update query problem - php

hi all i have a field "ammount" in mysql database which have "varchar(50)" type. When i insert data into that field e.g ammount= 4 kg its ok but when i update that field it gives me the following error.
Error in query: UPDATE ingredients SET ingredient_name='test recipe',ammount=4 gm where ingredient_id='59'. 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 'gm where ingredient_id='59'' at line 1
and my query is
$query="UPDATE ingredients SET ingredient_name='$ingredient',ammount=$ammount where ingredient_id='$ingredient_id'";

1) The correct spelling is "amount".
2) You should not be using variable interpolation like this for an SQL query. It is very unsafe. Use a prepared statement.
3) You didn't put quotes around $amount when defining $query, so they don't end up in the final substituted query string. Look closely at the error message: it shows you the query that SQL tried to process. Notice how it says ammount=4 gm? It can't handle that, because there are no quotes.
If you use prepared statements like you are supposed to, the quoting takes care of itself.

Your query has:
...,ammount=4 gm where...
which is incorrect. You need quotes around 4 gm.
Change
,ammount=$ammount where
to
,ammount='$ammount' where

Related

Error in the where part of my MYSQL query

Please help I have no clue what is wrong here. I have submitted the error and the actual query below.
Error:
UPDATE `WWM_Login` SET `Username`='RyzeAlchemist',`Email`='smadger#live.co.uk',`FirstName`='test',`MiddleName`='test',`LastName`='test',`DiscordID`='#RyzeAlchemist#6043',`P_openCompletedOrders`=1,`P_openCurrentOrders`=1,`P_openRequestedOrders`=1,`P_openCreateAnOrder`=1,`P_OpenEditUsers`=0,`P_CreateStaff`=1,`P_CreateClient`= WHERE UserID = 9
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'WHERE UserID = 9' at line 1
Query:
"UPDATE `WWM_Login` SET `Username`='$uid',`Email`='$email',`FirstName`='$firstname',`MiddleName`='$middlename',`LastName`='$lastname',`DiscordID`='$DiscordID',`P_openCompletedOrders`=$field[1],`P_openCurrentOrders`=$field[2],`P_openRequestedOrders`=$field[3],`P_openCreateAnOrder`=$field[4],`P_OpenEditUsers`=$field[5],`P_CreateStaff`=$field[6],`P_CreateClient`=$field[7] WHERE UserID = $id"
The error seems to be towards the end of your query:
`P_CreateClient`= WHERE
Shouldn't this be:
`P_CreateClient`='x' WHERE
Where x is the value you want to set
Encase your variables in single quotes, like so:
P_CreateStaff`= '$field[6]'
This way, if there is an empty to NULL value it will take it as such. I see some of your variables are encased in signle quotes, but that particular variable is not so when there is no value MySQL is trying to look for a value equal to WHERE, so it's taking your WHERE clause as the condition for P_CreateStaff. To avoid this, encase all your variables in single quotes.

Proper mySQL command for adding URLs

I'm having a problem when trying to add a URL to a mySQL database.
The string is a URL:
http://pbs.twimg.com/profile_images/1708867059/405000_10150426314376065_707061064_8645107_703731598_n_normal.jpg
The error I get is:
Error description: 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 '://pbs.twimg.com/profile_images/1708867059/405000_10150426314376065_707061064_86' at line 1
It seems as though it won't allow me to add a URL, I presume there is something wrong with some of the characters but I don't know what?
My SQL is:
INSERT INTO accounts (name,consumerkey,consumersecret,pic_url) VALUES ($twitterID,$consumerkey,$consumersecret,$picture_url)"
You cannot truly solve this kind of problem by adding a few characters (like ' or ") to your bespoke sql string!
Instead, get to know the real way to write sql in php (it's like a very badly kept secret), which is to use PDO statements. This will allow you to use placehoders like (:twitterID, :consumerKey, :consumerSecret, :pictureUrl) which will accept complex variables such as urls and any of the crap users send in much more gracefully.
In the long run, this will save you a lot of trouble and time.
You need to quote string values and any other character that SQL will complain about, in this case it's the colon; see further down below.
($twitterID,$consumerkey,$consumersecret,'$picture_url')
or
('".$twitterID."','".$consumerkey."','".$consumersecret."','".$picture_url."')
if you wish to quote all the values.
Sidenote: You can remove the quotes around the variables that are integers.
I.e.:
This based on, and without seeing how the rest of your code looks like:
$picture_url = "http://pbs.twimg.com/profile_images/1708867059/405000_10150426314376065_707061064_8645107_703731598_n_normal.jpg";
The error states that it is near : - near being just that, the colon.
...right syntax to use near '://pbs.twimg.com
^ right there
You can also use:
VALUES ($twitterID, $consumerkey, $consumersecret, '" .$dbcon->real_escape_string($picture_url) . "')";
$dbcon is an example of a DB connection variable and based on mysqli_ syntax.
Something you haven't stated as to which MySQL API you are using.
Plus, your present code is open to SQL injection.
Use prepared statements, or PDO with prepared statements.

