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PHP - Get Album Id, Upon Query With Browser Userid To Match Table User_id
There is something wrong with my statement it is not returning sql queries results based on matching the url based based userid to match with the table user_id and get the Ablum id and Album name: Please hlep: Code: [Select] if(isset($_GET['userid'])) { $db =& JFactory::getDBO(); $user =& JFactory::getUser(); $id = $user->id; $album_id = $_GET['userid']; $query = 'SELECT user_id, id, format_id, year, name FROM #__muscol_albums WHERE user_id = ' . $album_id; //$query = 'SELECT user_id FROM #__muscol_albums WHERE id = ' . $album_id ; $result = mysql_query($query) or die('Error, No Album Search failed'); list($name, $user_id, $id, $year) = mysql_fetch_array($result); echo $id; //exit; } Similar Tutorialsi do think is a small mistake i making can u please have a look if u can ok here's the problem i am trying to delete an album within the album should also delete the photos related to that album this what i tried gives this error Delete image failed. You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 3 my tables are table albums fields album_id, album_name, album_owner, sub_album table photos fields, photo_id, photo_name, photo_extension, photo_proper photo_owner, photo_date, photo_comments, photo_size, album_id photo_proper is the name image stored in folder <?php define('ROOT_DIR', './'); define('PROPER', TRUE); /** * include common files */ include_once(ROOT_DIR. 'includes/common.inc.php'); // No album id has been selected if (isset($_GET['albums'])) // get the album name since we need to display // a message that album 'foo' is deleted $result = mysql_query("SELECT album_id, album_name, album_owner, sub_album FROM albums WHERE album_id = $album_id") or die('Delete image failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); $album_id = $row['album_id']; $album_name = $row['album_name']; // get the image filenames first so we can delete them // from the server $result = mysql_query("SELECT photo_id, photo_id FROM photos WHERE album_id = $album_id") or die(mysql_error()); while ($row = mysql_fetch_assoc($result)) { define("GALLERY_IMG_DIR", "./photos/"); unlink(GALLERY_IMG_DIR . $row['photo_proper']); unlink(GALLERY_IMG_DIR . 'thumbs/' . $row['photo_proper']); } $result = mysql_query("DELETE FROM photos WHERE album_id = $album_id") or die('Delete image failed. ' . mysql_error()); $result = mysql_query("DELETE FROM album WHERE album_id = $album_id") or die('Delete album failed. ' . mysql_error()); // album deleted successfully, let the user know about it echo "<p align=center>Album '$album_name' deleted.</p>"; } else { echo "<p align=center>Cannot delete a non-existent album.</p>"; } ?> try 2 error line 5 <?php define('ROOT_DIR', './'); define('PROPER', TRUE); /** * include common files */ include_once(ROOT_DIR. 'includes/common.inc.php'); // No album id has been selected if (isset($_GET['albums'])) { // get the image file name so we // can delete it from the server $sql = "SELECT album_id, album_name, album_owner, sub_album FROM albums WHERE album_id = {$_GET['albums']}"; $result = mysql_query($sql) or die('Delete photo failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); // get the image filenames first so we can delete them // from the server $sql = "SELECT photo_id, photo_proper FROM photos WHERE photo_id = {$_GET['photos']}"; $result = mysql_query($sql) or die('Delete photo failed. ' . mysql_error()); if (mysql_num_rows($result) == 1) { $row = mysql_fetch_assoc($result); define("GALLERY_IMG_DIR", "./photos/"); // remove the image and the thumbnail from the server unlink(GALLERY_IMG_DIR . $row['photo_proper']); unlink(GALLERY_IMG_DIR . 'thumbs/' . $row['photo_proper']); // and then remove the database entry $sql = "DELETE FROM photos WHERE photo_id = {$_GET['photos']}"; $result = mysql_query("DELETE FROM album WHERE album_id = $album_id") or die('Delete album failed. ' . mysql_error()); // album deleted successfully, let the user know about it echo "<p align=center>Album '$album_name' deleted.</p>"; } else { echo "<p align=center>Cannot delete a non-existent album.</p>"; } } } ?> Hey Guys, A very noob question...Here it goes: I want to update the field user_id on my database table, there a lot of rows that the user_id is the same... "testID", I just want to update only ONCE. mysql_query("UPDATE `users` SET status = '$StatusCheck' WHERE user_id = 'testID'"); Right now it's updating ALL the rows that have the user_id "testID". I just want to update only one. Any ideas? Thanks in advance! Cheers! I need help making $_SESSION['code'] match variable $code when executed. They're both accessing rand() but with different results. Code: [Select] <?php function create_user($params) { db_connect_posts(); $code = rand(11111111,99999999); $_SESSION['code'] = $code; $query = sprintf("INSERT INTO users SET users.screen_name = '%s', users.user_email = '%s', users.user_pwd = '%s', users.image = '%s', created_at = NOW(), users.code = $code, users.active = '0'" , mysql_real_escape_string($params['screen_name']), mysql_real_escape_string($params['user_email']), md5($params['user_pwd']), mysql_real_escape_string($params['image']) ); $result = mysql_query($query); if(!$result) { return false; } else { return true; } } ?> Hi All,
I want to copy into a table values from another table that partially match a given value, case-insensitively. So far I do as follows but I wonder whether there is a quicker way.
