How to convert timestamp to normal time in php - php

I'm trying to convert timestamp from my mysql table in to normal time.
I've tried looking for information on the internet but there is only solutions for timestamps that you entered by hand
<?php
//setting header to
header('Content-Type: application/json');
//database
define('DB_HOST', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'bitcoin');
//get connection
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = sprintf("SELECT Kaina, Laiko_Kodas FROM localbtc LIMIT 200");
//execute query
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$data[] = $row /*("SELECT date_format(Laiko_Kodas) FROM localbtc")*/;
}
//free memory associated with result
$result->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);

while($row = $result->fetch_assoc()) {
$data = array();
$data2[]= $row['Kaina'];
$datetime=$row['Laiko_Kodas'];
$data2[] = date('M j Y g:i A', strtotime($datetime));
$dtata[]=$data;
}

Try the following code:
<?php
//setting header to
header('Content-Type: application/json');
//database
define('DB_HOST', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'bitcoin');
//get connection
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = "SELECT Kaina, Laiko_Kodas FROM localbtc LIMIT 200";
//execute query
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$row['Laiko_Kodas'] = date('H:i:s', strtotime($row['Laiko_Kodas']));
$data[] = $row;
}
//free memory associated with result
$result->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);

Related

Get ID for SQL command in PHP

I am using the chart.js library to create graphs. I can successfully create graphs using individual ID's from users. However, I would like to rewrite my code so that I could extract the user's ID from the URL and generate graph based on that specific ID.
Here is the code i'm using now to generate the points in the graph:
<?php
//setting header to json
header('Content-Type: application/json');
//database
define('DB_HOST', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'test');
//get connection
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = sprintf("SELECT treatment_log.bdi, treatment_log.date FROM treatment_log WHERE treatment_fk = 21 ORDER BY created_at");
//execute query
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$data[] = $row;
}
//free memory associated with result
$result->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);
Here is what the page outputs:
[{"bdi":"4","date":"2018-07-11"},{"bdi":"1","date":"2018-07-21"},{"bdi":"5","date":"2018-07-21"},{"bdi":"34","date":"2018-07-21"},{"bdi":"34","date":"2018-07-21"},{"bdi":"3","date":"2018-07-22"},{"bdi":"2","date":"2018-07-23"},{"bdi":"12","date":"2018-07-23"},{"bdi":"3","date":"2018-07-24"},{"bdi":"2","date":"2018-07-25"},{"bdi":"12","date":"2018-07-30"},{"bdi":"3","date":"2018-07-30"},{"bdi":"4","date":"2018-07-30"},{"bdi":"11","date":"2018-07-30"}]
In this case "21" is specific user. I am familiar with prepared statements, and have attempted rewrite this as a prepared statement.
Here is my attempt:
<?php
//setting header to json
header('Content-Type: application/json');
//database
define('DB_HOST', 'localhost');
define('DB_USERNAME', 'root');
define('DB_PASSWORD', '');
define('DB_NAME', 'test');
//get connection
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
$cid = htmlentities ($_GET['customer_id']);
//query to get data from the table
$sql = sprintf("SELECT treatment_log.bdi, treatment_log.date FROM treatment_log WHERE treatment_fk = ? ORDER BY created_at");
$stmt = mysqli_stmt_init($conn);
mysqli_stmt_prepare($stmt, $sql);
mysqli_stmt_bind_param($stmt, "i", $cid);
mysqli_stmt_execute($stmt);
$result = mysqli_stmt_get_result($stmt);
//execute query
$results = $mysqli->query($sql);
//loop through the returned data
$data = array();
foreach ($results as $row) {
$data[] = $row;
}
//free memory associated with result
$results->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);
Obviously my code doesn't seem to work. Help would be greatly appreciated.
Change this:
$result = mysqli_stmt_get_result($stmt);
//execute query
$results = $mysqli->query($sql);
//loop through the returned data
$data = array();
foreach ($results as $row) {
$data[] = $row;
}
into this:
$data = array();
mysqli_stmt_bind_result($stmt, $bdi, $date);
while(mysqli_stmt_fetch($stmt)) {
$data[]['bdi'] = $dbi;
$data[]['date'] = $date;
}