MySQL UPDATE syntax error with variables

Long story short, I'm trying to write a PHP code that will parse a text file into MySQL queries. Everything works fine except for the queries, which consist of UPDATE statements.
The entire code is kinda long, but if you want to see it - http://pastebin.com/xVR6ArD0
Here is just the part which is problematic :
while ($i<=$no_collumns)
{
$j = $i-1;
if (!
mysql_query
("UPDATE ResultsPredmet
SET ${parsed_collumns[$i]} = '${parsed_words[$j]}'
WHERE ${parsed_first_collumn} LIKE '${parsed_first_word}'")
)
{echo mysql_error()."\n"; break;}
// echo "\nUPDATE ResultsPredmet SET ${parsed_collumns[$i]} = '${parsed_words[$j]}' WHERE ${parsed_first_collumn} LIKE \"${parsed_first_word}\"";
$i++;
}
... where $parsed_collumns and $parsed_words are arrays of strings and $parsed_first_collumn and $parsed_first_word are strings.
I tried all combinations of quotes and escapes for the variables. I tried putting them in double quotes and escaping them, or double quotes and concatenating them, then maybe i thought it was the fact that I was comparing strings via the '=' operator so i tried with 'LIKE'. I googled for several hours and everywhere people said to use single quotation marks for variables so I tried that too and it didn't work.
In the end I echoed the queries and I get:
UPDATE ResultsPredmet SET grade = '10' WHERE name LIKE "Vildur"
UPDATE ResultsPredmet SET index = '117/2010' WHERE name LIKE "Vildur"
Updating table.
UPDATE ResultsPredmet SET grade = '6' WHERE name LIKE "NinoDoko"
UPDATE ResultsPredmet SET index = '132/2011' WHERE name LIKE "NinoDoko"
Updating table.
UPDATE ResultsPredmet SET grade = '10' WHERE name LIKE "Koco"
UPDATE ResultsPredmet SET index = '130/2011' WHERE name LIKE "Koco"
Done.
Which seem fairly fine to me. Other queries I got were the same only with the names with single quotes around them, or with no quotes or any other combinations.
The errors I get are :
Updating table.
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'index = '117/2010' WHERE name LIKE 'Vildur'' at line 1
Updating table.
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'index = '132/2011' WHERE name LIKE 'NinoDoko'' at line 1
Updating table.
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'index = '130/2011' WHERE name LIKE 'Koco'' at line 1
Apparently, the server that I'm using is MariaDB 5.5, but after a bit of research I figured it would be similar to just generic MySQL, though I might be completely off. The "Updating table." is just a random echo in my code. I've also tried the query without indenting it, still got the same errors. The values I get for grade and index are strings - or at least I hope so, since I'm getting them with explode().
index is a reserved word
UPDATE ResultsPredmet SET `index` = '10' WHERE name LIKE 'Vildur'
http://dev.mysql.com/doc/refman/4.1/en/reserved-words.html

