Hi guys, I need your help. I am checking on a database as I want to see if the records have been updated or not. Code: [Select] <?php session_start(); define('DB_HOST', 'localhost'); define('DB_USER', 'mydbuser'); define('DB_PASSWORD', 'mydbpass'); define('DB_DATABASE', 'mydbtable'); $errmsg_arr = array(); $errflag = false; $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } function clean($var){ return mysql_real_escape_string(strip_tags($var)); } $username = clean($_GET['user']); $password = clean($_GET['pass']); $test = clean($_GET['test']); $public = clean($_GET['public']); if (isset($_GET['user']) && (isset($_GET['pass']))) { if($username == '' || $password == '') { $errmsg_arr[] = 'username or password are missing'; $errflag = true; } } elseif (isset($_GET['user']) || (isset($_GET['test'])) || (isset($_GET['public']))) { if($username == '' || $test == '' || $public == '') { $errmsg_arr[] = 'user or others are missing'; $errflag = true; } } if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; echo implode('<br />',$errmsg_arr); } else { $qry="SELECT * FROM members WHERE username='$username' AND passwd='$password'"; $result=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if ($username && $password) { if(mysql_num_rows($result) > 0) { $qrytable1="SELECT images, id, test, links, Public FROM user_channel_list WHERE username='$username'"; $result1=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); while ($row = mysql_fetch_array($result1)) { echo "<p id='test'>"; echo $row['test'] . "</p>"; echo '<p id="images"> <a href="images.php?test=test&id='.$row['id'].'">Images</a></td> | <a href="http://' . $row["links"] . '">Link</a> </td> | <a href="delete.php?test=test&id='.$row['id'].'">Delete</a> </td> | <span id="test">'.$row['Public'].'</td>'; } } else { echo "user not found"; } } elseif($username && $test && $public) { $qry="SELECT * FROM members WHERE username='$username'"; $result1=mysql_query($qry) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result1) > 0) { $qrytable1="SELECT Public FROM user_channel_list WHERE username='$username' && test='$test'"; $result2=mysql_query($qrytable1) or die('Error:<br />' . $qry . '<br />' . mysql_error()); if(mysql_num_rows($result2) > 0) { $row = mysql_fetch_row($result2); mysql_query("UPDATE user_list SET Public=('$_GET[public]') WHERE username='$username' AND test='$test'"); echo "update!"; } else { echo "already updated!"; } } else { echo "user not found"; } } } ?> When I run debug the code on my php, if i input the data in a url bar while the records are the same as the data that I enter in the url, i should get the print out on my php "already updated", but I keep getting "update!". Do you know how i can check on mysql database to see if the records have been updated or not?? I'm creating a game where 2 people, on separate computers can battle each other. I'm doing it turn-based, like Final Fantasy but I came across a problem, how do I check if the other person has ended their turn? I was thinking that I would have an area in my database that would be true/false if it was their turn. So if the first person ends their turn, their 'turn' becomes false in the database, but how, on the other users computer, do I constantly check if the 'turn' has become false so that they may now play without having to keep updating the browser. Or is their a better way to do something like this? I think I've gotten as far as I can with my troubleshooting so hopefully someone can help me here. I have a constructor for a controller class that receives the $_SESSION array from index.php by reference. An instance variable is then created when references this references so that any changes made to this instance variable update the real $_SESSION variable. This was working until it mysteriously stopped after not touching it for a few hours. The local $this->session variable is working because it completes some tasks using it. I've also hardcoded values into the session from the index.php file and when they are hard coded the session works as you would expect so it seems to just not be connecting this local variable to the $_SESSION array anymore for some reason. index.php snippet: Code: [Select] session_start(); $action = 'login'; $controller = new UserController($_POST,$_SESSION); call_user_func(array($controller,$action)); constructor and login action from UserController (constructor is actually from a base controller but that shouldn't matter). Also the activeSession() function that checks if a session is valid and kicks the user if its not (which is what happens when anything is tried after logging in): Code: [Select] function __construct($post,&$session){ $this->post = $post; $this->session = &$session; $this->view = 'views/login.php'; $this->title = ''; $this->error = ''; } function login(){ try{ $user = $this->validateString($this->post['username'],'Username'); $pass = $this->validateString($this->post['password'],'Password'); $dao = new UserData(); $this->session['userid'] = $dao->login($user, $pass); $this->session['username']=$user; $dao = new PostData(); $this->posts = $dao->getFeedPosts($this->session['userid']); $this->view = 'views/posts.php'; $this->title = 'Post Feed'; } catch(Exception $e){ $this->handleException($e); } } function activeSession(){ if(!empty($this->session['userid'])) return true; else return false; } Any help would be amazing. I really don't know how this stopped working. Thanks in advance. I have been using an inventory application built on PHP/MySQL. Since this morning I could submit the data and they were perfectly reflected on the MySQL Table. However, for a few hours I cannot save the submitted data to the table and it doesn't show any error message. Please note no change have been made since it was successfully running. The developer of this application is not available right now.