Php code not giving response

The following code does show anything in the browser . All the database name and table name are correct . While adding a line echo json_encode($temp) in the while loop displays result as objects.
<?php
define('DB_HOST', 'localhost');
define('DB_USER', 'root');
define('DB_PASS', '');
define('DB_NAME', 'sample_userdata');
$conn = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
die();
}
$stmt = $conn->prepare("SELECT questions FROM sample_questions;");
$stmt->execute();
$stmt->bind_result($post);
$userdata = array();
while($stmt->fetch()){
$temp = array();
$temp['questions'] =$post;
array_push($userdata,$temp);
}
echo json_encode($userdata);
?>
When you use json_decode() function it creates object for array with keys
$x['test1'] = 1;
$x['test2'] = 2;
echo json_encode($x);
// Result: {"test1":1,"test2":2}
You can remove keys using array_values()
$x['test1'] = 1;
$x['test2'] = 2;
echo json_encode(array_values($x));
// Result: [1,2]

Why won't this php SELECT FROM query work?

Here is my simple php code:
<!DOCTYPE html>
<html>
<body>
<?php
$servername = "localhost";
$username = "root";
$password = "********"; //hiding my password
$dbname = "course";
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT name FROM tutors";
$result = $conn->query($sql);
if( $result === true ) {
echo "good";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
while($row = $result->fetch_assoc()) {
echo $row["name"];
}
?>
</body>
</html>
In my database called "course", I have a table called "tutors" which has a column called "name". I have two entries in that table with the names "deep thought" and "pyrenees" respectively.
When this code runs however, the only thing that prints out is:
Error: SELECT name FROM tutors
It is supposed to simply print out the two names that I mentioned before.
Does anyone know why this happens? I know for a fact that I have the two entries in my table!
I think the word "name" is a MySQL reserved word. Wrap your query variables in a tilde backticks like this:
$sql = "SELECT `name` FROM `tutors`";
This helps to escape those values from MySQL thinking you're trying to referencing a built in variable.
Why not use mysqli like so:
function getFollowers($link, $userid)
{
$sql = "SELECT users.id, username, profileImg FROM following INNER JOIN users ON users.id = following.userid WHERE followid = " . $userid;
$result = mysqli_query($link,$sql);
$resultsArray = [];
while($row = mysqli_fetch_assoc($result)) {
$resultsArray[] = $row;
}
mysqli_free_result($result);
return $resultsArray;
}
This is just a clean example, I am sure you get the idea.
Here is what $link is
function connection()
{
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'username');
define('DB_PASSWORD', 'password');
define('DB_NAME', 'databaseTable');
$link = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
return $link;
}
Or without the methods:
define('DB_SERVER', 'localhost');
define('DB_USERNAME', 'username');
define('DB_PASSWORD', 'password');
define('DB_NAME', 'databaseTable');
$link = mysqli_connect(DB_SERVER, DB_USERNAME, DB_PASSWORD, DB_NAME);
if($link === false){
die("ERROR: Could not connect. " . mysqli_connect_error());
}
$sql = "SELECT name FROM tutors";
$result = mysqli_query($link,$sql);
$resultsArray = [];
while($row = mysqli_fetch_assoc($result)) {
echo $row["name"];
}
mysqli_free_result($result);
To check if the query was successful you can do this:
if (mysqli_num_rows($result) > 0)
{
//has rows, so whatever you want with them.
}
You put the condition after defining $result.