I have a SQL Syntax error on my php page

Here is the mysql insert the I am running in php. I have removed the part giving the error but then I get a error on the next piece. I am not seeing what is diffrent to cause the error.
$fields="adv_exchange SET synum='".$synum."', worknum='".$_POST['worknum']."', user_id='".$current_user->ID."', f_name='".$current_user->user_firstname."', l_name='".$current_user->user_lastname."', email='".$current_user->user_email."', regnum=".$_POST['regnum'].", item='".$item."', qsver='".$_POST['qsver']."', flashrom='".$_POST['flashrom']."',expansion='".$_POST['board']."', rdisplay='". $_POST['rdisplay']."', screen_model='".$_POST['screen_model']."', p_hardware='".$_POST['cable']."', pcolor='".$_POST['pcolor']."', pname='".$_POST['pname']."', kboard='".$_POST['kboard']."', ip='".$_POST['ip']."', reg_name='".$_POST['reg_name']."', mem=".$_POST['mem'].", dt_server='".$_POST['dt_server']."', alert='".$_POST['alert']."', ows='".$_POST['ows']."', w_date='".$_POST['w_date']."', flashromver='".$_POST['flashromver']."', s_size='".$_POST['s_size']."', mag='".$_POST['mag']."', rcard='".$_POST['rcard']."', kvsid=".$_POST['kvsid'].", finger='".$_POST['finger']."', stand_alone='".$_POST['stand_alone']."', standards='".$_POST['standards']."', profile='".$_POST['profile']."', man_date='".$_POST['man_date']."', l_sn='".$_POST['l_sn']."', misc='".$_POST['misc']."', problem='".$_POST['problem']."'";
then $query = "insert into $fields";
I receive back
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 ' item='JS900CV', qsver='', flashrom='',expansion='', rdisplay='', screen_model='' at line 1
Blockquote
if I echo the $query I get this:
insert into adv_exchange SET synum='SY5135', worknum='123456', user_id='2', f_name='REMOVED', l_name='REMOVED', email='REMOVED', regnum=, item='JS900CV', qsver='', flashrom='',expansion='', rdisplay='', screen_model='', p_hardware='', pcolor='', pname='', kboard='', ip='192.168.1.16', reg_name='', mem=, dt_server='', alert='', ows='', w_date='', flashromver='', s_size='', mag='', rcard='', kvsid=3, finger='', stand_alone='', standards='', profile='', man_date='', l_sn='', misc='misc test\r\n', problem='gen test'
Depending on what I enter in the error is changing spots in my statement. Not all fields are used the form is dynamic that is supplying the data so the fields are dependent on what options are selected. On a side note in case of concern about using $_POST to insert directly into mysql, I sanitize the array first. Any help would be greatly appreciated.
Look at regnum=,. You don't provide a value for regnum. Either leave it out entirely or set it to an appropriate value.
You're using a very, very bad approach to MySQL databases: manually creating the queries. You should really use prepared statements instead: this issue will be resolved as well.
Don't use mysql_* functions, use PDO instead.
Your code would look like this (simplified):
// This holds the query
$statement = $pdo->prepare('INSERT INTO adv_exchange SET synum=?, worknum=?, etc=?, problem=?');
// This executes it with the given arguments. It's 100% injection-proof and safe. In fact, it's also faster.
$statement->execute(array($synum, $_POST['worknum'], $_POST['therest'], $_POST['problem']));
regnum=".$_POST['regnum']." is causing the problem. When it is undefined, you get regnum=, in the SQL query
A bigger concern is that you are not escaping your inputs. Either use mysql_real_escape_string around them, or better, use prepared statements.
You need to SET regnum=SOMETHING.
Currently it's empty.

SQL error when deleting from MySQL

I am coming across a problem when deleting data from my SQL data. I have tried various versions of my statement but to no avail. Below is the error I am presented with and the statement I am using.
$sql = "DELETE FROM `saved_holidays` WHERE (subscriberID= $user AND title= $check_value)";
//connect to database then execute the SQL statement.
$db->exec($sql);
and the error message is:
SQLSTATE[42000]: 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 '#xml119.com AND
title= Luxurious Jamaican holidays | 40% Discount On Accommodati' at
line 1
I can see that the correct data is being passed but the syntax is wrong. Can anyone help?
$check_value is a string, so you have to enclose it in ' in your query like this:
title = '$check_value'
For security purposes, you should also use mysql_real_escape_string on all string parameters you have. Or even better, use prepared statements: http://php.net/manual/en/pdo.prepared-statements.php
You need to put quotations around your variables. It doesn't like spaces.
Depending on the server you are using (MySQL or MSSQL) you have to use backticks, single quotes, or double quotes:
DELETE FROM saved_holidays WHERE (subscriberID="$user" AND title="$check_value")
Also, if you are using PDOs, you should consider using prepared statements:
$statment = $conn->prepare("DELETE FORM saved_holidays WHERE (subscriberID=? AND title=?)"); //$conn has to be your connection ceated by doing new PDO(...connection string...)
$statment->execute(array($user, $check_value));
Amit is correct your statement should look like this;
$sql = "DELETE FROM `saved_holidays` WHERE (subscriberID= '$user' AND title= '$check_value')";
the variable is a string so must be enclosed in single quotes.
This should then work for you.

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