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PHP - Query Function Not Working
Hello I have a general php insert function that I want to use to insert data into tables regulary, the problem is, each time I use it it creates a new row in the table instead of using certain columns to add information to, here is the function, can you tell me why does it insert a new row each time I use it?
class Query{
function add($table,$cols,$vals){
include 'conx.php';
$query = "
INSERT INTO $table($cols) VALUES('$vals');
";
$res = mysqli_query($con,$query) ;
if($res){
echo "Inserted value";
}else{
echo "Did not insert value";
}
}
}
Similar TutorialsI've followed the PHP Freaks pagination tutorial which you can find here. And I also got it to work, the only problem is that the script won't work when I use the query outside of the function. Here's the function: <?php function knuffix_list ($query, $dbc) { // find out how many rows are in the table: $query_row = "SELECT COUNT(*) FROM con"; $query_run = mysqli_query ($dbc, $query_row); $row = mysqli_fetch_row($query_run); $num_rows = $row[0]; // number of rows to show per page $rows_per_page = 5; // find total pages -> ceil for rounding up $total_pages = ceil($num_rows / $rows_per_page); // get the current page or set a default if (isset($_GET['current_page']) && is_numeric($_GET['current_page'])) { // make it an INT if it isn't $current_page = (int) $_GET['current_page']; } else { // default page number $current_page = 1; } // if current page is greater than total pages then set current page to last page if ($current_page > $total_pages) { $current_page = $total_pages; } // if current page is less than first page then set current page to first page if ($current_page < 1) { $current_page = 1; } // the offset of the list, based on current page $offset = (($current_page - 1) * $rows_per_page); echo "test " . $query; // SCRIPT ONLY WORKS IF I INSERT QUERY HERE $query = "SELECT * FROM con, user WHERE con.user_id = user.user_id ORDER BY contributed_date DESC LIMIT $offset, $rows_per_page"; $data = mysqli_query ($dbc, $query) or die (mysqli_error ($dbc)); // Loop through the array of data while ($row = mysqli_fetch_array ($data)) { global $array; // Variables for the table $con_id = $row['con_id']; $likes_count = $row['likes']; $dislikes_count = $row['dislikes']; $dbuser_name = $row['nickname']; $dbuser_avatar = $row['avatar']; $user_id = $row['user_id']; // The TABLE echo "<table padding='0' margin='0' class='knuffixTable'>"; echo "<tr><td width='65px' height='64px' class='avatar_bg' rowspan='2' colpan='2'><img src='avatar/$dbuser_avatar' alt='avatar' /></td>"; echo "<td class='knuffix_username'><strong><a href='profile.php?user=$dbuser_name' title='Profile of $dbuser_name'>" . $dbuser_name . "</a> ___ " . $user_id . "____ <form action='' method='POST'><button type='submit' name='favorite' value='fav'>Favorite</button></form>"; echo "</strong><br />" . $row['category'] . " | " . date('M d, Y', strtotime($row['contributed_date'])) . "</td></tr><tr><td>"; echo "<form action='' method='post'> <button class='LikeButton' type='submit' name='likes' value='+1'>Likes</button> <button class='DislikeButton' type='submit' name='dislikes' value='-1'>Dislikes</button> <input type='hidden' name='hidden_con_id' value='" . $con_id . "' /> </form></td><td class='votes'>Y[" . $likes_count . "] | N[" . $dislikes_count . "]</td></tr>"; echo "<tr><td class='knuffix_name' colspan='3'><strong>" . htmlentities($row['name']) . "</strong><br /></td></tr>"; echo "<tr><td colspan='2' class='knuffix_contribution'><pre>" . $row['contribution'] . "</pre><br /></td></tr>"; echo "</table>"; // POST BUTTONS inside the table $likes = $_POST['likes']; $dislikes = $_POST['dislikes']; $con_id = $_POST['hidden_con_id']; $favorite = $_POST['favorite']; $array = array ($likes, $dislikes, $con_id, $user_id, $favorite); } /********* build the pagination links *********/ // BACKWARD // if not on page 1, show back links and show << link to go back to the very first page if ($current_page > 1) { echo " <a href='{$_SERVER['PHP_SELF']}?current_page=1'><<</a> "; // get previous page number and show < link to go to previous $prev_page = $current_page - 1; echo " <a href='{$_SERVER['PHP_SELF']}?current_page=$prev_page'><</a> "; } // CURRENT // range of number of links to show $range = 3; // loop to show links in the range of pages around current page for ($x = ($current_page - $range); $x < (($current_page + $range) + 1); $x++) { // if it's a valid page number... if (($x > 0) && ($x <= $total_pages)) { // if we're on current page if ($x == $current_page) { // highlight it but don't make a link out of it echo "[<b>$x</b>]"; // if it's not the current page then make it a link } else { echo "<a href='{$_SERVER['PHP_SELF']}?current_page=$x'>$x</a>"; } } } // FORWARD // if not on the last page, show forward and last page links if ($current_page != $total_pages) { // get next page $next_page = $current_page + 1; // echo forward link for next page echo " <a href='{$_SERVER['PHP_SELF']}?current_page=$next_page'>></a> "; // echo forward link for last page echo " <a href='{$_SERVER['PHP_SELF']}?current_page=$total_pages'>>></a> "; } /***** end building pagination links ****/ mysqli_close($dbc); } ?> As you can see above, the script will only work if I put the query right below the $offset variable and above the $data variable. I put the test echo above the query to see how the query looks like when I induce the query from the outside through the function parenthesis into the function, and this is what I get printed out: test SELECT * FROM con, user WHERE con.user_id = user.user_id ORDER BY contributed_date DESC LIMIT , Obviously the $offset and the $rows_per_page variables are not set, when I induce the query from the outside into the function. So in that sense my question is: How can I induce the query from the outside into the function SO THAT the $offset and the $rows_per_page variables are set as well and NOT empty? p.s. I need the query outside of the function because I'm using a sort by category functionality. I have a script that runs at the end of a batch process specifically built to clean out empty records. How can I make it a function and then call the function. Here is my query; Code: [Select] $sql="DELETE FROM ".$dbname." . property WHERE property . id=''"; mysql_query($sql) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $sql . "<br />\nError: (" . mysql_errno() . ") " . mysql_error()); $sql="DELETE FROM ".$dbname." . agents WHERE agents . ip_source=''"; mysql_query($sql) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $sql . "<br />\nError: (" . mysql_errno() . ") " . mysql_error()); $sql="DELETE FROM ".$dbname." . settings WHERE settings . prop_id=''"; mysql_query($sql) or die("<b>A fatal MySQL error occured</b>.\n<br />Query: " . $sql . "<br />\nError: (" . mysql_errno() . ") " . mysql_error()); I would like this to be in a function that i could specifically call the function. Why? Well because I have 5 different sets of queries like this set above and I'm going to build a conditional. So like if condition 1 then run use this function, condition 2, use function ??? etc and so on. Hey guys I am stuck on how to best create a php function to do the following.
I need to echo out 4 restaurants. I have a database of numerous restaurant categories: breakfast, lunch, dinner, and coffee. In this database some categories have more results than others.
Basically I want to randomly choose results from each category query, Breakfast, Lunch, and dinner to create 4 echoed results. (I understand that one category will have an extra result as there are 3 categories and 4 echoed results.) I do not want one category of restaurant to bump out another category because it has more results. Also, if there are not enough or no results from one category, I want extra results from the other queries to make up for it so that there are always 4 echoed items.
Finally, if there are not enough results from breakfast, lunch, dinner to make 4 echoed items, I would like Coffee to fill the 4 required results.
Randomly echo 4 results from the below primary queries:
Primary Query: Breakfast Restraunts Query
Primary Query: Lunch Restraunts Query
Primary Query: Dinner Restraunts Query
Get results from secondary query if Primary queries can’t echo 4 results:
Secondary Query: Coffee Shops Query
I hope this makes sense. I understand how to do the mysql queries, I just don't know how to do the php function to echo the results as required. Any help would be appreciated!
Edited by jason360, 09 January 2015 - 06:53 PM. Can someone review this piece of code and advise what's missing that's preventing the SQL query from working? Issue: no results being echoed out from $results. Note: I confirmed the db connection established (only the query not working - no results pulled from db and displayed via echo statements). $db = mysql_connect($host,$user,$pw) or die("Cannot connect to MySQL."); mysql_select_db($database,$db) or die("Cannot connect to database."); echo "Success! Connected to database ".$database."<br /><br />"; // $jsearch is variable sent in from form "title" name $jsearch = $_POST['jobsearch']; // I confirmed that all the field names are correct //-query the database table $sql="SELECT id,title,details,status FROM jobs WHERE title = '$jsearch';"; //-run the query against the mysql query function $result=mysql_query($sql) or die("<br>Query string: $result<br>Returned error: " . mysql_error() ); //-create while loop and loop through result set while($row=mysql_fetch_array($result)){ $title =$row['title']; echo title."<br />"; $details=$row['details']; echo $details."<br />"; $status=$row['status']; echo $status."<br />"; $ID=$row['id']; I am using a query to update the db when a user logs in. It sets every value correctly but it doesnt set the colum u_online which is INT(1). $link->query("UPDATE ".TBL_PREFIX."users SET u_last_login = '".$config['time_now']."', u_online = 1, u_ip = '".$_SERVER['REMOTE_ADDR']."', u_activity_time = '".$config['time_now']."', u_points = u_points + '$login_points' WHERE u_username = '$username'") or die(print_link_error()); when i echo the query: UPDATE asf_users SET u_last_login = '1300310822', u_online = 1, u_ip = '127.0.0.1', u_activity_time = '1300310822', u_points = u_points + '1' WHERE u_username = 'doddsey65' which works because i tried it in phpmyadmin and it changed the column. But it doesnt work in the actual script. It changes every other value that its supposed to but the u_online. Anyone have any ideas why? Im using the following query to pull all topics that have a new reply since the users last visit and where the user has posted within them. Code: [Select] $topic_query = $link->query("SELECT t.*, u.*, p.* FROM ".TBL_PREFIX."topics as t LEFT JOIN ".TBL_PREFIX."users as u ON (u.u_username = t.t_poster) LEFT JOIN ".TBL_PREFIX."posts as p ON (t.t_tid = p.p_tid) WHERE t.t_last_post_time > u.u_activity_time AND p.p_poster = '".$user->user_name."' GROUP BY t.t_tid ORDER BY t.t_sticky DESC, t.t_time_posted DESC LIMIT $start_page, $topics_to_show") or die(print_link_error()); for some reason though it still shows two topics that have a last post time which is less than that of the users activity time. The last_post_time for the topic is 1300309160 compare that with the users last activity time 1300784679 as you can see the last activity time is more than the last post time. there are no records in the database that are more than the last activity time so i shouldnt be getting any results. Anyone have any ideas? Evening everybody, hoping someone can see my error here! $sql="INSERT INTO transactions (t_ref, m_affid, m_name, order_val, order_date, order_status, u_id, u_comm) VALUES('$t_ref', '$m_affid', '$m_name', '$order_val', '$order_date', '$order_status', '$u_id', '$u_comm')"; if (!mysql_query($sql)) { mysql_error(); } else { $newq = mysql_query("SELECT * FROM userbalance WHERE u_id = '$u_id'"); echo mysql_error(); $row = mysql_fetch_assoc($newq); $cpending = $row['pending']; $npending = $cpending + $u_comm; $updatepending = "INSERT into userbalance (pending) VALUES ('$npending') WHERE u_id = '$u_id'"; if (!mysql_query($updatepending)) { mysql_error(); } } Basically the 1st query works and inserts the data correctly, however once it gets to $newq = mysql_query("SELECT * FROM userbalance WHERE u_id = '$u_id'"); this query does not insert anything, no errors. ok Im building a search page. The user can enter a name, county, category, address into my search field. The results page will show the details based on that. It works great so here is a snippet of the working code. $result = "-1"; if (isset($_POST['searchField'])) { $result = $_POST['searchField']; } $result = sprintf("SELECT Clubs.clubID, Clubs.name, Clubs.county, Clubs.logo, Clubs.postcode, Clubs.intro, Clubs.thumbsup, Clubs.cat, Category.