$input_table=array('1'=>'toto','2'=>'tota','3'=>'hello','4'=>'TOTO','5'=>'toto');
$input_table_2 = array_map('strtolower', $input_table);
$value_to_look_for='Tot';
$value_to_look_for_2=strtolower($value_to_look_for);
$output_table=array();
foreach ($input_table_2 as $k=>$v)
{
if(false !== strpos($v, $value_to_look_for_2))
{
$output_table[]=$input_table[$k];
}
}
One drawback is that $input_table_2 is 'foreached' whereas there might be no occurrences, which would lead to a loss of time/resources for big arrays.
Thanks.
CREATE TABLE posts ( postId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, title VARCHAR(255) NOT NULL, author VARCHAR(24) NOT NULL, description TEXT NOT NULL, createdAt TIMESTAMP, PRIMARY KEY (postId) ); CREATE TABLE comments( commentId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, comment TEXT NOT NULL, postId INT(11), userId INT(11), createdAt TIMESTAMP, PRIMARY KEY (commentId), FOREIGN KEY (userId) REFERENCES users(userId), FOREIGN KEY (postId) REFERENCES posts(postId) ); CREATE TABLE replies ( repId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, reply TEXT NOT NULL, userId INT(11), commentId INT(11), createdAt TIMESTAMP, PRIMARY KEY (repId), FOREIGN KEY (userId) REFERENCES users(userId), FOREIGN KEY (commentId) REFERENCES comments(commentId) ); CREATE TABLE users ( userId INT(11) NOT NULL UNIQUE AUTO_INCREMENT, userName VARCHAR(100) NOT NULL,, email VARCHAR(100) NOT NULL, PRIMARY KEY (userId) ); how to retrive userName,comment, and createdAt from users and comments table while I have used userId as a Foreign key on the comment table if it isn't correct, correct me please This topic has been moved to MySQL Help. http://www.phpfreaks.com/forums/index.php?topic=317758.0 Although I am deciding on the best route to go on this, I was hoping for any feedback or ideas on the best and simplest approach to the below problem: Here is my table structu old_code_1 old_code_2 new_code_1 new_code_2 prod_code_1 prod_code_2 Each of the following fields has 5 digits e.g. 00045, 12654, etc. What I need to achieve is this: Concatenate each set (of the 3 sets above) data into one variable e.g. old_code_1 + old_code_2 = 0004512654. So old_code_1 & old_code_2 would become a combined data and the same for the othe 2 sets. Then. Once concatenated, I need to compa prod_code_1+prod_code_2 with old_code_1+old_code_2 and replace prod_code_1+prod_code_2 with new_code_1+new_code_2 where matches are found. The new_code_1 & 2 are in the same row as the old_code 1 & 2. So the data is in alignment that way for comparison. Thanks for any insight or suggestions on how to make this happen in the simplest form possible. In the PHP script I'm using, in the Upload Form the user selects an image to Upload, the Form renames it like so:
$allowedExts = array("gif", "jpeg", "jpg", "pdf", "png");
$temp = explode(".", $_FILES["file"]["name"]);
$extension = strtolower( end($temp) );
if (!in_array($extension,$allowedExts))
{
echo ("Error - Invalid File Name");
}
$length = 20;
$randomString = (time());
$thumbnail = $randomString . "." . $extension;
The random string works successfully, but I'd like to add the user_id to the beginning of it and a dash, like this:
user_id -
So, the new file name would be something like: user_id-randomString.extension
Can you please help me add that?
