I have to generate a random number which will be unique from the numbers stored in the mysql database. I have written the following function to generate. But it is not working.
function random(){
$invoice = rand(0,9999);
$check = mysql_query("SELECT order_ID FROM premises WHERE order_ID='$invoice'");
$match = mysql_num_rows($check);
echo $invoice;
echo "sdhfji";
if($match == 0){
mysql_query("INSERT INTO orders values('$invoice','$id','$name','$key_num')");
mysql_query("INSERT INTO persons values('$invoice','$fname','$lname','$address','$phone','$email'.'$business')") or die(mysql_error());
}else{
random();
}
echo $invoice;
}
It is not entering into the function. Please help me if there is any alternative way.
dFirst of, use mysqli since mysql is deprecated as of php 5.5 (to future proof your work).
I dont get what the purpose of this "random function" is, but I guess you simply want to insert a post into the database and insert "link" that post into another table?
/* cant figure out what the diffrenece in invoice and id is. Pls provice database-structure to improve this answer */
$query = "INSERT id (id, name, key_num )INTO orders values('$id','$name','$key_num')";
$mysqli->query($query);
//Get id from latest query
$latest_insert = $mysqli->insert_id;
$query = "INSERT INTO persons values('$last_id','$fname','$lname','$address','$phone','$email'.'$business')";
$mysqli->query($query);
Related
I've tried to follow several answers on this question but can't seem to get it to work for my specific problem.
I want to insert data but only if the flight_number doesn't exists already. How can I do that?
$sql = mysqli_query($con,
"INSERT INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`)
VALUES ('".$flight_number."', '".$mission_name."', '".$core_serial."', '".$payload_id."')"
);
Rob since you saying flight_number is a unique then you can use INSERT IGNORE
<?php
$sql = "INSERT IGNORE INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`) VALUES (?,?,?,?)";
$stmt = $con->prepare($sql);
$stmt->bind_param('isss',$flight_number,$mission_name,$core_serial,$payload_id);
if($stmt->execute()){
echo 'data inserted';
// INSERT YOUR DATA
}else{
echo $con->error;
}
?>
OR you could select any row from your database that equal to the provided flight number then if u getting results don't insert.
$sql = "SELECT mission_name WHERE flight_number = ? ";
$stmt = $con->prepare($sql);
$stmt->bind_param('i',$flight_number);
if(mysqli_num_rows($stmt) === 0){
// INSERT YOUR DATA
}
A unique index on flight number should do the trick.
CREATE UNIQUE INDEX flight_number_index
ON space (flight_number);
If you want to replace the existing row with the new one use the following:
$sql = mysqli_query($con,
"REPLACE INTO space (`flight_number`, `mission_name`, `core_serial`, `payload_id`)
VALUES ('".$flight_number."', '".$mission_name."', '".$core_serial."', '".$payload_id."')"
);
Make note that I just copied your code and changed INSERT to REPLACE to make it easy to understand. PLEASE PLEASE PLEASE do not use this code in production because it is vulnerable to injection.
If you don't want to replace the existing row, run an insert and check for errors. If there is an error related to the index, the row already exists.
Disclaimer: I haven't tested any of this code, so there may be typos.
When i inserting using this code it insert two datas and i downt know how to fix it
$sql = "SELECT Version_id FROM versions ORDER BY Version_id DESC LIMIT 1;";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$lastVersion =$row["Version_id"];
}
}
echo($lastVersion);
$lastVersion++;
$sql = "INSERT INTO versions (version)
VALUES ('v$lastVersion')";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
While I don't exactly understand what you mean with "two datas", I do see multiple issues with your code.
First of all it is horribly inefficient and prone to race conditions. It's also quite wrong, in that it doesn't do what you want it. Not to mention should be replaced with native database functionality.
Most of these can be fixed by simply changing the version_id field to a AUTO_INCREMENT. This will automatically give the new record the next available ID in the set, exactly as what you're trying to do. Then you can retrieve this ID by using "lastInsertId()"
That'll make all of the code in your post superflous, and only require you do do something like this when actually inserting data:
$sql = "INSERT INTO `version`(`setting`, `date`) VALUES (:setting, :date)";
$stmt = $db->prepare ($sql);
$res = $stmt->execute ($data);
$newID = $db->lastInsertId ();
After this the new version ID is stored in the $newID variable.
Of course, if you want to UPDATE the version ID for some reason, then INSERT is the wrong command to use. Also, why use an entire table for what's basically a simple version number? In short, your whole table doesn't make a whole lot of sense for me.
I recommend explaining the rationale behind it, so that we can possibly come up with some better solutions you can use.
I host multiple servers for multiplayer games and I am requiring some help with creating a PHP MySQL script.
I have made scripts for the game servers that output a few variables to a php script. These variables include the player name, a GUID number (Game User ID) and a couple other unimportant things. These are sent to the php script every time a player joins the server.
Anyway what I basically need it to do is every time a player joins the server it saves the player name, guid and join date/timestamp to a row in a MySQL table. The player will always have only one GUID code, which is sort of like their cd-key. What I have at this current time:
if ( $action == "save")
{
$name = mysql_real_escape_string($_GET['name']);
$guid = mysql_real_escape_string($_GET['guid']);
}
mysql_query("INSERT INTO `players` (`name`, `guid`) VALUES ('$name', '$guid') ON DUPLICATE KEY UPDATE `last_joined`=CURRENT_TIMESTAMP")or die(mysql_error());
echo "-10";
die();
Now, this works great as it is. But what I need it to do is; if the player comes on the server with a different name, it will log that instance into a new row and if they come on again with the same name it will update the same row with the current time stamp. And for instance, if they change their name back to the first name they use it will update the row that has that name recorded with the current time stamp.
