Cannot Create Database PHP - php

I am beginning to create my own Interface to use with MySql though I cannot seem to create a database with the code I have at the bottom. Everything else works also i can echo out the $ObtainDatabase variable to see that it does have a value stored. Any suggestions would be great.
<?php
session_start();
//define connection
$conn = new mysqli('localhost', 'over_watch','XXXXXXx','billing');
//Variables
$UserEmail = $_SESSION['email'];
$MysqlUserDataBaseCreate = $_POST['create_database'];
//CheckIfUserExists
$SeeIfUserExist = $conn->prepare("SELECT (email) FROM database_users WHERE email= ?;");
$SeeIfUserExist->bind_param('s',$UserEmail);
$SeeIfUserExist->execute();
$SeeIfUserExist->bind_result($ObtainedEmail);
$SeeIfUserExist->store_result();
$SeeIfUserExist->fetch();
$RowsReturnedFromPreparedStatment = $SeeIfUserExist->num_rows();
if($RowsReturnedFromPreparedStatment < 1){
$InsertIntoDatabase = $conn->prepare("INSERT INTO database_users(email,check_if_created) VALUES(?,?);");
$InsertIntoDatabase->bind_param('ss',$UserEmail,$MysqlUserDataBaseCreate);
$InsertIntoDatabase->execute();
$SelectDatabaseToCreate = $conn->prepare(" SELECT (check_if_created) FROM database_users WHERE email = ?;");
$SelectDatabaseToCreate->bind_param('s', $UserEmail);
$SelectDatabaseToCreate->execute();
$SelectDatabaseToCreate->bind_result($ObtainDatabase);
$SelectDatabaseToCreate->fetch();
$CreateDatabase = "CREATE DATABASE $ObtainDatabase ;";
$conn->query($CreateDatabase);
}else{
echo 'user permitted to one database';
}
?>

Can you try this Code.
Details about this error can be found in the mysql docs. Reading those details makes it clear that the result sets of a prepared statement execution need to be fetched completely before executing another prepared statement on the same connection.
Here is the doc where you can refer
<?php
session_start();
//define connection
$conn = new mysqli('localhost', 'over_watch','XXXXXXx','billing');
//Variables
$UserEmail = $_SESSION['email'];
$MysqlUserDataBaseCreate = $_POST['create_database'];
//CheckIfUserExists
$SeeIfUserExist = $conn->prepare("SELECT (email) FROM database_users WHERE email= ?;");
$SeeIfUserExist->bind_param('s',$UserEmail);
$SeeIfUserExist->execute();
$SeeIfUserExist->store_result();
$SeeIfUserExist->bind_result($ObtainedEmail);
$SeeIfUserExist->store_result();
$SeeIfUserExist->fetch();
$RowsReturnedFromPreparedStatment = $SeeIfUserExist->num_rows();
if($RowsReturnedFromPreparedStatment < 1){
$InsertIntoDatabase = $conn->prepare("INSERT INTO database_users(email,check_if_created) VALUES(?,?);");
$InsertIntoDatabase->bind_param('ss',$UserEmail,$MysqlUserDataBaseCreate);
$InsertIntoDatabase->execute();
$InsertIntoDatabase->store_result();
$SelectDatabaseToCreate = $conn->prepare(" SELECT (check_if_created) FROM database_users WHERE email = ?;");
$SelectDatabaseToCreate->bind_param('s', $UserEmail);
$SelectDatabaseToCreate->execute();
$SelectDatabaseToCreate->store_result();
$SelectDatabaseToCreate->bind_result($ObtainDatabase);
$SelectDatabaseToCreate->fetch();
$CreateDatabase = "CREATE DATABASE $ObtainDatabase ;";
$conn->query($CreateDatabase);
}else{
echo 'user permitted to one database';
}
?>

