I'm fairly new to stack overflow. i am creating a site were you type text in to 2 text boxes and it sends it to a database. i need it then to tell me what the id of that was save it as a session and then upload it to another database. sounds confusing. but I'm stuck of one part. its viewing the result thats from just that user. i have tried just showing the the last id of the last uploaded but it can be very unreliable if multiple people are trying to upload data and know there exact session. I'm also having trouble linking the session with the id. below is the code for the forum saving to the database. I'm pretty confident with sending the id of that users inputed data to another database. I'm just stuck on finding that users inputed texts id and creating a session holding the id number
<?php
header("Location:myscorenum.php");
$servername = "localhost";
$username = "root";
$password = "root";
$dbname = "score";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
session_start();
if (isset($_SESSION['username'])){
$username = $_SESSION['username'];
echo "working";
}else{
//3.2 When the user visits the page first time, simple login form will be displayed.
}
$value = $_POST['name'];
$value1 = $_POST['description'];
$sql = "INSERT INTO all_scores (name, description) VALUES ('$value','$value1')";
if ($conn->query($sql) === TRUE) {
echo "<a href=https://twitter.com/angela_bradley>My Twitter</a>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
ill have the processing to inset to another database in another file. I'm confident with uploading a specific session.
any questions don't hesitate to message me. thanks for your kind help.
You can use this:
mysqli::$insert_id -- mysqli_insert_id — Returns the auto generated id used in the last query
Like: echo $conn->insert_id;
Since you call it on $conn which is a mysqli instance that already has a connection, it will return your last inserted id, irregardless of other activities on the db (other queries do not affect the correctness of the output)
I edited my answer. This is the final part of your code and I've just added one line of code to it. Look if this is what you're looking for.
If this isn't working than make sure your database id field is AI.
if ($conn->query($sql) === TRUE) {
$id=$conn->insert_id; //this is where you get the last inserted id.
echo "<a href=https://twitter.com/angela_bradley>My Twitter</a>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
To use a session and set the ->insert_id to a session variable and use it in any other page, you can do this:
if ($conn->query($sql) === TRUE) {
session_start();
$_SESSION['id']=$conn->insert_id;
echo "<a href=https://twitter.com/angela_bradley>My Twitter</a>";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
and now in any other page do this to retrieve the session variable:
session_start();
$id=$_SESSION['id'];
here you go and you get the id.
Are you still confused?
Related
I am working with a coupon code website, it has a products section. Instead of pulling data from API server of shopping sites every time a visitor loads the page, I want to store the data in SQL and use it..
But the data should update every hour or only if an hour is passed.
cron job is not required as I dont want to happen it automatically.
If there is 100 webpage users in an hour, they will get the same data, but when the first user comes after 1 hour time, it should overwrite the products's information in SQL.
I got the following code from w3schools.com and it work fine for saving data in SQL.
I hope some can help me to overwrite data only if it the data older than 1 hour. Otherwise it should act as "data already exists".
<?php
$servername = "localhost";
$username = "nuser";
$password = "pass";
$dbname = "db";
$ProductId = "smartphone";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$check=mysqli_query($conn,"select * from testdb where productid='$ProductId'");
$checkrows=mysqli_num_rows($check);
if($checkrows>0) {
echo "Data Already Exists";
} else {
//insert results from the form input
$sql = "INSERT INTO Flipkart (productid, title, price)
VALUES ('productid','title','price')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();}
?>
The above code saves data in database, it can also detect duplicate data. but I want to add just one more function, which will overwrite data if the current data time is older than 1 hour. database has a TIMESTAMP "reg_date" column, which stores created date and time in YYYY-MM-DD HH:MM:SS format.
I am new to SQL and php, so pardon me if there is any problems with the question...
You can use a cron job.
This is the best explanation with code for PHP I have found so far for cron job:
https://code.tutsplus.com/tutorials/managing-cron-jobs-with-php--net-19428
I am currently working on a project which will sell a product to a user, and on checkout completion, the user will enter an email address. I have been trying (but to no avail so far) to use MySQL to put this data into a database. Only in a certain way. I need the input data to use the UPDATE method to be put into an already existing row. This row should match the following criteria. It should not have already been used, the email should not already exist within the database. I have tried so many different pieces of MySQL, I get all sorts from syntax errors to it updating every record in my table, I have gotten as far as it updating just one, but not checking if it already exists. I was hoping for a little insight as to how I can improve!
Code as follows:
<?php
$servername = "localhost";
$username = "Josh";
$password = "10584066";
$dbname = "customers";
// Get email address from input form
$Email = $_GET['keyword'];
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE list SET Email='$Email' , In_Use='1' WHERE In_Use='0' LIMIT 1";
if ($conn->query($sql) === TRUE) {
echo "Record updated successfully";
header('Location: /liteservers/logon/wood.php');
} else {
echo "Error updating record: " . $conn->error;
}
$conn->close();
?>
So, as you can see. The code is working almost as intended, I just want it to also check to see IF the email is already in the database and if it is, perhaps add the time that it attempted to record a new email within the same row as the already found existing email. I hope this actually makes sense if you need any more info that I haven't provided, let me know!
