Post date php code

Php format post date pdo mysql code example

Your output format is missing the time component — you need to add hours (in 24hr format), minutes and seconds to the string. there is no such datatype in PHP If you’re trying to format the date, then use PHP’s standard date formatting functions and methods EDIT example Solution: As I suggested in my comment you can use this code: To convert your variable contents to a format which MySQL needs Now when you say that you have done it before and it was not working, I assume somewhere between this conversion and binding them to the you where changing them in your code, so either you can convert them immediately before binding them or as you suggested just convert them in the bind like this:

DATE and TIME with PDO PHP MYSQL

1. Your input string is in an unusual format. Therefore you need to use the DateTime::createFromFormat method to parse it ( strtotime() can’t do the job) using the correct format string.
2. Your output format is missing the time component — you need to add hours (in 24hr format), minutes and seconds to the string.

$dp = DateTime::createFromFormat("m/d/Y - H:i a", $_POST["dp"]); $dpstr = $dp->format('Y-m-d H:i:s'); $query->bindParam(':dp', $dpstr); 

Assuming $_POST[«dp»] contains «08/31/2020 — 05:04 pm» then $dpstr will be 2020-08-31 17:04:00 .

DATE and TIME with PDO PHP MYSQL, Your input string is in an unusual format. Therefore you need to use the DateTime::createFromFormat method to parse it (strtotime() can’t do the job) using the correct format string. Your output format is missing the time component — you need to add hours (in 24hr format), minutes and seconds to the string. Here’s a working example:

Php/mysql DateTime Format using PDO

'Release Date' => (date) $r['Release date'], 

You’re Typecasting to a PHP datatype of date . there is no such datatype in PHP

If you’re trying to format the date, then use PHP’s standard date formatting functions and methods

foreach($result as $r) < $releaseDate = new DateTime((string) $r['Release date']); $temp[] = array('Ticker' =>(string) $r['Ticker'], 'Release Date' => $releaseDate->format('Y-m-d'), 'Price' => (string) $r['Price'], 'Amount' => (string) $r['Amount']); > 

Json — php/mysql DateTime Format using PDO, Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more

Querying DATE in MySQL using PHP PDO

As I suggested in my comment you can use this code:

$date_start = date('Y-m-d', strtotime($start_date)); $date_end = date('Y-m-d', strtotime($date_end)); 

To convert your variable contents to a format which MySQL needs

Now when you say that you have done it before and it was not working, I assume somewhere between this conversion and binding them to the sql statement you where changing them in your code, so either you can convert them immediately before binding them or as you suggested just convert them in the bind like this:

$this->bind(':start_date', date('Y-m-d', strtotime($start_date))); 

And then your query will look like this

$this->sql = "SELECT COUNT(id) AS number_of_items FROM item_table WHERE id > :id AND date_visited BETWEEN :start_date AND :end_date"; 

So to sum everything up, you can use one of these two ways. either of them should work, but you can use any of them you are more comfortable with:

1. Convert the variables before binding them

$date_start = date('Y-m-d', strtotime($start_date)); $date_end = date('Y-m-d', strtotime($date_end)); $this->sql = "SELECT COUNT(id) AS number_of_items FROM item_table WHERE id > :id AND date_visited BETWEEN :start_date AND :end_date"; $this->prepare($this->sql); $this->bind(':id', 0); $this->bind(':start_date', $date_start); $this->bind(':end_date', $date_end); // rest of your code 

2. Convert the variables during binding

$this->sql = "SELECT COUNT(id) AS number_of_items FROM item_table WHERE id > :id AND date_visited BETWEEN :start_date AND :end_date"; $this->prepare($this->sql); $this->bind(':id', 0); $this->bind(':start_date', date('Y-m-d', strtotime($start_date))); $this->bind(':end_date', date('Y-m-d', strtotime($date_end));); // rest of your code 

Querying DATE in MySQL using PHP PDO, Teams. Q&A for work. Connect and share knowledge within a single location that is structured and easy to search. Learn more

Creating a Form to POST into MySQL using PDO properly?

Your form is posting to success.php, so make sure that the insert code is in the success.php file:

setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // Set SQL $sql = 'INSERT INTO mytable (FName, LName, Age, Gender) VALUES (:first, :last, :myage, :gen)'; // Prepare query $query = $db->prepare($sql); // Execute query $query->execute(array(':first' => $first, ':last' => $last, ':myage' => $myage, ':gen' => $gen)); > catch (PDOException $e) < echo 'Error: ' . $e->getMessage(); > 

setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); // Set SQL $sql = 'INSERT INTO mytable (FName, LName, Age, Gender) VALUES (:first, :last, :myage, :gen)'; // Prepare query $query = $db->prepare($sql); // Execute query $query->execute(array(':first' => $first, ':last' => $last, ':myage' => $myage, ':gen' => $gen)); $db= null; > catch (PDOException $e) < echo 'Error: ' . $e->getMessage(); $db= null; > 

Mysql — date format in PDO PHP, First things first; you can do the same query efficiently using MySQL’s join statements: SELECT se.*, s.*, sea.*, sn.* FROM show_episode se INNER JOIN show_episode_airdate sea ON se.episode_id = sea.episode_id INNER JOIN show_network sn, shows s ## This part might be a bit wrong.

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Php how to post a date in php

Solution 2: Don’t change your PHP code, but change your database table: add a column named, say, , of type TIMESTAMP with as default value CURRENT_TIMESTAMP. Solution: To show date in format in HTML page, again convert it into [ Note : I’m assuming your fetch array as and column as .

How do I post date in PHP?

There are multiple ways to send date in your current json:

You can send date in hidden fields from your html like :

You can add date separately in your $event variable.