PLEASR HELP I AM A NOVICE IN PHP/MYSQL.
Hi all, i want delete old updated images with considering its index in array. <?php global $oldimg; $oldimg = array(); //action: edit news if (isset($_GET['id'])) { $NewsID = (int)$_GET['id']; if ($NewsID == 0) { $rdir = '<META HTTP-EQUIV="Refresh" CONTENT="1.4;URL=panel-news.php">'; die($rdir);} //get user from the database and put data into $_POST variables. //Include database connection details require_once('../config.php'); //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"); } $rs = mysql_query("SELECT newsimg1, newsimg2, ". " newsimg3 FROM news WHERE id = $NewsID"); if (mysql_num_rows($rs) == 0) die('no such a newsID!'); $row = mysql_fetch_assoc($rs); $oldimg[0] = $row['newsimg1']; $oldimg[1] = $row['newsimg2']; $oldimg[2] = $row['newsimg3']; } ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>" enctype="multipart/form-data"> <input type=file name="file[]" size=20 accept="image/jpg,image/jpeg,image/png"> <input type=file name="file[]" size=20 accept="image/jpg,image/jpeg,image/png"> <input type=file name="file[]" size=20 accept="image/jpg,image/jpeg,image/png"> <input type="hidden" name="MAX_FILE_SIZE" value="2097152" /> <input type="hidden" name="NewsID" value='<?php echo (isset($NewsID))?$NewsID:"0";?>'> <input type="submit" value="edit" id="save" name="save"/> </form> <?php if (isset($_POST['save']) && isset($_POST['NewsID'])){ $NewsID = (int)$_POST['NewsID']; //Include database connection details require_once('../config.php'); //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"); } //max fle size value $max_file_size = 2097152; //Global newsimg global $newsimg;global $ctr; $ctr = 0; //timestamp to make files unique names $timestamp = time(); //destination folder path $destpath = "../Dir/Newsimg/"; //looping each file or image from the form while(list($key,$value) = @each($_FILES["file"]["name"])) { //check if any empty or errors if(!empty($value)){ if ($_FILES["file"]["error"][$key] > 0) { $edir ='<div id="fail" class="info_div">'; $edir .='<span class="ico_cancel"><strong>'; $edir .="err : ". $_FILES["file"]["error"][$key] .'</strong></span></div>'; echo($edir); } else { //add the timestamp to filename $file_name = $timestamp.$_FILES['file']['name']; //temp name from upload form, key of each is set $source = $_FILES["file"]["tmp_name"][$key] ; //getting the file type $file_type = $_FILES["file"]["type"][$key]; //placing each file name into a variable $filename1 = $_FILES["file"]["name"][$key]; //lowering the file names $filename = strtolower($filename1); //adding timestamp to front of file name $filename = "$timestamp$filename"; ++$timestamp; if ($file_type != "image/jpeg" && $file_type != "image/png" && $file_type != "image/jpg") { die(" <div id='fail' class='info_div'> <span class='ico_cancel'><strong> Invalid Format!</strong></span><br /> <a href=\"javascript: history.go(-1)\">Retry</a> </div>"); } //moving the file to be saved from the temp location to the destination path move_uploaded_file($source, $destpath . $filename); //the actual final location with timestamped name $final_file_location = "$destpath$filename"; if (file_exists($final_file_location)) { if (isset($oldimg[$ctr])) { chdir('../Dir/Newsimg/'); echo $oldimg[$ctr]; unlink($oldimg[$ctr]);} $newsimg[$ctr] = $filename; $ctr++; ?>TNX. I'm trying to get a simple javascript popup script going anytime the database is updated in real-time. I'm not sure as to what to do next, because I'm still a newbie with jQuery and ajax, but the following code is what I have right now:
PHP MySQL query page:
<?php $con = mysqli_connect('localhost','root','root','mydb'); if (!$con) { die('Could not connect: ' . mysqli_error($con)); } mysqli_select_db($con,"mydb"); $sql="SELECT * FROM incoming_calls"; $result = mysqli_query($con,$sql); while($row = mysqli_fetch_array($result)) { $callArray[] = array('phonenumber' => $row['phone_number'], 'id' => $row['phone_login_id']); } if (!empty($callArray)) { //echo json_encode($callArray); for ($i = 0; $i < count($callArray); ++$i) { print $callArray[$i]['id'] . ": " . $callArray[$i]['phonenumber'] . "<br>"; } } mysqli_close($con); ?>MySQL "incoming_calls" db table (I have a node.js script, that draws SMDR data from a phone system and posts it to this table): id phone_number phone_login_id date_created 3 000-000-0000 1225 12/15/2014 14:53 6 000-000-0000 1222 12/15/2014 14:53 9 000-000-0000 1202 12/15/2014 14:53 12 000-000-0000 1201 12/15/2014 14:55 18 000-000-0000 1232 12/15/2014 14:55 21 000-000-0000 1222 12/15/2014 14:57 24 000-000-0000 1222 12/15/2014 14:57 27 000-000-0000 1201 12/15/2014 14:58 30 000-000-0000 1207 12/15/2014 14:58 33 000-000-0000 1212 12/15/2014 14:59HTML ajax call page: <!DOCTYPE html> <html lang="en"> <head> <meta charset="utf-8"> <title>Phone calls</title> <script src="http://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> </head> <body> <script language="javascript" type="text/javascript"> function getCall(){ $.get("phonecall.php", function(data){ $("#call").html(data); }); } setInterval(getCall,5000); </script> <div id="call"></div> </body> </html>Any help with this is greatly appreciated! Thanks ahead of time! Edited by Juan1989, 16 December 2014 - 11:07 AM. I am creating a stream page that updates automatically without the page reloading by reloading a div on the page using a timer. However, I only want it to run the update and fade in/fade out when there is a new status in the table. I tried doing it through PHP getting the number of rows in the table on the page itself and then checking the number of rows on an action page, however, the number of rows didnt update when the div was reloaded. I then tried doing it using jquery/javscript but had the same problem. How can I acheive this? Thanks in advance if ($count==1){ header("Location:store.php"); }very simple I have issolated it and it doesn't redirect maybe u can see where my mistake is My script is finally working as intended, but I want to add some additional data results. I am trying to resolve how I can display a count of ONLY the records updated by my query: // START :: Query to replace matches mysql_query("UPDATE orig_codes_1a AS a JOIN old_and_new_codes_1a AS b ON concat(a.orig_code_1, a.orig_code_2) = concat(b.old_code_1, b.old_code_2) SET a.orig_code_1 = b.new_code_1, a.orig_code_2 = b.new_code_2") or die(mysql_error()); // END :: Query to replace matches In this query I count ALL records selection: // START :: Create query to be displayed as final results of original codes table. $result = mysql_query("SELECT * FROM orig_codes_1a") or die(mysql_error()); I want to display a count on this part of the query: ....concat(a.orig_code_1, a.orig_code_2) = concat(b.old_code_1, b.old_code_2).... Well I have a script that executes a scan on a system set to run infinitely, and I need it to echo out a message each time it loops through, but I don't want it to echo out the message with the next loop message below it, and the next one below that etc... I've tried using the flush(); function and been messing around with that with no luck. For security reasons I don't want to release any of the processing code, but here is the basic construction of the script: <?PHP ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** $RepeatIt = -1; for($g=1; $g!=$RepeatIt+1; $g++) { ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** $ScanMessage = ":.:.: SCANNING THE HITLIST FOR MOBSTER: ".$MobName." (SCAN #$g) :.:.:"."<br/><br/>"; echo $ScanMessage; ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** ***PROCESSING AND SCAN CODE*** } ?> At the moment it's returning: :.:.: SCANNING THE HITLIST FOR MOBSTER: DEUS EX DESTROYER (SCAN #1) :.:.: :.:.: SCANNING THE HITLIST FOR MOBSTER: DEUS EX DESTROYER (SCAN #2) :.:.: :.:.: SCANNING THE HITLIST FOR MOBSTER: DEUS EX DESTROYER (SCAN #3) :.:.: :.:.: SCANNING THE HITLIST FOR MOBSTER: DEUS EX DESTROYER (SCAN #4) :.:.: So what I want it to do is just delete the scanning message and replace it with the next scan message so while running this script you would see just the number increment on the same line. Any suggestions? Thanks. I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP require_once("./include/membersite_config.php"); if(!