I've written the below query:
At the moment this pulls back everything from the MarketingCampaigns table, regardless of which user created the record.
I need to be able to return the result, which counts only the records created by that user.
<?php
$servername = "localhost";
$username = "root";
$password = "doimkr943k3f";
$dbname = "crm4";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT format(count(id),0) as id12 FROM MarketingCampaigns";
$result2 = $conn->query($sql);
$row = $result2->fetch_assoc();
echo $row["id12"];
?>
The below is a query I can see has been 'auto-generated' by the tool I use, which checks which fields the user should see in table view. I'm just really unsure how to convert this into the simple, single value SQL queries I have above.
// mm: build the query based on current member's permissions
$DisplayRecords = $_REQUEST['DisplayRecords'];
if(!in_array($DisplayRecords, array('user', 'group'))){ $DisplayRecords = 'all'; }
if($perm[2]==1 || ($perm[2]>1 && $DisplayRecords=='user' && !$_REQUEST['NoFilter_x'])){ // view owner only
$x->QueryFrom.=', membership_userrecords';
$x->QueryWhere="where `Complaints`.`id`=membership_userrecords.pkValue and membership_userrecords.tableName='Complaints' and lcase(membership_userrecords.memberID)='".getLoggedMemberID()."'";
}elseif($perm[2]==2 || ($perm[2]>2 && $DisplayRecords=='group' && !$_REQUEST['NoFilter_x'])){ // view group only
$x->QueryFrom.=', membership_userrecords';
$x->QueryWhere="where `Complaints`.`id`=membership_userrecords.pkValue and membership_userrecords.tableName='Complaints' and membership_userrecords.groupID='".getLoggedGroupID()."'";
}elseif($perm[2]==3){ // view all
// no further action
}elseif($perm[2]==0){ // view none
$x->QueryFields = array("Not enough permissions" => "NEP");
$x->QueryFrom = '`Complaints`';
$x->QueryWhere = '';
$x->DefaultSortField = '';
}
I have a table called membership_userrecords which includes the below fields.
You can see the PK value in the table and which user owns it.
I'm just not sure how to do the SQL query.
Can you help?
EDIT: I really need to work on my PHP syntax lol. Thanks #aynber
Assuming the username and the memberID are the same, this should be your query.
$sql= $conn->prepare("SELECT format(count(*),0) as id12 FROM MarketingCampaigns where memberID = ?");
$sql->bind_param("s", $username);
$sql->execute();
$sql->bind_result($row);
$sql->fetch();
echo $row;
I'm really unsure about your data here.
Related
Not sure if this is even possible, I am trying to get it to where "ItemA" has a value of 100 in the database. Which I have assigned a value to problem, what I really want is if a user inputs 50 into an input field, that "ItemA" value will now show 150 when it is called back in a select query. I am unsure of how to add the value in a column and a users input together.
I am new to PHP and SQL, so please forgive me. Any help is greatly appreciated and I thank you for your time!
The PHP that I have:
<?
$payout = $_POST["payout"];
$payouts = $_GET["payout"];
$withdraw = $_POST["bank_withdraw"];
$cashspent = $_POST["cash_spent"];
$servername = "localhost";
$username = "*****";
$password = "*****";
$db = "*****";
$conn = new mysqli($servername, $username, $password, $db);
if ($conn->connect_error){
die("Connection failed: ". $conn->connect_error);
}
$sql = "UPDATE finances SET payout = '$payouts' + ? WHERE finance_id = 1";
$stmt = mysqli_stmt_init($conn);
if ( ! mysqli_stmt_prepare($stmt, $sql)){
die(mysqli_error($conn));
}
mysqli_stmt_bind_param($stmt, "ii",
$payout,
$payouts
);
mysqli_stmt_execute($stmt);
In this example, I am trying to have it where a user is able to add the two values from what is already in the database.
Item in database (payout) = 100
Input from user = 50
Item from database + input from user = 150, showing in database
if I understand your question correctly there are 2 parts:
Select payout from database and add it to the user input.
update the value in the database
For 1) select the payout from the database $select_sql = "SELECT payout FROM finances WHERE finance_id = 1"; and you could use a variable such as $increasedPayout and assign this new variable $increasedPayout = $payouts + $select_sql;
and then for 2) use the $increasedValue variable in your SQL UPDATE statement.
$sql = "UPDATE finances SET payout = '$increasedPayout' WHERE finance_id = 1";
I want to select the password data of a user so they can log in on my website (for a member only website). I have a hash of the password and the username written to a table called "users" upon account creation. I do not know how to select a row on the table, so I get the error when the code looks for, something?
I found this on w3, but I don't understand what each part of the code means.
I tried to edit the code so it would match my user case, but I don't know how to.
$servername ="127.0.0.1";
$dbusername = "root";
$dbpassword = "";
$dbname = "users";
//create connection to db
$conn = new mysqli($servername, $dbusername, $dbpassword, $dbname);
$sql = "SELECT id, username, password FROM users";
$result == $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row == $result->fetch_assoc()) {
echo $userid = $row["id"] && $serverpassword = $row["password"] && $serverusername = $row["username"];
}
} else {
echo "User Lookup Failed";
}
$conn->close();
You don't need to select all records from database and then iterate all of them to check correct user. Besides, you should only select user by username and password as below:
$sql = "SELECT id, username, password FROM users WHERE username = '".$serverusername."' AND `password` = '".serverpassword."' ";
Apart, you should use data binding instead of variable to avoid SQL injection.
