what is wrong with this code? It generates no erros, but doesn't work either:
$query = "UPDATE $usertable SET dl=dl+1 WHERE id=$id";
$result = MYSQL_QUERY($query);
There is a unique id number for every row in my table. There is also a column called "dl" which holds the number of downloads certain files have gotten. I'm trying this code on my download page so that every time someone downloads a file (identified by its id number) the "dl" column for that file is incremented.
When i try and update the database table i get this error.
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'desc='My name is ***** and I'm a 17-yr. old junior. I play electric ba' at line 1
what i am doing is having the user fill in a large text area and saving it to a database. So i think the problem could possibly be that they are using ' and " in textarea to mess it up? Code:
$sqlUpd = "UPDATE employee SET fname = '$fname', lname = '$lname', title = '$title', email = '$email', id = '$id', pass = '$pass', all='$all', prepare_user_quotes = '$prepare_user_quotes', manage_jobs='$manage_jobs', manage_users='$manage_users', add_employee='$add_employee', add_process='$add_process', add_news_story='$add_news_story', add_dyk_tip='$add_dyk_tip', create_mail_list='$create_mail_list' WHERE unique_id='$val'" OR die(mysql_error());
but always seem to get a syntax error. I've tried all the available methods, but still syntax error pops up. This is the error I'm getting: Code:
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 'all=Ƈ', prepare_user_quotes = Ƈ', manage_jobs=Ɔ', manage_users=Ɔ', add_emplo' at line 1
I'm not sure why this code is not working and throwing the following error:
You have an error in your SQL syntax, check the manual that corresponds to your MySQL server version for the right syntax to use near 'AND role != 7 AND role != 4' at line 3
$sql = 'SELECT * FROM tblusers INNER JOIN permissions on tblusers.usrID = permissions.user_id WHERE permissions.team_id='.$team_id.' AND role != 7 AND role != 4'; require("connection.php"); $result = mysql_db_query($DBname,$sql,$link) or die(mysql_error()); code....
I know that the variable $team_id is working fine, because if I "echo" it, it works fine.
I keep getting this error:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 2
When I use the script below. I'm finding it a bit confusing because everything about it continues to work, it's just it gives me an error. When the script runs, the outcome is Followed by the error, which cuts the remainder of the page off. Does anybody know why it's doing this?
$dogyay = $_POST['dogid'];
$checkenergy = "SELECT energy FROM dogs WHERE id=$dogyay"; $energylevel = mysql_query($checkenergy) or die(mysql_error());[code].....
The issue is in the MySQL query and I get the error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(pmobile, pemail) SET pmobile = '', pemail = '' WHERE parentName = ''' at line 1
ZFDebug output error: Doctrine_Connection_Mysql_Exception: SQLSTATE: Syntax error or access violation: 1064 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 'sdfg sdfg , active = 1, description =...
But I look at this code and I see no errors in syntax.Tabla adds structure to the database:
CREATE TABLE IF NOT EXISTS `intelektualnie`.`CoachsTownsLangs` ( `id` INT NOT NULL AUTO_INCREMENT , `Coachs_Users_id` INT NULL [code]....
I'm having some trouble getting my database to update. This is my first attempt at PHP and MySQL and so far I'm doing well. I just used some of the code from the tutorials on this site and changed them to suit my needs. Everything works except updating the database. Clicking edit swaps a row in a table with a form that should allow me to update the entry. Here is the error it throws.
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 I have followed the correct syntax as far as I can tell, but like I said, this is my first time with PHP and MySQL. Following is all my code (minus CSS which has nothing in it really) with the appropriate sections highlighted.
This is a code from A-Mod, but doesn't see like no one really cares..so thought I'll post it here and maybe I'll get some help. When I'm on activity.php and click next page I get the following error: Couldn't obtain game data DEBUG MODE
SQL Error : 1064 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 5 Code:
I'm trying to update multiple columns at once in my database. The updated information is coming from another form page. This is my code and below it is the error. I have also tried this with single quotes and no brackets and I receive the same error message without the brackets with quotes. Code:
Well after scratching my head for too long, I give in, I've got to ask or I'll go crazy! I've been fiddling with PHP/MySQL again recently and although I successfully get a search of the db via a form field and have found a way* of adding a checkbox and update multiple records (a popular request I think!), my next hurdle is combining them.
I was hoping I could mash the script together (logically with my limited knowledge) to produce a 3 stage script. Page1 would send Page2 the search term (works so far) and then Page2 would send itself(and the db) those recorded selected for updating.
So far, when I try to I'm getting many errors, and I believe they occur because the script is trying to get information that's not there (from the search). Also it's because I've tried to modify existing code and I'm really guessing now :-(
Here's the code on the second page, the first page simply has a $_POST in it and then the second page collects that info:
It might be worth knowing that security isn't an issue as this script is run offline via WAMPServer (a brilliant freebee for testing!). Well, I really look forward to any comments,
well i have this messages table with sample values like these:
msg_id recipient_id read locked new 0 1 N Y Y 2 1 Y N N
ok, so lets just say this is a messaging table, and i want to reset all messages addressed to recipient with id=1 i was wondering why UPDATE `messages` SET `new`='Y',`read`='N',`locked`='N' where `recipient_id`=1;
doesn't work, MYSQL always returns 0 affected rows.. to robert gamble: yes, im sure the values were changed, since my purpose for this update query is to reset the data i was using for the testing phases :D
I use a function, myrandomPIN (), to generate random PIN numbers.
The following sql query updates records with the SAME PIN number but. I want to generate DIFFERENT pin numbers for every record. The function is ok but I can't figure out how to run it individually for each record. In other words I do not know the correct syntax to use UPDATE in a loop (if necessary) so that a different call to the function is done every time or ecah record ends up with a different PIN.
Or may be this can be done with a single mysql_query($sql2)?
-------------------------------------------------------------------------- $qr = @mysql_query($sqlb1) or die("ERROR"); $rs1 = mysql_fetch_array($qr); $thepin=myrandomPIN(); $sql2="UPDATE clients SET pin='$thepin' WHERE code='".$thecode."'"; $qr = @mysql_query($sql2) or die("ERROR""); --------------------------------------------------------------------------
Whenever I try to update any piece of PHP code to update a MySQL database, nothing happens. I have tried copying in some of the working codes of a website and tried the same, but no success. I recently installed XAMPP. I am connecting using the correct user id and pass to the database. The scripts are not giving me any error, but just not connecting, that's all.
While making such a usage as noted below <FORM name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>" >
I get the following error Firefox can't find the file at /C:/xampp/htdocs/="<?php.. so on
I have a basic db that I access with MySQL query browser. Everything seems fine to me but I am using this db as part of a php shopping basket and when I try to add an item I get:
Notice: 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 '>function.extract]: First argument should be an array in functions.inc.php on line 31 Notice: Undefined variable: price in functions.inc.php on line 36 Notice: Undefined variable: price in functions.inc.php on line 39 Notice: Undefined variable: total in unctions.inc.php on line 39
I'm assuming the last three are caused by this problem as price should be passed to the cart, and total is worked out using it. However although I know mySQL code it was the MySQL query browser that actually generated the code and I cannot see a way to view or debug the code.
The db has one table in it which is made up of id, name, subname, desc, and price.
The code in the php file that is being referred to is: