I'm inserting some values to a table using a sql statement. All the values get inserted, except the 'qty' column. When I do an echo for sql statement, it shows the value has been assigned to the sql statement. Table column data type is INT. Can anyone help me to spot the problem?
$it_code = $_POST['id'];
//Prompt user to enter required qty
echo '<script>';
echo 'var qty = prompt("Enter Qty", "");';
echo '</script>';
//Get the item details from item table
$sqlString = "Select * from items where it_code='".$it_code."'";
$objItem = new Allfunctions;
$result = $objItem->retriveData($sqlString);
//Get the selected qty from java script to php variable
$qty = "<script>document.writeln(qty);</script>";
$user_id =$_SESSION['user_id'];
//Insert to orders table
$sqlStringInsert = "INSERT INTO orders(user_id,it_code,qty,status,tr_timestamp) VALUES ('$user_id','$it_code','$qty','PENDING',CURRENT_TIMESTAMP())";
echo $sqlStringInsert;
$objItem->addToTable($sqlStringInsert,'Succsessfully added.','Error adding'); // This is a custom built method to insert data in to a table by taking the sql statement as the parameter
Following is the sql statement generated,
Following is the table data. Qty is always 0 eventhood the sql statement had a value at qty column.
Obligatory warning
The code as you have shown it should not be used in production as it is vulnerable to SQL-injection attacks. Preventing this is well covered on this site in other answers.
Answer
The issue is that the value of $qty you are injecting into your SQL is the string "<script>document.writeln(qty);</script>". You just can't see it because you are likely echoing it out to the browser to test it.
Wrap your echoing of the SQL statement in a call to htmlentities() (docs) to see what's actually happening.
Depending on the version and settings, MySQL is very forgiving of bad data being injected to a column. In this case, it sees a string of text being inserted to a numeric column, and just truncates it to 0. See this in action here:
CREATE TABLE Foo (
Id INT,
Qty INT
);
INSERT INTO Foo(Id, Qty) VALUES (1, 'a string of text');
SELECT * FROM Foo; -- returns a row (1, 0)
https://www.db-fiddle.com/f/dEoaYGEyXEjs6ocVBwyyyr/1
Related
$ResultLine = $db->query("SELECT customerid FROM mydb WHERE surname= Holmes ORDER BY customerid DESC LIMIT 10")->fetchArray();
$res = $ResultLine[0];
echo json_encode($res);
Let's say there are 3 customers with the same surname in the database and different customerid. How can I structure the code above to return all the pairs customerid/surname as an array to my html page? The code above obviously only returns one customerid.
The call to fetchArray() returns one row at a time. So this snippet gets the first row and sends the first element ([0]) to json_encode. Remove ->fetchArray() from the set statement, then iterate over $ResultLine, something like:
while ($row = $ResultLine->fetchArray()) {
//do something with $row values
}
Doc found here
I am trying to make value +1 in database every time use use button.
my function in HTML:
function onClick(arg){
alert("thx fo click");
$.ajax({
type: 'POST',
url: 'data.php',
data: {
'arg': arg,
},
success: function(response) {
}
});
}
arg means value of button and it is a ID for a row in the database
and PHP:
<?php
$link = mysql_connect($servername, $username, $password);
$id = $_POST['arg'];
$sql = "UPDATE Buttons(SUMA) SET SUMA = SUMA + 1 WHERE ID = '$id'";
$conn->query($sql);
mysql_close($link);
?>
And that make nothing. How can i fix it?
You have several syntax errors here.
First and foremost though, check out mysqli_ (or PDO) and start using that instead of mysql_
For why to use mysqli_ - MySQL vs MySQLi when using PHP
Comparing mysqli_ and PDO - https://websitebeaver.com/php-pdo-vs-mysqli
With that out of the way....
You're defining your database connection without selecting a schema, but don't reference your schema in the query, meaning mysql won't know what to update. Either reference your schema in the connection or in each query. Also check on your table name, is it really Buttons(SUMA)?
You defined your database connection as $link, but are using $conn to attempt the query. Probably a 'typo' from copy and paste. Be careful of this...
As Artistic Phoenix mentioned, you have to make sure you're column cannot be set to NULL, and starts at 0 to begin. While you're at it and we're going through, make sure your datatype is set to int for the increment count.
After making those changes if you don't have success, I'd try running your query outside your code in a DB manager to ensure that portion is having the intended affect on your data, before looking at the errors in your code.