I have an url as domain.com/abc?orderstatus=cancel
Now, when someone reaches this link, I want to run a query that deletes the last record from the database.
So this is what I tried:
<?php
// Code here for the way to connect to database and insert records which works
// Now I added this code so that only if its on the domain.com/abc?orderstatus=cancel url, it will delete the last record.
$orderstatus = $_GET['orderstatus'];
if($orderstatus == 'cancel') {
$sql3 = "delete from table order by CustomerID desc limit 1";
}
?>
However, this is not working for me. May I know what am I doing wrong?
ps: I tried to cut out as many sql codes which work so that it makes reading easy. If there is any info that I am missing, please do let me know and I'll put it in.
You can use MAX() for MySQL if you have autoincremented on the ID or whatever. MAX() will delete the highest number on the field you specify.
$sql3 = "DELETE FROM table_name
WHERE CustomerID = (SELECT x.id FROM (SELECT MAX(t.CustomerID) AS id FROM table_name t) x)";
//Execute that query
$query3 = mysqli_query($db_conn, $sql3);
If you want to perform DELETE on the basis of ORDER BY then you may have to write nested query. You will get a SQL syntax error if you go with delete from table order by CustomerID desc limit 1
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
$orderstatus = $_GET['orderstatus']; // check for sql injections or XSS
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to delete a record
$sql = "DELETE FROM {YOUR TABLE_NAME} WHERE {YOUR WHERE CLAUSE} ";
if ($conn->query($sql) === TRUE) {
echo "Record deleted successfully";
} else {
echo "Error deleting record: " . $conn->error;
}
$conn->close();
?>
I have a SQL table, a php page. Both set up on a website using cPanel hosting.
The SQL table is like this. Table name is "dbase".
It has two columns, id and name and each column has two rows, say.
Id -> 01, 02
Name -> Name1, Name2.
What I want to do is.
If my page URL is examp.le/page.php?id=01 , or something like that which has a parameter 'id' valued as 01.
How do I assign value for php variable in that page say, $name as the corresponding value for 'name' column in the '01' row?
You have to get the parameter form the url using $_GET[] and then you have to query from the table to get the Name.
Just try this. Hope it helps you.
$id = $_GET['id'];
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "dbname";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT name FROM dbase where Id=".$id;
$result = $conn->query($sql);
$name;
if ($result->num_rows > 0) {
// output data of each row
$row = $result->fetch_assoc();
$name = $row['Name'];
} else {
echo "0 results";
}
Now just access $name wherever you want
I am new in PHP. I use session first time. I have two tables in db. First table with name pacra_teams with column id and title. Second table is og_users with multiple column but i use team_title as foreign key as store id against team title.
Now i want to create a session and want to display team name from table pacra_teams and user name from table og_users.
I try following code but i failed.
<?php
// starts session
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pacra1";
$conn = new mysqli($servername, $username, $password, $dbname);
$sql="SELECT *
FROM og_users
LEFT JOIN pacra_teams
ON og_users.id = pacra_teams.id
LIMIT 1
";
// setting variable values during session
$_SESSION['og_users.username']=$username;
$_SESSION['pacra_teams.title']=$title;
?>
call these variables
<?php
session_start();
?>
<?php
print_r($_SESSION);
?>
Please help me how i can do this?
One Thing More. if i run seesion.php page it display undefine variable "title"
and if i run print code. It display username "root" but i dont have any user name root in my db
You already defined a query but didn't execute it.
// starts session
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pacra1";
$conn = new mysqli($servername, $username, $password, $dbname);
$sql="SELECT *
FROM og_users
LEFT JOIN pacra_teams
ON og_users.id = pacra_teams.id
LIMIT 1
";
$result = $conn->query($sql);
$row = $result->fetch_object();
// setting variable values during session
$_SESSION['og_users.username'] = $row->USER_NAME; // Change to correct column name in table og_users
$_SESSION['pacra_teams.title'] = $row->TITLE_COLUMN_NAME; // Change to correct column name in table pacra_teams
The result will be the same every time without a WHERE clause in your sql statement. It's only going to return the first row it finds. It looks like you're trying to set user information in a session variable so you can call the data throughout your application so here's a possible solution assuming you grab an ID for the user somewhere (IE web form).
This is a simple answer to explain a concept, not a tutorial.
<?php
//Setup your connection stuff here
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "pacra1";
//Get a user's name from a form
$userName = $_POST['username'];
// Perform your query
$db= new mysqli($servername, $username, $password, $dbname);
$sql = "SELECT *
FROM og_users
LEFT JOIN pacra_teams
ON og_users.id = pacra_teams.id
WHERE og_users.username = {$userName} LIMIT 1";
if(!$result = $db->query($sql)){
die('Error [' . $db->error . ']');
}
// Setting variable values during session
while($row = $result->fetch_assoc()) {
$_SESSION['ogUsername'] = $row['USERNAME']; // USERNAME is a placeholder for the example
$_SESSION['pacraTeamsTitle'] = $row['TITLE']; // Same here
}
It's not perfect, but hopefully it helps explain the concept and helps you complete your task.