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PHP - Problem With Form Date Picker
Hi,
Im having some troubles with a date picker on a form. It enters the date into the date field like this : 21 Sep 2011 I want to be able to insert it into a date field in the mysql database but it just enters it as 0000-00-00 Any ideas ? Many thanks, Scott. Similar TutorialsHi is it possible to make the Icon for my input activate the date picker (I'm using Materialize for my class) A lot of users seem to instinctively try to click the icon rather than the input field on mobile devices <div class="input-field col s12 m8 l7 xl7"> <i class="material-icons prefix">calendar_today</i> <input type="text" name="r_date" class="datepicker"> <label for="r_date"></label> </div>
<script>
$(document).ready(function(){
$('.sidenav').sidenav();
$('select').formSelect();
$('.datepicker').datepicker({
format:'yyyy-mm-dd'
});
$(".dropdown-trigger").dropdown();
});
</script>
This topic has been moved to JavaScript Help. http://www.phpfreaks.com/forums/index.php?topic=316616.0 I have tried a large number of "solutions" to this but everytime I use them I see 0000-00-00 in my date field instead of the date even though I echoed and can see that the date looks correct. Here's where I'm at: I have a drop down for the month (1-12) and date fields (1-31) as well as a text input field for the year. Using the POST array, I have combined them into the xxxx-xx-xx format that I am using in my field as a date field in mysql. <code> $date_value =$_POST['year'].'-'.$_POST['month'].'-'.$_POST['day']; echo $date_value; </code> This outputs 2012-5-7 in my test echo but 0000-00-00 in the database. I have tried unsuccessfully to use in a numberof suggested versions of: strtotime() mktime Any help would be extremely appreciated. I am aware that I need to validate this data and insure that it is a valid date. That I'm okay with. I would like some help on getting it into the database. I have a calendar select date function for my form that returns the date in the calendar format for USA: 02/16/2012. I need to have this appear as is for the form and in the db for the 'record_date' column, but I need to format this date in mysql DATE format (2012-02-16) and submit it at the same time with another column name 'new_date' in the database in a hidden input field. Is there a way to do this possibly with a temporary table or something? Any ideas would be welcome. Doug Hello all, Can anyone help me , the below code which creates a drop down list for a user works fine on a desktop but on a mobile seems to be random the course it selects
Thanks in advance
<?php
// database connect
include('../db_connect.php');
// We need to use sessions, so you should always start sessions using the below code.
session_start();
// If the user is not logged in redirect to the login page...
if (!isset($_SESSION['loggedin'])) {
header('Location: login.php');
exit();
}
// if submit button selected run code
if(isset($_POST['submit'])){
$_SESSION['district'] = $_POST['district'];
header('Location: add_d_choice.php');
}
?>
<!DOCTYPE html>
<html>
<!-- Compiled and minified CSS -->
<link rel="stylesheet"
href="https://cdnjs.cloudflare.com/ajax/libs/materialize/1.0.0/css/materialize.min.css">
<!-- Compiled and minified JavaScript -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/materialize/1.0.0/js/materialize.min.js"></script>
<!--Import Google Icon Font-->
<link href="https://fonts.googleapis.com/icon?family=Material+Icons" rel="stylesheet">
<header>
<?php
include('../menu.php');
?>
</header>
<body bgcolor=#fafafa>
<br>
<br>
<div class="container">
<div class="row">
<div class="card col s10 m8 l6 xl4 push-l3 push-m2 push-s1 push-xl4 z-depth-5">
<div class="card-content">
<div class="input-field col s12 m12 l12 xl12">
<form action="add_choice.php" method="POST">
<select name = "district">
<label>Select District you raced at</label>
<option value="" disabled selected>Choose your district</option>
<option value="Z">Zwift TT - Z</option>
<option value="">--------------------</option>
<option value="A">Central A</option>
<option value="B">East - B</option>