$filename = 'data.json'; $handle = @fopen($filename, 'r+'); if ($handle) < fseek($handle, 0, SEEK_END); if (ftell($handle) >0) < fseek($handle, -1, SEEK_END); fwrite($handle, ',', 1); fwrite($handle, json_encode($event) . ']'); >else < fwrite($handle, json_encode(array($event))); >fclose($handle); > ?> 

Well, it looks like I’ve figured out a better solution. Quite simple, really.

Just add a hidden input in HTML.

And replace the $_POST value.

@MohdSayeed Thanks for your inspiration anyway.

PHP Front To Back [Part 8]

In this video we will look at the PHP date() function as well as mktime() and strtotime() to create Duration: 11:21

Post date format and select date format from the database

$date = $_POST['date'] ; $date = date("Y-m-d",strtotime($date)); 

To show date in d-m-Y format in HTML page, again convert it into

[ Note : I’m assuming your fetch array as $row and column as date .]

For more info, please have a look on strtotime — php manual

Update Code (As question is edited)

$sql = "SELECT * FROM table WHERE ORDER BY id DESC LIMIT 1"; $result = $conn->query($sql); if ($result->num_rows > 0) < // output data of each row while($row = $result->fetch_assoc()) < echo "" . $row["id"]. ""; echo "" . date("d-m-Y",strtotime($row['date'])). ""; > > else "; > $conn->close(); 

PHP date_format() Function, Required. Specifies the format for the date. The following characters can be used: d — The day of the month (from 01 to 31)

How to get the date and time of a post?

You could get the date and time at the point of the $_POST of your current system by using:

This would return the date/time in the following format: 2015-01-11 13:17:52 — Which you can then store in the database with your INSERT INTO query.

Make sure you have the correct timezone set within PHP as by default it may return UTC time. http://php.net/manual/en/function.date-default-timezone-set.php

Don’t change your PHP code, but change your database table: add a column named, say, time , of type TIMESTAMP with as default value CURRENT_TIMESTAMP.

You don’t need to change your PHP, the database will automatically put the current time in the new time column when you insert a new record.

PHP date() format when inserting into datetime in MySQL, $sql = «CREATE TABLE date_test (. id INT AUTO_INCREMENT PRIMARY KEY, · )»; ; $sql = «INSERT INTO date_test( created_at ). VALUES( ‘2018-12-05 12:39:16’ );»;. if (

Change PHP POST variable into date format for msql database

You need to use strtotime() to convert the data to timestamp . date() needs the second parameter be a timestamp value. Try with —

$dateRequired = date("Y-m-d", strtotime($dateRequired)); 

Dates in the m/d/y or d-m-y formats are disambiguated by looking at the separator between the various components: if the separator is a slash (/), then the American m/d/y is assumed; whereas if the separator is a dash (-) or a dot (.), then the European d-m-y format is assumed.

Change date to dd-mm-yyyy using php $_POST,

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PHP: Как правильно отправить дату в POST запросе?

Прошу помочь с задачей.
Делаю скрипт на PHP для отправки POST запроса на официальный сайт Украинской ЖД для получения инфы о наличии поездов.
Но постоянно получаю в ответ «Введена неверная дата отправления»

Скрипт парсит id сессии с куками и токен тоже на сайте с билетами

Там же я посмотрел, что браузер отправляет на сервер.
И ввел те же параметры и хедер.

Короче вот таким запросом:

// Пост запрос $url = 'http://booking.uz.gov.ua/ru/purchase/search'; $data = array('station_id_from' => '2200001', 'station_id_till' => '2208001', 'station_from' => '%D0%9A%D0%B8%D0%B5%D0%B2', 'station_till' => '%D0%9E%D0%B4%D0%B5%D1%81%D1%81%D0%B0', 'date_dep' => '20.06.2015', 'time_dep' => '00%3A00', 'time_dep_till' => '', 'another_ec' => '0', 'search' => ''); $options = array( 'http' => array( 'header' => "Host: booking.uz.gov.ua\r\n". "User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:38.0) Gecko/20100101 Firefox/38.0\r\n". "Accept: text/html,application/xhtml+xml,application/xml;q=0.9,*;q=0.8\r\n". "Accept-Language: uk,ru;q=0.8,en-US;q=0.5,en;q=0.3\r\n". "Accept-Encoding: gzip, deflate\r\n". "DNT: 1\r\n". "Content-Type: application/x-www-form-urlencoded; charset=UTF-8\r\n". "GV-Ajax: 1\r\n". "GV-Referer: http://booking.uz.gov.ua/ru/\r\n". "GV-Screen: 1280x800\r\n". "GV-Token: $uz->token\r\n". "GV-Unique-Host: 1\r\n". "Referer: http://booking.uz.gov.ua/ru/\r\n". "Content-Length: 208\r\n". "Cookie: HTTPSERVERID=$uz->cookie_gv_server_n;_gv_sessid=$uz->cookie_gv_sessid;_gv_lang=ru;__utma=31515437.1166281925.1433453305.1433453305.1433453305.1;__utmb=31515437.1.10.1433453305;__utmc=31515437;__utmz=31515437.1433453305.1.1.utmcsr=(direct)|utmccn=(direct)|utmcmd=(none); __utmt=1\r\n". "Connection: keep-alive\r\n". "Pragma: no-cache\r\n". "Cache-Control: no-cache\r\n", 'method' => 'POST', 'content' => http_build_query($data), ), ); $context = stream_context_create($options); $result = file_get_contents($url, false, $context); // Обработка результатов $inp = $result; $s = preg_replace('/\\\u0([0-9a-fA-F])/','&#x\1;',$inp); $s = html_entity_decode($s, ENT_NOQUOTES, 'UTF-8'); $s = json_decode($s); print_r($s->value);

Пробовал все тоже на http-master.com в респонсе снова не та дата.

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