I'm guessing the arugment is passing correctly to your script, but to confirm, you can always echo it on the backend, and to be doubly sure alert() it in JS before it's passed through.
Take the time to go through that reading, update your script to use mysqli_ or PDO, and if you're still having troubles, I'm more than happy to jump back in here and help you further.
What is 1 + NULL it's still NULL.
IF you didn't default the column to '0' , then you can't increment it.
This can best be shown in a simple DB fiddle
Starting Null:
CREATE TABLE t(
id INT(10),
v INT(10)
);
INSERT INTO t (id)VALUES(1);
SELECT * FROM t;
UPDATE t SET v = v+1 WHERE id=1;
SELECT * FROM t;
In Both selects you will get a value of null for v as seen below in the fiddle:
https://www.db-fiddle.com/f/m1vgKpov1oiRJEfZEgmk1j/0
In simple terms, you cannot add 1 (or any number) to a NULL value. Well you can but it's still null. null + 1 = null
Starting 0:
CREATE TABLE t(
id INT(10),
v INT(10) default 0
);
INSERT INTO t (id)VALUES(1);
SELECT * FROM t;
UPDATE t SET v = v+1 WHERE id=1;
SELECT * FROM t;
In this case the first Select return 0 for v and the second returns 1 for v. As seen in the modified fiddle.
https://www.db-fiddle.com/f/m1vgKpov1oiRJEfZEgmk1j/1
Also (SQLInjection)
As I said in the comments:
What if arg = "' OR 1 --"
Or in other words don't inject user variables (or any clientside data) into your SQL or it winds up looking like this:
UPDATE `Buttons(SUMA)` SET SUMA = SUMA + 1 WHERE ID = '' OR 1 --'"
Which will increment every row in your DB that is not null. Basically the ' closes the other quote, then OR 1 is always true(for every row). Unfortinalty I cant show the -- comment part in the fiddle (it comments out the second select), but here is the end result.
https://www.db-fiddle.com/f/m1vgKpov1oiRJEfZEgmk1j/3
This is why we use prepared statements. This example is somewhat trivial but I have seen login code on here I was able to bypass simple by putting "' OR 1 LIMIT 1 --'" in, and with offset you could iterate though each user. They were looking for 1 row to be returned on the match of a username and password.
A few other things:
Table name Buttons(SUMA) is that really the name, as it will only work if escaped with the backtic. As I did in the above SQLInjection example.
$link = mysql_connect($servername, $username, $password); are these defined, the open tag is right above them. I generally chock that up to simplified example code. But it's worth asking. Obviously you can't connect to the DB if those are undefined.
Cheers!
I'm having trouble creating php code that would insert values into MySQL database but only if they don't already exist.
I send array from javascript to PHP file using $.ajax type POST.
Do I need additional 'SELECT' query to check if values already exist?
PHP File(Works, inserts values):
<?php
SESSION_START();
include('config.php');
if(isset($_POST['predictedMatches'])&&$_SESSION['userid']){
$predictedMatches=$_POST['predictedMatches'];
$userid=$_SESSION['userid'];
}else die("ERROR");
$sql="";
foreach($predictedMatches as $predictedMatch){
$sql.="INSERT INTO predictions(result,userFK,matchFK,tournamentFK) VALUES('".$predictedMatch['result']."','".$userid."','".$predictedMatch['id']."','".$predictedMatch['tourid']."');";
}
if($conn->multi_query($sql) === TRUE){
echo "OK";
}else{
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
Use the ON DUPLICATE KEY UPDATE feature. It won't insert, if the primary key exists. But you have to update some value, so use the column which is in no index or in the least indexes () in your case probably result). Your primary key has to be composted out of the three FKs:
ALTER TABLE `predictions` ADD PRIMARY KEY( `userFK`, `matchFK`, `tournamentFK`);
PHP-Code, just the SQL statment (I'm a Java Guy, so i tried my best)
$sql.="INSERT INTO predictions (result, userFK, matchFK, tournamentFK) "
."VALUES('".$predictedMatch['result'] ."','".$userid."','"
.$predictedMatch['id']."','".$predictedMatch['tourid']."') "
."ON DUPLICATE KEY UPDATE result = result ;";
To know if the query was inserted you have to look at the affected row count:
1 Row - Insert
2 Rows - Update
Take a look at $conn->affected_rows after the query.