Hello, I've been racking my brains (and spending sleepless nights) trying to get a login system to work by where the member will insert their email address as [username] and password (already stored in the DB) - then the page to divert to an administration panel with their User_id for them to only edit their information. The Code I have so far..... The login_form.php Code: [Select] <?php //Start session session_start(); //Unset the variables stored in session unset($_SESSION['SESS_CLIENT_EMAIL']); unset($_SESSION['SESS_MAIN_ID']); ?> <!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> <meta http-equiv="content-type" content="text/html; charset=utf-8" /> <title>Client Admin Panel</title> <link href="style.css" rel="stylesheet" type="text/css" media="screen" /> </head> <body> <div id="wrapper"> <div id="header"> <h1>CLIENT LOGIN</h1> <h2>CLIENT ADMINISTRATION PANEL</h2> version 2.10 </div> <div id="menu"> </div> <div id="content"> <div id="right"> <div class="post"> <h2>CLIENT ADMINISTRATION PANEL - CLIENT LOGIN</h2><br /> <h3><span class="err"><strong><font color="#800000">PLEASE LOGIN</font></strong></span></h3><form id="loginForm" name="loginForm" method="post" action="login-exec.php"> <table width="315" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td width="150"><b>Email Address:</b></td> <td width="157"><input name="login" type="text" class="textfield" id="client_email" /></td> </tr> <tr> <td><b>Secret Word:</b></td> <td><input name="password" type="password" class="textfield" id="client_password" /></td> </tr> <tr bgcolor='#f1f1f1'> <td> </td> <td><input type="submit" name="Submit" value="Login" /></td> </tr> <tr> <td colspan="2"><hr /></td> </tr> <tr> <td><b>Forgot SecretWord?:</b></td> <td><font face='tahoma, arial, helvetica' size='2' ><a href='forgot-password.php'>Click Here</a></font></td> </tr> <tr> <td colspan="2"><hr /></td> </tr> <tr> <td><b>New Client?:</b></td> <td><font face='tahoma, arial, helvetica' size='2' ><a href='../dhsite/webpages/reg_1.php'> Register Here</a></font></td> </tr> </table> <br /> </form></p> </div> </div> </div> <div id="footer"> <p class="copyright">Copyright © *****************</p> </div> </div> </body> </html> And the handler: login_exec.php Code: [Select] <?php //Start session session_start(); $_SESSION['var'] = $val; //Include database connection details require_once('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 $client_email = clean($_POST['login']); $client_password = clean($_POST['password']); //Input Validations if($client_email == '') { $errmsg_arr[] = 'Email Address missing'; $errflag = true; } if($client_password == '') { $errmsg_arr[] = 'Password missing'; $errflag = true; } //If there are input validations, redirect back to the login form if($errflag) { $_SESSION['ERRMSG_ARR'] = $errmsg_arr; session_write_close(); header("location: login-form.php"); exit(); } //Create query $qry="SELECT client_email, client_password, main_id FROM users WHERE client_email='$client_email' AND client_password='$client_password'"; $result=mysql_query($qry); //Check whether the query was successful or not if($result) { if(mysql_num_rows($result) == 1) { //Login Successful session_regenerate_id(); $member = mysql_fetch_assoc($result); $_SESSION['SESS_CLIENT_EMAIL'] = $member['client_email']; $_SESSION['SESS_MAIN_ID'] = $member['main_id']; session_write_close(); header("Location: test_admin_panel.php?user_id=".$main_id.""); exit(); }else { //Login failed header("location: login-failed.php"); exit(); } }else { die("Query failed"); } ?> Any help would be VERY much appreciated!! Hi guys, why am i getting this error: Illegal string offset 'user_id' but when echo $value it brings the correct output. Thanks
$user_id = 5;
$user_name = "obodo";
$_SESSION['test'] = array('user_id' => $user_id, 'user_name' => $user_name);
foreach( $_SESSION['test'] as $value )
{
echo $value['user_id']; //give error
/* echo $value //works */
}