<option value="C">Lincolnshire - C</option>
<option value="D">Liverpool - D</option>
<option value="E">London East - E</option>
<option value="F">London North - F</option>
<option value="G">London South - G</option>
<option value="H">London West - H</option>
<option value="J">Manchester - J</option>
<option value="K">Midland - K</option>
<option value="L">North - L</option>
<option value="M">North East - M</option>
<option value="N">South East Midlands - N</option>
<option value="O">North Midlands - O</option>
<option value="P">South - P</option>
<option value="Q">South East - Q</option>
<option value="R">South Wales - R</option>
<option value="S">South West - S</option>
<option value="T">Teeside - T</option>
<option value="U">West - U</option>
<option value="V">Yorkshire - V</option>
<option value="W">Scotland - W</option>
<option value="Y">National - Y</option>
</select>
<input class="btn teal accent-4" type="submit" name="submit" value="Submit">
<br>
<br>
</form>
</div>
</div>
</div>
</div>
</div>
</div>
<!-- Javacript -->
<script src=https://code.jquery.com/jquery-3.4.1.min.js></script>
<!-- Compiled and minified JavaScript -->
<script src="https://cdnjs.cloudflare.com/ajax/libs/materialize/1.0.0/js/materialize.min.js"></script>
<script>
$(document).ready(function(){
$('.sidenav').sidenav();
$('select').formSelect();
$('.datepicker').datepicker();
$(".dropdown-trigger").dropdown();
});
</script>
</body>
<footer>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<br>
<?php
include('../footer.php');
?>
</footer>
</html>
$value0 = $_POST['date1'];
$value1 = $_POST['date2'];
$value2 = $_POST['date3'];
$sql = "INSERT INTO Datetable (startdate,enddate, total)
VALUES ('$value0','$value1', '$value2')" ;
$result = mysqli_query($sql);
Hi, I was wondering is it possible to not insert values in MySQL if form entry is left blank? Right now if I dont enter any values in form for my dates than MySQL entry shows 0000-00-00 but I need it not to show anything.
I am working on a form with php generated values. What I have right now is as follows:
<tr>
<td><p>ENTRY TERM: <select name="entry_term" >
<option value="">Choose One:</option>
<?php
$year=date("Y");
for ($y=0;$y<5;$y++) {
echo '<option value="Fall '.($year+$y).'">Fall '.($year+$y).'</option>';
echo '<option value="Spring '.($year+$y+1).'">Spring '.($year+$y+1).'</option>';
}
?>
</select>
</p></td>
This currently generates a list of :
Fall 2014
Spring 2015
Fall 2015
Spring 2016
Fall 2016
...etc for next 5 years
What I need to do is have this time sensitive to exclude once a specific month has passed. For example once August 2014 has started "Fall 2014" should be excluded... or when Jan 2015 has started "Spring 2015" should be excluded.
Can someone help me solve this as I am unsure of how to move forward with this.
Thanks,
Hi People. I am building a logbook application and have a form field where the user will enter a date. However, I would like to simplify it for the user and give them a button which would add "todays" date into the box for them. Please could someone tell me the code to do so. The date format in my DB is YYYY-MM-DD. The form would also need to take input from the user if they were adding info from a different day instead of today. Thanks in advance. Code: [Select] <input type="text" name="date" id="date" size = "25"/> </label> <input type="submit" name="today" id="today" value="Add Today" /> 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 || $currentdate == 2014-02-08) { } Thanks very much, Jack Hi, I have the date stored in my datrabase like this: 2010-04-03 I am then outputting the date like this: Sat 13 Nov This is done using this code: list($year,$month,$day) = explode('-',$Coursedate); $time_stamp = mktime(0,0,0,$month,$day); $Coursedate2 = date("D d M",$time_stamp); The problem I have come across is when listing a date for next year. The database has a value of 2011-01-29 which is a Saturday but it is outputting it as a Friday: Fri 29 Jan The 29th of January 2010 is a Friday but 2010 is a Saturday. Any idea what is wrong with the code? Code: [Select] <?php $dayNames = array("Nedelja", "Ponedeljak", "Utorak", "Srijeda", "Četvrtak", "Petak", "Subota"); $dan = $dayNames[date('N')]; echo $dan.", ".date('d.m.Y.') ; ?