Related

Fetch db data into variable, and insert into another db

I need to fetch "UID" row data from db_tags where "Contador" is selected, then insert it into db_rfid. I've tried the fetch commands, but doens't seems to work.
require_once "config.php";
require_once "session.php";
if($_SERVER["REQUEST_METHOD"] == "POST" && isset($_POST['submit'])){
$contador = trim($_POST['Contador']);
$nome = trim($_POST['nome']);
if($query = $db->prepare("SELECT * FROM db_tags.tags2 WHERE CONTADOR = ?")){
$query->bind_param('s', $contador);
$query->execute();
$query->store_result();
$insertQuery = $db->prepare("INSERT INTO db_rfid.tbl_rfid (UID, Nome) VALUES (?,?)");
$insertQuery->bind_param("ss", $uid, $nome);
$result = $insertQuery->execute();
header("inserir.php");
$insertQuery->close();
$query->close();
exit;
}
}
mysqli_close($db);
?>
because a default database is selected in your config.php file.
$db=PDO("...;dbname=XXXX;...");
For this you need to create 2 PDO objects
$dbConnection1=PDO("...;dbname=db_tags;...");
$dbConnection2=PDO("...;dbname=db_rfid;...");
usage:
$query1 = $dbConnection1->prepare("SELECT * FROM tags2 ...");
$query1->execute(...)
$query2= $dbConnection2->prepare("INSERT INTO tbl_rfid ...");
$query1->execute(...)

Have problem with write the code to check database and insert into different table

I wrote some code to check database table name is hms_history if type = commission then will add another row of data into hms_history but change amount_balance to -amount_balance then will add another data to table hms_deposit. I try to write the code but when I check it show only connect successful, but nothing changes in the database at all.
This is my code
<?php
$servername = "localhost";
$username = "xxxx";
$password = "xxxx";
function setChecked($conn,$params) {
$s = $conn->prepare("UPDATE `hm2_history`
SET ref_id=-1
WHERE id=:id
");
$s->execute($params);
}
try {
$conn = new PDO("mysql:host=$servername;dbname=xxxx", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
echo "Connected successfully";
}
catch(PDOException $e)
{
echo "Connection failed: " . $e->getMessage();
}
$stmt = $conn->prepare("SELECT * FROM hm2_history
WHERE id NOT IN(SELECT ref_id FROM hm2_history WHERE ref_id > 0) AND
ref_id=0 AND commission=0");
//deposits
$s = $conn->prepare("INSERT INTO `hm2_deposits`
SET `user_id`=:user_id,
`type_id`=:type_id,
`deposit_date`=:deposit_date,
`last_pay_date`=:last_pay_date,
`status`=:status,
`q_pays`=:q_pays,
`amount`=:amount,
`actual_amount`=:actual_amount,
`ec`=:ec,
`compound`=:compound,
`dde`=:dde,
`unit_amount`=:unit_amount,
`bonus_flag`=:bonus_flag,
`init_amount`=:init_amount,
`ref_id`=:ref_id
");
$v['type_id']= 9;
unset($v['id']);
$lastDepositId = $conn->lastInsertId();
$date = date('Y-m-d H:i:s');
$s->execute($v);
//history1
$s = $conn->prepare("INSERT INTO `hm2_history`
SET `user_id`=:user_id,
`amount`=:amount,
`type`=:type,
`description`=:description,
`actual_amount`=:actual_amount,
`date`=:date,
`ec`=:ec,
`deposit_id`=:deposit_id,
`rate`=:rate
");
$x = array(
'user_id'=>$v['user_id'],
'type'=>'bonus',
'amount'=>$v['amount'],
'description'=>'Bonus note',
'actual_amount'=>$v['actual_amount'],
'date'=>$date,
'ec'=>68,
'deposit_id'=>0,
'rate'=>1.0,
);
$s->execute($x);
//history2
$s = $conn->prepare("INSERT INTO `hm2_history`
SET `user_id`=:user_id,
`amount`=:amount,
`type`=:type,
`description`=:description,
`actual_amount`=:actual_amount,
`date`=:date,
`ec`=:ec,
`deposit_id`=:deposit_id,
`rate`=:rate
");
$x = array(
'user_id'=>$v['user_id'],
'type'=>'depoit',
'amount'=> -$v['amount'],
'description'=>'Deposit from plan',
'actual_amount'=> -$v['actual_amount'],
'date'=>$date,
'ec'=>999,
'deposit_id'=>$lastDepositId,
'rate'=>1.0,
);
$s->execute($x);
?>