Performance
INSERT ... ON DUPLICATE KEY UPDATE is definitively faster than a SELECT and INSERT but it's slower than an INSERT of just the needed datasets. The update is done in the database, even if it is the same value. Unfortunately there is no ON DUPLICATE KEY UPDATE INGNORE. If you have a lot of inserts, that will result in updates, than it may be better to use a cache, lookup values in an array and compare with the array before inserting. Only use the ON DUPLICATE KEY UPDATE as fallback.
I am using AJAX to post to a highScores table in a sqlite database, the total of the high score has to be calculated by a query and retrieved back through the database. How can I send the score through AJAX to the PHP script as an integer so I can perform mathematical calculations on it? St the moment I cant seem to do anything with it because the value gets to the php script as a string.
PHP is a loosely-typed language so you can easily cast the received value.
See the following answer for more information: How do I convert a string to a number in PHP?
http://www.php.net/manual/en/function.intval.php
or typecast
$something = (int) $var * 5;
Shouldn't you just remove the single quotes around your $score variable?
instead of:
$query = "UPDATE User SET Score = '$score' + Score WHERE Username = '$player'";
this:
$query = "UPDATE User SET Score = $score + Score WHERE Username = '$player'";
You do need to do your own sql-injections checks.
Otherwise you'll make it a string in your sql-query.
It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.
Closed 10 years ago.
I am trying to update a table row using the below code
t.executeSql('UPDATE flatcomments SET BuildingCode = ?, FlatNo = ?, Comment = ?, Closed = ?, New = ?)',
[buildingcode, flatdescription, flatcomment.toUpperCase(), 1, 1]);
with no success. Where am i wrong
This is a nice PHP tutorial.
All SQLite commands will work for you.
<?php
# CREATING A TABLE
$dbname = 'base';
$mytable ="tablename";
$base= new SQLiteDatabase($dbname, 0666, $err);
if ($err) exit($err);
$query = "CREATE TABLE $mytable(
ID bigint(20) NOT NULL PRIMARY KEY,
post_author bigint(20) NOT NULL,
post_date datetime,
post_content longtext,
post_title text,
guid VARCHAR(255)
)";
$results = $base->queryexec($query);
# The SQL command CREATE TABLE defines the columns. It is sent to the SQLite manager by the PHP method queryExec() which returns true or false, depending on whether the operation is successful or not.
# See the code of the script sqlite-create-table.php.
# ADDING POSTS
# The table of posts that we just created will be filled, like in Wordpress, with the posts we write, each post corresponding to a row of the table.
# The SQL command: INSERT INTO allows to store the data.
$number = 1;
$title="My first post";
$content="The content of my post...";
$date = strftime( "%b %d %Y %H:%M", time());
$author=1;
$url = "http://www.lantian.eu";
$query = "INSERT INTO $mytable(ID, post_title, post_content, post_author, post_date, guid)
VALUES ('$number', '$title', '$content', '$author', '$date', '$url')";
$results = $base->queryexec($query);
# For the purposes of the tutorial, we place the contents of the post directly into variables. In practice these variables will be assigned from an t online or local ext editor, as shown in the CMS tutorial.
# The author is represented by the number 1 because Wordpress does not put the names in the posts table but in a separate table instead.
# The guid column contains the URL of the post that also serves as an ID unique.
# The INSERT command jas for first parameter the table name and in parentheses the list of columns to fill, then the parameter VALUE provides a list of values corresponding to the columns in the same order.
# Thus, post_title, which contains the titles, will has for value $title, the variable that was assigned the title of the post.
# The same queryExec method is used to send the request.
# Source code of the script sqlite-write.php.
# READING A POST
# You can access the contents of the database with the SELECT command.
$query = "SELECT post_title, post_content, post_author, post_date, guid FROM $mytable";
$results = $base->arrayQuery($query, SQLITE_ASSOC);
$arr = $results[0];
if($results)
{
$title = $arr['post_title'];
$content = $arr['post_content'];
$user = $arr['post_author'];
$date = $arr['post_date'];
$url = $arr['guid'];
}
# To the SELECT command is given the list of columns that you want get the content and lines will be assigned to $results array. Indeed, the arrayQuery PHP method returns an array of arrays, each representing a line of the table.
# In practice we will use instead other commands that we will see later to limit the resources usage.
# Data are retrieved in the associative array $arr, where the column names are keys and their content the values.
# Source code of the script sqlite-read.php.
# DELETING A TABLE
# The deletion of a table is made by the DROP TABLE command of SQL.
$query = "DROP TABLE $mytable";
$results = $base->queryexec($query);
# See the code sqlite-delete-table.php.
?>