*, County.*FROM Clubs INNER JOIN Category ON Clubs.cat = Category.catID INNER JOIN County ON Clubs.county = County.countyID WHERE ((Clubs.name Like %s) OR (Clubs.cat LIKE %s) OR (County.county LIKE %s) OR (Clubs.area LIKE %s) OR (Clubs.postcode LIKE %s) OR (Category.categorys LIKE %s)) " , String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text")); This code works great no matter what I enter. the problem Im having is I only want to show results based on the logged in users county. I have a variable stored under $user['county'] which is the ID found in the County table under countyID I have tried writing the code so many different ways. Here is my latest attemped but it doesnt work. It breaks the whole query and nothing is displayed. $result = "-1"; if (isset($_POST['searchField'])) { $result = $_POST['searchField']; } $County= "-1"; if (isset($user['county'])) { $County= $user['county']; } $result = sprintf("SELECT Clubs.clubID, Clubs.name, Clubs.county, Clubs.logo, Clubs.postcode, Clubs.intro, Clubs.thumbsup, Clubs.cat, Category.*, County.* FROM Clubs INNER JOIN Category ON Clubs.cat = Category.catID INNER JOIN County ON Clubs.county = County.countyID WHERE ((Clubs.name Like %s) OR (Clubs.cat LIKE %s) OR (County.county LIKE %s) OR (Clubs.area LIKE %s) OR (Clubs.postcode LIKE %s) OR (Category.categorys LIKE %s)) AND County.countyID = %s" , String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text"), String($result . "%", "text") String($County. "%", "int")); Thanks Danny Hey, I am coding this forum and the following PDO prepare query calls the topic details. The prepare query works perfectly well, however when I add it into a function within another class and then call it, the script does not seem to work. I get errors like the following: Fatal error: Call to a member function fetch() on a non-object in C:\wamp\www\new\sources\forum\new.php on line 117 The prepare query fails to excute, this is because I do not know how to add it into a function. This is what I tried: Code: [Select] class FUNC { function userDetails($table, $column, $cVaulue, $oBy, $ASDSC) { global $dbh; $sth = $dbh->prepare('SELECT * FROM `'.$table.'` WHERE `'.$column.'` = '.$cVaulue.' ORDER BY `'.$oBy.'` '.$ASDSC.''); $sth->bindParam(':id', $id, PDO::PARAM_INT); $sth->execute(); } } The FUNC class is called as the $load variable, and in my forum class I have decalred it in my global so the variable can be passed: Code: [Select] $id = $forum_data['id']; $load->userDetails('db_topics', 't_forum_id', ':id', 't_whenlast', 'DESC'); $coll=$sth->fetch(PDO::FETCH_ASSOC); $this->html->RightForum($coll); The code is very silly, but the fact is I did not know how to do this.. Please help me put this query into a function because I need to make use of it in many other parts of the forum and I don't want to constantly declare this query. Hi, Doing a Query SELECT * FROM table WHERE field BETWEEN low_number AND high_number If i have three rows with three numbers: IE: 27, 50, 80. and i run the query above, it returns a result of: 27 and 50. How can i get it to return a result of: 27,50, and 80.?? WHY: age range search: i want to return an age range of people = and between the age of say: 20 and 70 but i want to include the 20 and 70. How can i do this? Hiya peeps! I have this code; $q3 = "SELECT `data` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $q4 = "SELECT `data_2` FROM `firm_text_strings` WHERE `id` = '100' AND `languages_id` = 'EN'"; $r3 = mysql_query($q3); $r4 = mysql_query($q4); if($r3 && $r4) { while($w3 = mysql_fetch_assoc($r3) && $w4 = mysql_fetch_assoc($r4)) { $resultStr = $w3['data'] . '<br />' . $w4['data_2']; } } else { $resultStr = 'error'; } everything is working up to the while function, it just wont do the loop for some reason. Anyone have any ideas. Many thanks, James. Hello all, I am using the Lynda.com PHP course to further learn php. On one of the examples, they are using a SQL Query that works for them but is erroring out for me. It is the query below for the $page_set variable, particularly the use of the WHERE subject_id = {$subject["id"]}". Code: [Select] <?php require_once ('includes/conf.php'); ?