$fgmembersite->CheckLogin()) { $fgmembersite->RedirectToURL("login.php"); exit; } ?> <form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST"> <input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br> Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br> Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br> Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br> <button>Submit</button>my editform.php <?php $con = mysqli_connect("localhost","root","user","pass"); if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'"); header("Location: test.php"); ?> I have a table with a list of users and an edit button and delete button. When the edit button is pressed on a site it passes the user_id as p_id to the page that catches it and displays the data. The problem is when I click on the "update user" button, I get the following error:
Warning: Undefined variable $the_user_id in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php on line 10 The weird thing is I had another update user page with a table I created that ran the query to update the table in the database just fine. But as I created it, it didn't look all that great so I recreated the page and used a bootstrap table because of the much cleaner look. Both pages have the exact same PHP code, the only difference is the bootstrap table I added in. So I'm really at a loss with this. Other than the table and PHP code, there is a script at the bottom of the page for the table itself to allow for searching within the table, i'll include that as well. The PHP code is as follows:
<?php //THE "p_id" IS BROUGHT OVER FROM THE EDIT BUTTON ON VIEW_ALL_USERS if (isset($_GET['p_id'])) { $the_user_id = $_GET['p_id']; } // QUERY TO PULL THE SITE INFORMATION FROM THE p_id THAT WAS PULLED OVER $query = "SELECT * FROM users WHERE user_id = $the_user_id "; $select_user = mysqli_query($connection,$query); //SET VALUES FROM ARRAY TO VARIABLES while($row = mysqli_fetch_assoc($select_user)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_email = $row['user_email']; $user_phone = $row['user_phone']; //$user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id']; } THE UPDATE QUERY CODE....................................................................................................................
<?php if(isset($_POST['update_user'])) { $user_id = $_POST['user_id']; $user_firstname = $_POST['user_firstname']; $user_lastname = $_POST['user_lastname']; $username = $_POST['username']; $user_email = $_POST['user_email']; $user_phone = $_POST['user_phone']; //$user_image = $_POST['user_image']; $user_title_id = $_POST['user_title_id']; $user_role_id = $_POST['user_role_id'];
$query = "UPDATE users SET "; $query .= "user_id = '{$user_id}', "; $query .= "user_firstname = '{$user_firstname}', "; $query .= "user_lastname = '{$user_lastname}', "; $query .= "username = '{$username}', "; $query .= "user_email = '{$user_email}', "; $query .= "user_phone = '{$user_phone}', "; //$query .= "user_image = '{$user_image}', "; $query .= "user_title_id = '{$user_title_id}', "; $query .= "user_role_id = '{$user_role_id}' "; $query .= "WHERE user_id = '{$the_user_id}' "; $update_user = mysqli_query($connection,$query); if(! $update_user) { die("QUERY FAILED" . mysqli_error($connection)); } } ?> THE "UPDATE USER" BUTTON THE USER CLICKS ON TO UPDATE....................................................................................................................
<div class="col-1"> <button class="btn btn-primary" type="submit" name="update_user">Update User</button> </div>
Any Help is Greatly Appreciated! Edited March 23 by ZserenePlease delete. |