Hello all...fairly new to this php/mysql thing... working on my final project thats due in about 24 hours... and i hit a rut... im making a pretty basic, online classifieds site. users can sign up, login, post new listings and view others listings by clicking on different categories. the problem i am having right now is this...When the user clicks on "My listings" i need it to pull only the listings that were created by that users user_id, which is the primary key in my user_info table...my professor suggested storing it in hidden field through the login submit button...very confused and frustrated... any help is much appreciated... here is my code: Code: [Select] $query2 = "SELECT * FROM members_copy WHERE RSUSER = ".$_SESSION["rsUser"].""; $result2 = mysql_query($query2); $UserID = $result2['USERID']; echo $UserID; When I echo out UserID it doesn't show a number?! Please help?! Hi there, I must say this forum looks great and its good to be here. Now I'll get straight to the point. I am working on a website where supplier and a member should be able to login with a single UserID. That is, a member can be a supplier at the same time (logged in with the same ID) and if he is the supplier, he should be given a link taking him to the supplier admin area. If he is just a member, that supplier link will not be there. Having a separate ID for member and supplier is not what is required. If the member is a supplier there will be a supplier admin link. Looking forward to hearing from you guys! Any inputs/feedback is always welcome. Cheers! Sid. Hey,
I'm really new to PHP and having some difficulties with $_SESSION and getting userid from the database. I've managed to put content to my database and also a login script. Though, adding sessions has been a pain. Here's what I got so far:
$sql = "SELECT username, password FROM users WHERE username = '$username' and password = '$pas'";
$query_login = $db->prepare($sql);
$query_login->execute(array('userid' => $userid, 'username' => $username, 'password' => $pas));
$result = $query_login->rowcount();
if ($result>0)
{
session_start();
$_SESSION['username'] = $username;
$_SESSION['logged'] = 1;
$_SESSION['userid'] = $result['userid'];
header('Location: ../user/user.php');
}
When I try to compile the following code, I receive a warning: Notice: Indifined index: userid $letters = array_merge(range("A", "Z"), array("&#198", "&#216", "&#197")); mysql_select_db("okern", $link_id); $userid = $_POST['userid']; echo "$userid"; foreach($letters as $letter) { if(isset($_POST[$letter])) { $product = $letter . '1'; $antall = $letter . '2'; $enhet = $letter . '3'; $melding = $_POST['melding']; $query = "INSERT INTO orders VALUES ('farhad', '$product', '$antall', '2011-04-18', '$melding')"; $result = mysql_query($query) or die(mysql_error()); } } When I try to compile my php file in phpDesigner I accept Notice. But, when I run the same php file from browser echo "$userid"; code workes. And I see the userid transferred from another php file on the screen. So I do not understand why I get this notice in compilation process. I am creating a football predictor game and have a table in the database to hod all my users i have written a query to pull all out the users that have made a prediction that day with their predictions made i have put this into an array so my array will hold all the user details 1 being their userId I have set the order by on the query to userId so I get all the same users predictions in 1 block. I now want to extract all the predictions for each user and put these into their own array but I don't know which users the array will hold, how many users their will be, or how many predictions each user would have made so i'm not sure how to create this array i'm quite new to php can anyone help me with this This topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=326412.0 Hello, I was not sure on which board to put this thread, so please move it if there is a more appropriate board. I am wanting to create a website that displays images which I have put into a database. I would like the number of images on one page not to exceed 10, for example, and when the number of images in the database exceeds this number, a new page will be created. I would like it to function similar to Google Images where the number of pages is dictated by the number of images in the database. I cant figure out how to achieve this, so any pointers or thoughts would be great Thanks, Matlab |
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