Null return from localhost php coding

I set up my localhost, I have all my relevant files in place. I run the php where I run the sql query, it returns NULL values. But it should not. I visit to the directory of the link via browser.
Can someone point me in the return direction, why it is returning NULL? I ran the SQL query, and works fine on my phpmyAdmin
<?php
header('Content-Type: application/json');
define('DB_HOST', '127.0.0.1');
define('DB_USERNAME', 'myusername');
define('DB_PASSWORD', 'Ihavenopassword');
define('DB_NAME', 'mytablename');
$mysqli = new mysqli(DB_HOST, DB_USERNAME, DB_PASSWORD, DB_NAME);
//$conn = mysqli_connect($db_host, $db_user, $db_pass, $db_name);
if(!$mysqli){
die("Connection failed: " . $mysqli->error);
}
//query to get data from the table
$query = sprintf("SELECT playerid, score FROM score ORDER BY playerid");
//execute query
$result = $mysqli->query($query);
//loop through the returned data
$data = array();
foreach ($result as $row) {
$data[] = $row;
}
//free memory associated with result
$result->close();
//close connection
$mysqli->close();
//now print the data
print json_encode($data);
$result is not what you expect. You need to fetch every item, with fetch_assoc for example:
//loop through the returned data
$data = array();
while ($row = $result->fetch_assoc()) {
$data[] = $row;
}

is it possible select Database 2 simultaneously in the same conection

I have problem with 2 select Databases from the same connection simultaneously, the code is:
#$dbmssSQLGestasa_conn = mssql_connect($servidor, $usuario, $contra);
if(!$dbmssSQLGestasa_conn){
header("Location:nobase.php");
exit();
}
mssql_select_db('GESTASA', $dbmssSQLGestasa_conn);
the code of the other connection:
#$dbmssSQLTasa_conn = mssql_connect($servidor, $usuario, $contra);
if(!$dbmssSQLTasa_conn){
header("Location:nobase.php");
exit();
}
//Apertura de la base de datos
mssql_select_db('TASA', $dbmssSQLTasa_conn);
it dont work and give me error "(severity 16)".
is It possible do 2 or more selections databases en la same connection mssql?
Dont use mysql_select_db();
In Mysqli :
$servername = "localhost";
$username = "username";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM test123.posts"; // databse.tablename
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$mack[] = $row;
}
}
$sql1 = "SELECT * FROM portal.abouts"; // databse.tablename
$result1 = $conn->query($sql1);
if ($result1->num_rows > 0) {
// output data of each row
while($row1 = $result1->fetch_assoc()) {
$mack1[] = $row1;
}
}
echo "<pre>";
print_r($mack);
print_r($mack1);
echo "</pre>";
$conn->close();
You can easily use 2 databases in same time with Below Codes
I'm Using variables as Capital word But you can use your own variable words.
<?php
define('HOST', "YOURHOSTNAME");
define('USER', "YOURHOSTNAME");
define('PASS', "YOURHOSTNAME");
define('DATABASE1', "NAMEOFDATABASE1");
define('DATABASE2', "NAMEOFDATABASE2");
$DATABASE1 = mysqli_connect(HOST, USER, PASS, DATABASE1);
$DATABASE2 = mysqli_connect(HOST, USER, PASS, DATABASE2);
if(!$DATABASE1){
die("DATABASE1 CONNECTION ERROR: ".mysqli_connect_error());
}
if(!$DATABASE2){
die("DATABASE2 CONNECTION ERROR: ".mysqli_connect_error());
}
$sql = "SELECT * FROM TABLE"; /* You can use your own query */
$DATABASE1_QUERY = mysqli_query($DATABASE1, $sql);
$DATABASE2_QUERY = mysqli_query($DATABASE2, $sql);
$DATABASE1_RESULT = mysqli_fetch_assoc($DATABASE1_QUERY);
$DATABASE2_RESULT = mysqli_fetch_assoc($DATABASE2_QUERY);
/* SHOW YOUR RESULT HERE WHICH DATABASE YOU WANT FROM */
echo $DATABASE1_RESULT['id'];
echo $DATABASE2_RESULT['id'];
/*After complete your all work don't forgot about close database connections*/
mysqli_close($DATABASE1);
mysqli_close($DATABASE2);
?>

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