The only thing I have tried is making the 'name' column, a primary key and on a duplicate entry it would update it. However if I did that and another player came on the server with the same name it would just update the last player's data.
So it needs to record every username a player uses.
There's probably quite a simple solution but I've never had the time to learn to MySQL and I need this done soon.
Thanks for any help.
Make the GUID the primary unique key.
Then instead of just inserting the row, check if that guid exists in the database first and then if it does, update the row. If it doesn't then you can insert it.
You can take a shot for this:
$guid = mysqli_real_escape_string($conn, $_GET["guid"]);
$name = mysqli_real_escape_string($conn, $_GET["name"]);
if (!empty($guid) && !empty($name)) {
//Check if the user exists
$sql = "SELECT COUNT(*) AS cnt FROM players WHERE guid = " . $guid;
$res = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($res);
if ($row['cnt']) {
//If yes, update
$sql = "UPDATE players SET `last_joined` = NOW()
WHERE `guid` = " . $guid;
} else {
//If no, insert
$sql = "INSERT INTO players (`guid`, `name`, `last_joined`)
VALUES (" . $guid . ", '" . $name . "', NOW())";
}
mysqli_query($conn, $sql);
echo "-10";
die();
} else {
echo 'Missing parameter';
die();
}
NOTE:
I am using mysqli fucntions, because mysql functions are deprecated. You can use PDO also.
I'm building a simple bug tracking tool.
You can create new projects, when you create a project you have to fill in a form, that form posts to project.class.php (which is this code)
$name = $_POST['name'];
$descr = $_POST['description'];
$leader = $_POST['leader'];
$email = $_POST['email'];
$sql="INSERT INTO projects (name, description, leader, email, registration_date)
VALUES ('$name', '$descr', '$leader', '$email', NOW())";
$result = mysql_real_escape_string($sql);
$result = mysql_query($sql);
if($result){
header('Location: ../projectpage.php?id='.mysql_insert_id());
}
else {
echo "There is something wrong. Try again later.";
}
mysql_close();
(It's not yet sql injection prove, far from complete...)
Eventually you get redirected to the unique project page, which is linked to the id that is stored in the MySQL db. I want to show the name of that project on the page, but it always shows the name of the first project in the database.
(here I select the data from the MySQL db.)
$query = 'SELECT CONCAT(name)
AS name FROM projects';
$result = mysql_real_escape_string($query);
$result = mysql_query ($query);
(here I show the name of the project on my page, but it's always the name of the first project in the MySQL db)
<?php
if ($row = mysql_fetch_array ($result))
echo '<h5>' . $row['name'] . '</h5>';
?>
How can I show the name of the right project? The one that is linked with the id?
Do I have the use WHERE .... ?
Yes, You have to use the WHERE to specify which project You want to get. I'm also not sure why are You using CONCAT function when You want to get only one project.
Other important thing is that You have to use mysql_real_escape_string() function on parameters before You put them in the query string. And use apropriate functions for specific type of data You receive.
So Your statement for getting the project should look like this:
SELECT name FROM projects WHERE id = ' . intval($_GET['id'])
Also when before You use the mysql_fetch_assoc() function, check if there are any records in the result with
if(mysql_num_rows($result) > 0)
{
$project = mysql_fetch_assoc($result);
/* $project['name'] */
}
try this
// first get the id, if from the url use $_GET['id']
$id = "2";
$query = "SELECT `name` FROM `projects` WHERE `id`='".intval($id). "'";
$result = mysql_query(mysql_real_escape_string($query));
use mysql_fetch_row, here you'll not have to loop through each record, just returns single row
// if you want to fetch single record from db
// then use mysql_fetch_row()
$row = mysql_fetch_row($result);
if($row) {
echo '<h5>'.$row[0].'</h5>';
}
$row[0] indicates the first field mentioned in your select query, here its name
The might be of assistance:
Your are currently assing a query string parameter projectpage.php?id=
When you access the page the sql must pick up and filter on the query string parameter like this:
$query = 'SELECT CONCAT(name) AS name FROM projects WHERE projectid ='. $_GET["id"];
$result = mysql_real_escape_string($query);
$result = mysql_query ($query);
Also maybe move mysql_insert_id() to right after assigning the result just to be safe.
$result = mysql_query($sql);
$insertId = mysql_insert_id();
Then when you assign it to the querystring just use the parameter and also the
header('Location: ../projectpage.php?id='.$insertId);
I have this function in a Code Igniter model that creates a new "video."
// Creates a new video.
public function newVideo($title, $description) {
$this->db->query("INSERT INTO videos VALUES (NULL, '$title', '$description')");
return // id of this new row
}
How do I obtain the ID of this new row in my MySQL table? I could get the last row and add 1, but there could be concurrency bugs I believe.
Very simple answer doesn't really need more than 30 chars does it?
$this->db->insert_id();
maybe you need something like this
$this->db->insert_id();
Try something like that :
$lastID = -1; //That's where it'll be stored
$query = "SELECT LAST_INSERT_ID()";
$result = mysql_query($query);
if ($result)
{
$row = mysql_fetch_row($result);
$lastID = $row[0];
}
Code for an insert is:
$sql = "INSERT INTO videos (title) VALUES(NULL, '$title', '$description')";
$this->db->query($sql);
More information on queries available at: http://codeigniter.com/user_guide/database/queries.html
Another workaround is that you can have a look into videos table in your database, and make you video id as autonumber, this will automatically generate an id for your video and then you can try this:
//on your model
$sql = "INSERT INTO videos (title,description) VALUES('$title', '$description')";
$this->db->query($sql);
//retrieve new video id
$this->db->insert_id()