> <?php include ('includes/header.php'); ?> <?php require_once ('includes/functions.php'); ?> <table id="stucture"> <tr> <td id="navigation"> <ul class="subjects"> <?php $subject_set = mysql_query("SELECT * FROM subjects", $connection); if (!$subject_set) { die ("Database query failed: " . mysql_error()); } while ($subject = mysql_fetch_array($subject_set)) { echo "<li>{$subject["menu_name"]}</li>"; } $page_set = mysql_query("SELECT * FROM pages WHERE subject_id = {$subject["id"]}"); // This appears to be the problem area, unless it is fooling me. It works with this exact statement in their code. if (!$page_set) { die ("Database query failed: " . mysql_error()); } echo "<ul class='pages'>"; while ($page = mysql_fetch_array($page_set)) { echo "<li>{$page['menu_name']}</li>"; } echo "</ul>"; ?> </ul> </td> <td id="page"><h2>Content Area</h2> </td> </tr> </table> <?php require ('includes/footer.php'); ?> The error listed is this: Database query 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 1 Thanks, you all are the best! BtD Hi there, I have this query, which outputs the records properly in PhpMyAdmin. But when using this query inside my PHP code, the result is incomplete. Instead of retrieving the expected 15 results, I only get the first one twice. Also, I don't get any errors. Code: (php) [Select] <?php $get_files = "SELECT `id` FROM `files` WHERE `category` = 'W';"; include_once('connection.inc.php'); $con = mysql_connect(MYSQL_SERVER, MYSQL_USERNAME, MYSQL_PASSWORD) or die ("Connection failed: " . mysql_error()); $dbname = "myDB"; $select_db = mysql_select_db($dbname, $con) or die ("Can't open database: " . mysql_error()); $result_files = mysql_query($get_files) or die ("Query failed: " . mysql_error()); $files = mysql_fetch_array($result_files); $W_files = "'".implode("', '", $files)."'"; // echo $W_files; outputs the first number twice, instead of the expected 15 different ones $get_checklist = "SELECT * FROM `checklist` WHERE `id` IN ($W_files) ORDER BY `id` ASC;"; $result_checklist = mysql_query($get_checklist) or die ("Query failed: " . mysql_error()); // .... code for displaying ?> Hi, i am currently designing a new website and am testing a few different compnents of it. One part gathers two variables of the url using the GET method and selects the relevant data from the database. It then performs an UPDATE of a field called balance before INSERTING a new record. Everything works and gets inserted except for the UPDATE. Here is the code Code: [Select] <?php include ("connect.php"); $id = ""; $offerid = ""; $id = $_GET['user_id']; $offerid = $_GET['offer_id']; $sql_user = mysql_query("SELECT * FROM user_info WHERE id = $id"); $sql_offer = mysql_query("SELECT * FROM offers WHERE offerid = $offerid"); $balance += $payout; mysql_query("UPDATE user_info SET balance = $balance WHERE id = $id"); mysql_query ("INSERT INTO offer_credited (id, offerid, date_credited, payout_to_user) VALUES ('$id', '$offerid', CURDATE(), '$payout')") or die(mysql_error()); echo "Your Offer has been recorded and Updated" ?> The balance field is a float that i want to increase by the amount of the payout variable which is from the offers table. However, the balance never gets updated and remains at 0. Any ideas what it could be, i am thinking it is something to do with, Code: [Select] $balance += $payout; But i can't think of what else to do. Hopefully this isn't too confusing. Thanks for the help I would like to edit a record that I already have in my database but my query is not working for some reason. Hope someone can help. The errors i get are Notice: Undefined variable: first_name in C:\chemicaluser\www\editinfo.php on line 40 Notice: Undefined variable: last_name in C:\chemicaluser\www\editinfo.php on line 40 Notice: Undefined variable: log_number in C:\chemicaluser\www\editinfo.php on line 40 .... line 40 = my query $query = "UPDATE table1 SET first_name = '$first_name', last_name = '$last_name' WHERE log_number = '$log_number' "; Very simple database, 1 table w/ 2 fields table name = table1 fields = first_name, last_name my php file Code: [Select] <?php require ('/menu.php'); require_once('sqlconnect.php'); if (isset($_GET['log_number']) && isset($_GET['first_name']) && isset($_GET['last_name']) ) { $log_number = $_GET['log_number']; $first_name = $_GET['first_name']; $last_name = $_GET['last_name']; } else if (isset($_POST['log_number']) && isset($_POST['first_name']) && isset($_POST['last_name']) ) { $log_number = $_POST['log_number']; $first_name = $_POST['first_name']; $last_name = $_POST['last_name']; } if (isset($_POST['submit'])) { $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); $query = "UPDATE table1 SET first_name = '$first_name', last_name = '$last_name' WHERE log_number = '$log_number' "; $result = mysqli_query($dbc, $query) or die ('Error querying database'); mysqli_close($dbc); } ?> my form Code: [Select] <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <label for="first_name">First name</label> <input type="text" id="first_name" name="first_name" value="<?php if (!empty($first_name)) echo $first_name; ?>"/> <label for="last_name">Last Name</label> <input type="text" id="last_name" name="last_name" value="<?php if (!empty($last_name)) echo $last_name; ?>" /> <input type="submit" value="Edit Information" name="submit" /> </form> I'm having a problem with adding icon/avatar to my database..this is a submit form where I just submit the avatar url. I don't know how to change the submit link from avatar_add.php?mode=submit to index.php?mode=submit&x=avatar_add. http://www.graphics.allmerk-noplay.com/index.php?x=avatar_add Code: [Select] <?php $host =""; // host name $user =""; // user $pass =""; // password $dbase =""; // database name // connect to the server mysql_connect("$host","$user","$pass"); // and select the database mysql_select_db($dbase); $mode=$_GET["mode"]; // get the mode $QUERY_STRING = 'x=avatar_add'; if(!isset($mode) || empty($mode)) // if mode is empty, switch to case index { $mode='index'; } switch($mode) { case 'index'; // displays the form ?> <form method="post" action="avatar_add.php?mode=submit"> Avatar URL:<br> <input type="text" name="avatar_url" size="30"><br><br> <input type="submit" value="Submit"> <?php break; case 'submit'; // form processing mode $avatar_url=$_POST["avatar_url"]; // check for empty fields if(empty($avatar_url)) { echo "Error! Please go back and fill in all the required fields!"; // if you have a footer include, insert it before the die statement die; // stops the script from executing the rest of the code } // let's get the date in YYYY-MM-DD format $date = date("Y-m-d"); $query="INSERT into icons_db (id,avatar_url) VALUES ('','$avatar_url')"; $result=mysql_query($query) or die ("Couldnot execute query: $query." . mysql_error()); if($result) // if mysql query successful { echo "Thank you! The avatar has been added to or database."; } break; } ?> Basically I need to input data into two tables. I am running 3 different query's but only 2 of them work. The other one doesn't. None working query: mysql_query("INSERT INTO users(username, password, email, pin, key) VALUES('$username', '$password', '$email', '$key', '$pin')"); Working querys: mysql_query("DELETE FROM beta_keys WHERE keys_new='$key'"); **and** mysql_query("INSERT INTO beta_keys(keys_used) VALUES('$key')"); So any ideas why the top one doesn't work but the bottom two do? The following mysql query is not returning rows like I expect it to. '$update_field' is a variable, matching an actual field name in table 'users'. 'user_task[1]' is an integer value. What am I missing here? Code: [Select] $query_update_user = "UPDATE users SET ".$update_field." = 'Y' WHERE user_no = '".$user_task[1]."'"; my SQL Query wont execute on on following lines: Code: [Select] $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); i get the following error: 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 ''gallery' ('image', 'memberid', 'caption') VALUES ('gallery/9074849_1.jpg', '1',' at line 1 here is my full code: Code: [Select] <?php require_once('connect.php'); $rand = mt_rand(1,9999999); $rand2 = mt_rand(1,9999999); $member_id = $_SESSION['SESS_MEMBER_ID']; $caption = $_POST["caption"]; if(isset($_FILES['uploaded']['name'])) { $allowed_filetypes = array('.jpg','.gif','.bmp','.png','.jpeg'); $max_filesize = 524288; // Maximum filesize in BYTES (currently 0.5MB) $fileName = basename($_FILES['uploaded']['name']); $errors = array(); $target = "gallery/"; $fileBaseName = substr($fileName, 0, strripos($fileName, '.')); // Get the extension from the filename. $ext = substr($fileName, strpos($fileName,'.'), strlen($fileName)-1); //$newFileName = md5($fileBaseName) . $ext; $newFileName = $target . $rand . "_" . $member_id.$ext; // Check if filename already exists if(file_exists("gallery/" . $newFileName)) { $errors[] = "The file you attempted to upload already exists, please try again."; } // Check if the filetype is allowed. if(!in_array($ext,$allowed_filetypes)) { $errors[] = "The file you attempted to upload is not allowed."; } // Now check the filesize. if(!filesize($_FILES['uploaded']['tmp_name']) > $max_filesize) { $errors[] = "The file you attempted to upload is too large."; } // Check if we can upload to the specified path. if(!is_writable($target)) { $errors[] = "You cannot upload to the specified directory, please CHMOD it to 777."; } //Here we check that no validation errors have occured. if(count($errors)==0) { //Try to upload it. if(!move_uploaded_file($_FILES['uploaded']['tmp_name'], $newFileName)) { $errors[] = "Sorry, there was a problem uploading your file."; } } //Lets INSERT database information here //Here we check that no validation errors have occured. if(count($errors)==0) { $result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')") or die (mysql_error()); { $errors[] = "SQL Error."; } } //If no errors show confirmation message if(count($errors)==0) { echo "<div class='notification success png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> The file {$newFileName} has been uploaded<br>\n </div> </div>"; //echo "The file {$fileName} has been uploaded"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } else { //show error message echo "<div class='notification attention png_bg'> <a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a> <div> Sorry your file was not uploaded due to the following errors:<br>\n </div> </div>"; //echo "Sorry your file was not uploaded due to the following errors:<br>\n"; echo "<ul>\n"; foreach($errors as $error) { echo "<li>{$error}</li>\n"; } echo "</ul>\n"; echo "<br>\n"; echo "<a href='gallery.php'>Go Back</a>\n"; } } else { //Show the form echo "Use the following form below to add a new image to your gallery;<br />\n"; echo "<form enctype='multipart/form-data' action='' method='POST'>\n"; echo "Please choose a file: <input name='uploaded' type='file' /><br />\n"; echo "Caption: <input name='caption' type='text' /><br />\n"; echo "<input type='submit' value='Upload' />\n"; echo "</form>\n"; //Echo Tests! echo "<br /><br />Random FileName: "; echo $rand; echo "<br />"; echo "member ID: #"; echo $member_id; } ?> any help appreciated. its prob something simple. my table has the following fields: "gallery" id (primary Key, AUTO_INC) memberid (fetched from session) image (will store image name including extension) caption (from "caption" text field in form) I've got to be missing something pretty basic here.. considering the query is pretty basic. I'm trying to figure out how to pull a query as an array so I can compare it against another array (array_diff) I'm doing a mysql_fetch_array, and I'm getting an error ( mysql_fetch_array(): supplied argument is not a valid MySQL result resource): Quote $checker = "SELECT ID FROM edible_uses"; $result2 = mysql_fetch_array($checker) or die(mysql_error()); //echoing to see if I'm getting what I need. echo $row['ID']; I've done a mysql_query and I get results. The table name and all that is correct. I'm stumped. This seems like a pretty simple query? I tried mysql_fetch_assoc as well. Same result? I tried it with an extra set of parenthesis around it. nope. |
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