> This is my code for showing days in my language, but it works on every day except sunday! Does anybody have any clue? Thanks! Hi Guys...i have this problem... I have this page where user can type in their date of birth...I would like to convert the date of birth input by the user to the DATA format which is defined by phpmyadmin (YYY-MM-DD) so that i can put into the database... My main objective of this is to calculate age... I have this form to allow use to type date of birth: Code: [Select] <label for="dob">D.O.B.: </label> <input type="text" name="dob" id="dob" class="regfields"/> This is to store the input to a variable: Code: [Select] $adddob = $_POST['dob']; and finally this query to insert all the input into the database: Code: [Select] $query = "insert into emp (FNAME, LNAME, CDSID, PAYNO, MAIL , TELNO , DOB , BRANCH , LICNO , CLASS , VFROM , VTO , EID , PASS , PIN) values ('$addfname','$addlname','$addcdsid','$addpayno','$addmail','$addtelno','$adddob','$addbranch','$addlicno','$addclass','$addvfrom','$addvto','$addeid','$addpassword','$addpin')"; Thanks in advance...If this is possible then I will set my dob column to the DATA format... Hi, I'm preparing a code for an automated email, but the date time is not working correctly. I wanted to do a cron scheduler that emails people twice a day - 9am and 9pm. 9am sends email about records logged starting 9pm yesterday to 9am, while 9pm sends email about records logged from 9:01am to 9:00pm. The database uses date/time, so I was doing an if statement of date time, but I wasn't getting the result I wanted. Sometimes the $now time is not within the yesterday 9pm to the 9pm today range. I don't know what I'm doing wrong, but I'm sure that it has something to do with the if statement, as the some of the $now time enters the else statement Here is my code: <?php $today = date('Y-m-d 00:00:00'); $nineyesterday = date('Y-m-d H:i:s', mktime(date("H") - (date("H") + 6), date("i") - date("i"), date("s") - date("s"), date("m") , date("d"), date("Y"))); $now = date('Y-m-d H:i:s', mktime(date("H") + 13, date("i"), date("s"), date("m") , date("d"), date("Y"))); $nineam = date('Y-m-d 09:00:00'); $nine30 = date('Y-m-d 09:30:00'); $ninepm = date('Y-m-d 18:00:00'); echo '9pm yesterday: '.$nineyesterday; echo '<br/>'; echo '9am: '.$nineam; echo '<br/>'; echo '9pm: '.$ninepm; echo '<br/>'; echo 'now: '.$now; echo '<br/>'; if ((strtotime($now) > strtotime($nineyesterday)) and (strtotime($now) <= strtotime($nineam))) { //count the answers by yes or no echo 'am'; } elseif ((strtotime($now) > strtotime($nineam)) and (strtotime($now) <= strtotime($ninepm))) { //count the answers by yes or no echo 'pm'; } else { echo 'error'; } ?> Thanks in advance I have such a little code. If I echo $order_date, I receive normal date in YYYY-MM-DD format. But when I use $order_date in mysql query it is not showing me anything. I do not receive any error or warning also. Please help me. $query = "SELECT Product,Amount FROM orders WHERE Date='$order_date'"; $result = mysql_query($query) or die(mysql_error()); while($products_list = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $products_list["Product"] . "</td>"; echo "<td>" . $products_list["Amount"] . "</td>"; echo "</tr>"; }
//$start_point = '2014-07-22';
$lastday = date('t',strtotime($start_point));
$month_val = date('n',strtotime($start_point));
echo "Last day: ". $lastday . "<br/>";
echo "Month: ". $month_val . "<br/>";
I have a problem with a date value read from a MySQL database table into a PHP variable but when manipulating the vale to determine the month of the date, unexpected results appear. I am not sure what causes this but I suspect PHP is not properly handling date to string conversion.
The output from the code shows 1 as the month when it should be 7. Hower the line of code with the comment "//$start_point = '2014-07-22' if added in give the correct result. When the value of the variable "start_point" is printed, it gives the value "2014-07-22".