Use INSERT ON DUPLICATE KEY UPDATE syntax and mysql_affected_rows will tell you if its and insert or update.
I am trying to design a scoreboard for a project for my local youth club and would like to be able to click a button onto a wordpress page that will run a PHP script to update a value by 1 in a sql db table, then i can grab the value and display it else where.
Not too worried about passwords being used in scripts at this will only be used within the local network that nobody else has access to, have looked around and found a few bits of code but im not able to actually get it working, here's what i've got so far.
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "sot";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = UPDATE wp_sotstats SET CaptainChest=CaptainChest+1 WHERE id=1
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
$conn->close();
?>
The database name is 'sot' the table i want to update is called 'wp_sotstats' and the field within the table is 'CaptainChest' i only need this to really work with just the one entry which the id is '1'
Any thoughts?
$sql = UPDATE wp_sotstats SET CaptainChest=CaptainChest+1 WHERE id=1
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
Is not creating record, it updating it. For me, I will take current value from database as Select, then do ++ to it and do Update query.
$sql = UPDATE wp_sotstats SET CaptainChest=CaptainChest+1 WHERE id=1
Please wrap it in quotes and finish statement with semicolon
I am working on an academy website here at www.grmaedu.com
Web Specs: Built on Wordpress with the following plugins, Visual Form Builder Pro & Revolution Slider
So i have managed to do 90% of the work. My query is I want to assign automatic roll numbers to students who are submitting the application form here at www.grmaedu.com/application
Here are the remaining things I want to do:
Automatically assign Roll Number to students "after or on " form submission
Submit the form to the concerned mySql Database, Right now it emails correctly to the designated email address with no issues. All thanks to Visual Form Builder Pro
The date picker field is not working in the application form(I even updated my jQueryUI file)
I hope the provided details are enough for the solution.
I finally figured a way out. Here are the steps which i performed.
I installed a WP Plugin named Contact Form DB. This plugin saves all the form submitted values and stores them in a mySQL database.
I opened up phpmyadmin to see how it was stored and found it in one of the tables. I noticed that each entry had an unique id number. That is what I was trying to do. This plugin made it easier for me.
Then i created a Page template file name rollnumber.php and inserted the following code in
<?php
/* Define Connection properties */
$servername = "localhost";
$username = "grmaedu_wp2";
$password = "******";
$dbname = "grmaedu_wp2";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT entries_id FROM `wp_vfb_pro_entries` ORDER BY entries_id DESC LIMIT 1";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$a = $row["entries_id"];
$a++;
echo "<h3>Your Roll Number assigned is: " . $a . "</h3>";
//echo "id: " . $row["entries_id"] . "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
So now when the page loads it automatically fetches the unique id and adds 1 to the last submitted id and voila a new roll number is generated.
Index key is the best option to set as roll number. because it is unique and then you can get user data directly form this id.
<?php
/* Define Connection properties */
$servername = "localhost";
$username = "grmaedu_wp2";
$password = "******";
$dbname = "grmaedu_wp2";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "SELECT entries_id FROM `wp_vfb_pro_entries` ORDER BY entries_id DESC LIMIT 1";
$result = mysqli_query($conn, $sql);
echo "<h3>Your Roll Number assigned is: " . mysqli_insert_id() . "</h3>";
//echo "id: " . mysql_insert_id() . "<br>";
}
} else {
echo "0 results";
}
mysqli_close($conn);
?>
Basically, I have some rows that look like the following image:
Each ExampleUser has an ID and that ID isn't going to change at all.
I need the sql to read what ever $sentid is set to. To better explain, I have this lua code. (The lua code is working perfectly fine)
HS:PostAsync("http://examplesite0wwa/PresCand.php","sentid="..script.Parent.ExampleUser1.USERD.Value, 2)
And this sends as Variabel1 = Data1
And then I have this sql code.
<?php
$vote = $_POST["votes"];
$sentid = $_POST["sentid"]
$conn = new mysqli("localhost","anexapleomf1","","my_anexapleomf1");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "UPDATE `my_anexapleomf1`.`President Candidates` SET votes = votes + 1 WHERE `President Candidates`.`id` = $sentid";
if ($conn->query($sql) === TRUE) {
echo "<br>Record updated successfully <br>";
} else {
echo "<br>Error updating record: <br>" . $conn->error;
}
$conn->close();
?>
And what this should be doing, is reading what ever the sent $sentid was, and
then setting the $sentid in WHERE, buuut it's not doing that.
An example of this would be if $sentid was sent with the value of 2, and then the
WHERE `President Candidates`.`id` = $sentid
would be read as
WHERE `President Candidates`.`id` = 2
You don't have any problem of reading the post variable in your code, try to echo the post variable to see if it has been posted effectively, if not, check out the action method of your form.