PHP - How Do You Find The Next Year A Day Falls On A Date?

Hi all,   I have a search page where I'd like members to be able to search for races that they and their friends have entered into a database. When entering the data, users put the race date into the race_date field (Which is a DATETIME field) in the table.   I've set up the search and display pages, so users can search by a number of other criteria (first name, last name, race distance, time, etc), but am having trouble with searching by year. On my search form, is a field called race_year.   So, obviously I need some change in the code below, specifically in relation to the line $where .= " AND race_date='$race_year'"; because I want to extract the year part of the race_date column from the table and see if it matches $race_year .   Any advice would be appreciated.   Cheers, Dave  

$race_year = $_POST['race_year'];
if ($race_year != '')
{ // A year is selected
  $where .= " AND race_date='$race_year'";
}

Hi,

I've got a date picker on a form which puts data into a database in the YYYY-MM-DD format. Just wondering how I could also put the name of the month (extracted from that) as well as just the year into separate columns. Ie: To use the field race_date from the form to also fill the 'race_month' and 'race_year' columns in the database.

This code obviously only fills the 'race_date' column so far:
Code: [Select]

global $_POST;

$race_date = $_POST["race_date"] ;


....



$query = "INSERT INTO 10k_races (race_date, race_month, race_year)" . "VALUES ( '$race_date', '$race_month',  '$race_year')";

Hi i have a drop down menu for date which is meant to insert all 3 values into a date column bt is only sending the year how can i fix it

<select name="date_of_birth">

    <option value="1">January

    <option value="2">February

    <option value="3">March

    <option value="4">April

    <option value="5">May

    <option value="6">June

    <option value="7">July

    <option value="8">August

    <option value="9">September

    <option value="10">October

    <option value="11">November

    <option value="12">December

</select>

<select name="date_of_birth">

    <option value="1">1

    <option value="2">2

    <option value="3">3

    <option value="4">4

    <option value="5">5

    <option value="6">6

    <option value="7">7

    <option value="8">8

    <option value="9">9

    <option value="10">10

    <option value="11">11

    <option value="12">12

    <option value="13">13

    <option value="14">14

    <option value="15">15

    <option value="16">16

    <option value="17">17

    <option value="18">18

    <option value="19">19

    <option value="20">20

    <option value="21">21

    <option value="22">22

    <option value="23">23

    <option value="24">24

    <option value="25">25

    <option value="26">26

    <option value="27">27

    <option value="28">28

    <option value="29">29

    <option value="30">30

    <option value="31">31

</select>

<select name="date_of_birth" id="year">

<?php

$year = date("Y");

for($i=$year;$i>$year-50;$i--)

{

if($year == $i)

echo "<option value='$i' selected>Current Year</option>";

else

echo "<option value='$i'>$i</option>";

}

?>

Hello, i have two fields.  a beginning year and an ending year.  How can i make new fields out of the years in between the beginning and ending years.  i hope that makes sense. 

I have another question related to string searches.
What I have is a string of couple thousand characters of all types but the date format is allways the same: dd-mth-year (two digits for day-three letters for month-four digits for year). Can I use the format alone to determine where the date is located which in turn would make it very easy to store that date into variables for inserting into database, etc.
Is there anyway to tell php the format to look for or do I have to have actual numbers/letters to perform the search?

My app just found an error in my login.php file on accident. The app is supposed to get credentials from the user then submit them to my server for verification. I accidentally submitted null credentials (empty strings) and the server returned auth=true. I can't figure out why. 

Server code:

<?php
require_once '../../includes/db_connect.inc';
require_once 'user.php';
$header = "Content-Type: application/json";
$_SESSION['error'] = array();
if (!isset($_GET['secureSubmit']) || $_GET['secureSubmit'] != true){
    die(header("Location: ../access_denied.php"));
}
// check requirements
$required = array('username', 'password');
foreach ($required as $requiredField){
  if (!isset($_GET[$requiredField]) || $_GET[$requiredField] == ""){
    $_SESSION['error'][] = $requiredField . " is incomplete or missing.";
    }
}
  if (count($_SESSION['error']) > 0){
    $errors = array();
    for ($i = 0; $i < count($_SESSION['error']); $i++){
      $errors[]=array(
        'num' => $i,
        'desc' => $_SESSION['error'][$i]
      );
  }
  print json_encode($errors);
  exit;
}else{
  $user = new User;
  if ($user->authenticate($_GET['username'], $_GET['password'])){
    print json_encode(array(
      'auth' => true,
      'call' => "login",
      'sid' => session_id(),
      'credits' => $_SESSION['credits']
      )
    );
  }else{
    print json_encode(array('auth' => false,));
  }
}
?>
	

I went to my browser and entered

http://localhost:10509/login.php?secureSubmit=true&username=%22%22&password=%22%22

and recieved

{auth":true,"call":"login","sid":"29e81981a4709407e2fd8a8f734ad9bc","credits":null}

as a response, can anyone find where the invalid positive is coming from? The initial check for empty username or password should be failing before it ever gets to the check.