Please see the code
Telly
I am trying to get this script to display the date like Dec 17, 2010 and have the search function work too. If I try to change the date the search malfunctions and if I take the date function away the date is 2010 12 17 I thought we had this nailed down the other day but in my excitement I neglected to test the city search. Here is the script. <?php $find = trim($_GET['find']); $field = $_GET['field']; if($find && $field) { // we have search form submitted // check for values to prevent sql injection $valid_fields = array("venue_state", "venue_city", "start_date"); if(!in_array($field, $valid_fields)) die("Error: Invalid field!"); //connect mysql_connect("localhost", "arts_cshow", "TrPh123Yuo") or die(mysql_error()); mysql_select_db("arts_shows") or die(mysql_error()); echo "<h2>Search Results for $find in $field</h2>\n"; $find = addslashes($find); $result = mysql_query("SELECT * FROM craft_shows WHERE $field LIKE '%$find%'"); if(mysql_num_rows($result) == 0) { echo "<p>0 matches found.</p>"; } else { echo "<table><tr><td class=\"shows\">"; $sql = "SELECT *, DATE_FORMAT(`start_date`, '%b %e, %Y') AS s_date FROM craft_shows"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "<a href='/show_submits/show_detail.php?id={$row['id']}'>Details</a>\n"; echo $row['venue_state'] . " {$row['s_date']} {$row['show_name']} {$row['venue_city']}<br>\n"; } echo "</td></tr></table>\n"; } } ?> thanks Hi all.. I'm grabbing the following date format from an rss feed: Sun, 16 Jan 2011 00:00:00 -0800 What's the best way to strip away the first 5 characters (Sun, ) and the last 15 characters ( 00:00:00 -080)... but still keeping this as a date (not a string) to store in my database? Thanks in advance for any ideas! hy every one i have two files one is index.php and second insert.php as follow, index.php Code: [Select] <html> <body> <form action="insert.php" method="post"> Time <input type="text" name="time" /> <input type="submit" /> </form> <?php $ctime = date('h:i:s A'); echo "Current time is : " . $ctime ; echo "<br>"; $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("time", $con); $result = mysql_query("SELECT * FROM time"); while($row = mysql_fetch_array($result)) { echo date('h:i:s A', strtotime($row['time'])); echo "<br />"; } mysql_close($con); ?> </body> </html> here is my insert.php file code, Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("time", $con); $sql="INSERT INTO time (time ) VALUES ('$_POST[time]')"; if (!mysql_query($sql,$con)) { die('Error: ' . mysql_error()); } echo "1 record added"; echo "<br>"; echo "<a href='index.php'>Home</a>"; mysql_close($con) ?> this code work fine but the form which submit the value of time in database whos code is in index.php file above, i have to enter time in the form text field like this 17:00 and it save the value in database but, i want to enter time in the form text field like this 05:00 PM which automatically save the value in database as 17:00 what should i do with the code help me please thanks I have a form where you can pick all the days in the week. When the form is posted my code finds the date of the day you picked that most recently went by. Then the code takes that date, and echos the date of every day of the week that you picked that comes after that date for a year. But for some reason he thinks we're in 2009, even though i know the date puts out that we're in 2010. Here's the code: <?php //function if(isset($_POST['submit'])) { function nextWeeksDay($date_begin,$nbrweek) { $nextweek=array(); for($i = 1; $i <= $nbrweek; $i++) { // 52 week in one year of course $nextweek[$i]=date('d-m-y', strtotime('+'.$i.' week',$date_begin)); } return $nextweek; } //Get what user posted as day, and find the date of the past day gone. $roday = $_POST['day']; echo $roday; echo date('d-m-y',strtotime('last ' . $roday)); $sistdag = date('d-m-y',strtotime('last ' . $roday)); /// end function /// example of a select date // var $date_begin = strtotime($sistdag); //D Day Month Year - like function format. $nbrweek=52; // call function $result=nextWeeksDay($date_begin,$nbrweek); // Preview for($i = 1; $i <= $nbrweek; $i++) { echo '<br> - '.$result[$i]; } } ?> All help appreciated! |
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