Updating database using dropdown without using a submit button

I'm trying to update the table status value whenever I make a selection from the dropdown list.
The problem is I'm having a syntax error on my update query. I've read stuff about syntax error and I can't quite understand it. I think I'm gonna need a more specific help. Here's what I've done:
<?php
$hostname = "localhost";
$username = "root";
$password = "";
$databasename = "companydb";
try
{
$conn = new PDO("mysql:host=$hostname;dbname=$databasename",$username, $password);
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
if(isset($_POST["status"]))
{
$query = "UPDATE tickets SET status = '$status' WHERE id = $id";
$statement = $conn->prepare($query);
$statement->execute(array('status' => $_POST["status"]));
$count = $statement->rowCount();
if($count > 0)
{
echo "Data Inserted Successfully..!";
}
else
{
echo "Data Insertion Failed";
}
}
else
{
echo "unknown index: 'status'";
}
}
catch(PDOException $error)
{
echo $error->getMessage();
}
?>
And here's my table schema:
You are not performing prepared statements properly. You need to add the placeholder in the query and not the variables. The variables should be added in the execute() line.
$query = "UPDATE tickets SET `status` = :status WHERE `id` = :id";
$statement = $conn->prepare($query);
$statement->execute(array(':status' => $_POST["status"],':id' => $id));
Also FYI, $id is undefined.
Try Changing this:
$query = "UPDATE tickets SET status = $status WHERE id = $id";

how to perform value calculation in mysql database using php

i have a column inside my user table named pairCount which has a value of 2 by default, but whenever an action occur, i want the value to be minus by one (1). so i wrote the following code
<?php
session_start();
require_once './include/Constants.php';
require_once './include/DatabaseConn.php';
require_once './vendor/autoload.php';
require_once './include/User.php';
require_once './include/Level.php';
use \phputil\JSON;
$user = User::getCurrentUser();
$db = new DatabaseConn();
$link = $db->connect();
$sql = "update users set `payee-1` = NULL, status = ?,`pairCount` = ? WHERE `payee-1` = ?";
$stmt = $link->prepare($sql);
$stmt->bind_param('sss', $status, $pairCount, $payee1);
$status = PAIR;
//here is were i did the calculation, but it given me error
$pairCount = $pairCount (-1);
$payee1 = ($user->getUserName());
$res = $stmt->execute();
$stmt->store_result();
if ($res) {
echo "<h1>Congrate you have successfully unsubscribe to a pay this person";
header('Refresh: 3;url=icant3.php');
} else {
echo 'undone';
}
It is also possible to do that in your SQL query.
update users set `payee-1` = NULL, status = ?, `pairCount` = pairCount - 1 WHERE `payee-1` = ?
Then you don't need to do anything with $pairCount variable in your PHP code.

mysql update not working.. needs help on update code

I'm using $_POST to data to my php page to update the mysql database. I'm trying to use the customer id to pick the row and update the company name and fname(firstname). When I figure this out, I'll add the rest to be updated. I've also included what I've tried via the "//" Thank you
-----dbconnect-----
$id= $_POST['id'];
$company= $_POST['company'];
$fname = $_POST['fname'];
echo $id;
echo $company;
echo $fname;
//$sql = mysqli_query($con,"UPDATE customer SET company = $company WHERE id= '.$id.'")
//$sql = "UPDATE customer SET company ='".mysql_real_escape_string($_POST['company'])."WHERE id='".mysql_real_escape_string($_POST['id'])."'";
$sql = "UPDATE customer SET company = $company WHERE id= '1'";
mysqli_select_db('customer');
$retval = mysqli_query( $sql, $con );
if(! $retval )
{
die('Could not update data: ' . mysqli_error());
}
echo "Updated data successfully\n";
mysqli_close($conn);
}
Making a few assumptions here but try this out...
// make mysqli throw exceptions on error
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
// assuming your connection looks something like this
// you can pass the database name as the fourth argument instead of using select_db()
$con = new mysqli('localhost', 'user', 'pass', 'customer');
// $id = $_POST['id'], etc
// use a prepared statement with parameter placeholders.
// for more info see http://php.net/manual/mysqli.quickstart.prepared-statements.php
$stmt = $con->prepare('UPDATE customer SET company = ? WHERE id = ?');
// bind parameter variables and execute
$stmt->bind_param('si', $company, $id);
$stmt->execute();
echo 'Updated data successfully', PHP_EOL;

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