Edited March 19, 2019 by Karaethon
Code cleanup for readability

I have spent almost 2 days looking for solution to the strangest error I have ever meet.

Whenever I do commenting and my php script, the sricpt falls running.
Code: [Select]
<?php 

error_reporting(E_ALL | E_STRICT);
echo 'test1';
//echo 'testi2';
echo 'test3';
flush();

?>
trigger_error("Here we are!", E_USER_ERROR);
 

This will output: test1.

I have tried lot of things to find out the problem but I can't get any error message and whenever I remove the comments, the script works absolute fine. The script works also with comments in my local server, but whenever I download it on the other server, the problem appears. The file encoding is a UTF-8 format without BOM, and the file is one part of my Yii application. I haven't relasized similar behaving in other scripts.

I wonder if someone has some idea what could cause this kind of behaving?

I'm very thankful to man who can give me ideas...

Hey,

I'm using a script which allows you to click on a calendar to select the date to submit to the database.

The date is submitted like this: 2014-02-08

Is there a really simple way to prevent rows showing if the date is in the past?

Something like this:
if($currentdate < 2014-02-08 || $currentdate == 2014-02-08) {

}

Thanks very much,
Jack

Why does this work:
$dateAllowedSelect = date('Y') -18;

But this doesn't:
$dateAllowedSelect = date('Y') +5;

I nee to get the current year plus 2010 - 2015 in a select box.

Thank you.

Gud Pm!
Im trying to display a list of years (last year, current year and next year) with these simple code i made..
It displays the current and next year correctly except last year, which gives me "1970".
I don't get why it gives me that since may next year works properly.. 

Code: [Select]
$currentyear = date('Y');
$subtract_year = strtotime(date('Y', strtotime($currentyear)) . '-1 year');
$lastyear = date('Y', $subtract_year);
$addyear = strtotime(date('Y', strtotime($currentyear)) . '+1 year');
$newyear = date('Y', $addyear);
$year_array = array($lastyear, $currentyear, $newyear);
foreach($year_array as $year)
{
echo "<option value='{$year}'>";
echo $year;
echo "</option>";
}

I'm not familiar with dates yet.. so i'm gonna read the manual while w8ting for your kind replies. 

So I have decided to do research as my project and gain something from this. But also I will have to include programming which i prefer PHP. and will be testing these for my research. Does this sound like a good project and also If it is good sound.. Where do I start?...  need a starting point and then I could go on to doing it..

I am using the following code to sort by Year. It displays the current year fine and current year also displays in selection but does give the option for the previous years selection.

Code: [Select]
<form action="archivednews.php" method="post">
<select name="year" id="year">
<?PHP for($i=date("Y");
$i<=date("Y");
$i++)  
if($year == $i)
echo "<option value='$i' selected>$i</option>";
else
echo "<option value='$i'>$i</option>";
?>
</select>
<input type="submit" value="GO">
</form>

Okay I am trying to list out years in a dropdown box. It should show say how many years old. I want to start from 16 though and stop at 100. So the year has to be 1995 for 16 but then I would have to change the code every year. so I was wondering how do I modify this code to do what I explained?
Code: [Select]
<?php
 $start_year = ($start_year) ? $start_year - 1 : date('Y') - 100;
    $end_year = ($end_year) ? $end_year : date('Y');
for ($i = $end_year; $i > $start_year; $i -= 1) {

$date=date(Y);
$age = $i - $date;

        echo '<option value="'.$i.'">'.$age.'</option>';
    }
?>

Thanks in advanced

I need help with a query.   I've got a table that stores references to calls that come in to our office. It has a datetime field, and for the purposes of the query, it's the only thing in this table that is important.   I've got another table that stores appointment information. The appointment start time is also a datetime field. For this query, this start time is the only thing in this table that is important.   I need to know how many times in the last year there was a day when we had at least 150 calls come in and made at least 1 appointment.

SELECT COUNT(*)
FROM appointments a
LEFT JOIN (
  SELECT DATE(time)
  FROM calls
  WHERE <there are at least 150 calls on a day>
) c 
ON DATE(a.start_time) = c.time
WHERE a.start_time >= CURDATE() - INTERVAL 1 YEAR 

To make things difficult, I guess the time would have to be converted to a day, because I'm looking for all days where there was an appointment, and the datetime field is too specific.   I've been looking online for 30 mins or so, and was hoping somebody here could point me in the right direction. I'm not making much progress by myself.
Edited by sKunKbad, 06 October 2014 - 01:17 PM.

Hello,

I seem to be having a problem.  I am trying to extract the year from a date

Code: [Select]
2012-03-01
echo "2012";

I have tried this and it only displays 1969

$dateorig = "2012-03-01";
$new_year = date("Y", strtotime($dateorig));
echo $new_year;