PHP - How Do Insert Date() Into Mysql Db?

Hi, new to php so please excuse the simplicity of my questions.

I have a menu list that inserts data into a field, is it possible with php to insert a date into another field dpendent on value? For example:

MenuList - Status1, Status, Status3 inserted into tblfieldStatus when selected

3 other fields in same table are
Status1DateNow
Status2DateNow
Status3DateNow

If a user selected Status2 from menu then Status2Date would be populated with the date today?

Kind regards
     

Hi
I have started my error checking for my form.  I am understand that the date must be inserted in the format yyyy/mm/dd but on my form i need it to be inserted in the format dd/mm/yyyy.  I have wrote some code but have only been doing php about a week properly so I can't see what my error is.  Whenever I try to insert the date, the form its self does not error but if i check the database the date is simply 0000/00/00.   Can anyone help?


if(isset($_POST['submit'])) {
     $first_name = mysql_real_escape_string($_POST['first_name']);
  $last_name = mysql_real_escape_string($_POST['last_name']);
   $DOB = mysql_real_escape_string($_POST['DOB']);
  $sex = mysql_real_escape_string($_POST['sex']);
 $email = mysql_real_escape_string($_POST['email']);
  $username = mysql_real_escape_string($_POST['username']);
   $password = mysql_real_escape_string($_POST['password']); 
   $agree = mysql_real_escape_string($_POST['agreed']);
    $creation_date = mysql_real_escape_string($_POST['creation_date']);
$user_type = mysql_real_escape_string($_POST['member_type']);
 $access_level = mysql_real_escape_string($_POST['access_level']);
  $validation = mysql_real_escape_string($_POST['validation_id']);
     $club_user = mysql_real_escape_string($_POST['user_type']);
  
 $date_check = "/^([0-9]{2})/([0-9]{2})/([0-9]{4})$/";
 if($first_name == "") {
$message = "Please enter a first name";
$success = 0;
}
else if($last_name == "") {
$message = "Please enter a surname";
$success = 0;
}

else if($DOB =="") {
$message = "Please enter your date of birth.";
$success = 0;
}
else if(!(preg_match("/^([0-9]{2})\/([0-9]{2})\/([0-9]{4})$/", $DOB))) {
$message = "Please enter your birthday in the format dd/mm/yyyy";
$success = 0;
}
else if($email =="") {
$message = "Please enter a correct email.";
$success = 0;
}
else if($username =="") {
$message = "Please enter a username.";
$success = 0;
}
else if($password =="") {
$message = "Please enter a password greater than 6 characters long.";
$success = 0;
}


else {


$DOB = $explode[2]."-".$explode[1]."-".$explode[0];
  
  $password_md5 = md5($password);
$insert_member= "INSERT INTO Members (`first_name`,`last_name`,`DOB`,`sex`,`email`,`username`,`password`,`agree`,`creation_date`,`usertype`,`access_level`,`validationID`) 
VALUES 
('".$first_name."','".$last_name."','".$DOB."','".$sex."','".$email."','".$username."','".$password_md5."','".$agree."','".$creation_date."','".$user_type."','".$access_level."', '".$validation."')";

$insert_member_now= mysql_query($insert_member) or die(mysql_error());

$url = "thankyou.php?name=".$_POST['username'];
header('Location: '.$url);

What I'm basically trying to accomplish is, that the datetime of the user posted information is registered with the Now() function in the query, BUT that the user entered information gets echo'd out with the Date showing ONLY.

What I have is this:
-Just as an example-

INSERT INTO db (date_created) VALUES (now())

 while ($row = mysqli_fetch_array($data)) {

echo '<tr><td>' . $row['date_created'] . '</td></tr>';

}
mysqli_close($dbc);

With this way of course I get Date and Time showcased just like it's inserted into the DB.

What would be a legit way of showcasing just date.

Thanks for help.

EDIT: or is there a better way than using now() to showcase the date when the user entered information was posted?

i have been trying to insert date with when users post there comment but when i echo the date() with the comments..it just display 0000.00.00.00 just like that and when i checked my DB it was like that too..please what can i do.this is my code
Thankd in advance Code: [Select]
<?php
include"header.php";
if(isset($_POST['submit']))
{
$postdate=mktime(0,0,0,date("m"),date("d")+1,date("y"));
$comment=mysql_real_escape_string($_POST['comment']);
if($comment!=='')
{
$ins="INSERT INTO post(post_content,post_date)VALUES('$comment','$postdate')";
mysql_query($ins) or die(mysql_error());
}
else
{
echo"You can not post an empty page";
}
}

Does anybody know how to take the queried standard date format - 2012-04-02 for example - and print it to the page as April 2, 2012? Or at the very least, to switch to 04-02-2012? Trying to find some tutorials online.

Hi. Im searching for a way to enter multiple records at once to mysql databased on a date range. Lets say i want to insert from date 2010-11-23 to date 2010-11-25 a textbox with some info and the outcome could look in database like below.

----------------------------------------------------------
| ID   |   date                |   name   |    amount   |
-----------------------------------------------------------
| 1     |   2010-11-23    |   Jhon     |        1         |
| 2     |   2010-11-24    |   Jhon     |        1         |
| 2     |   2010-11-25    |   Jhon     |        1         |
-----------------------------------------------------------

The reason i need the insertion to be like this is because if quering by month and by user to see vacation days then vacations that start in one month and end in another month can be summed by one month. If its in one record then it will pop up in both months.

Help is really welcome.

Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated

I am trying to use text boxes to insert numbers into the database based on what is inputed.
If I have a string, like this for example:
$variable = 09385493;

And I want to insert it into the database like this:
mysql_query("INSERT INTO integers(number)
VALUES ('$variable')");

When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as
9385493

Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too.

Thanks

I need help badly!

What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more).

Thanks!
Code: [Select]
<?php
   require_once('Connections/strana.php'); 
 mysql_select_db($database_strana, $strana);
 
 ?>
<link href="css/styles.css" rel="stylesheet" type="text/css" />

<table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0">
<tr>
<td align="center">
<form action="" method="post" enctype="multipart/form-data" name="form1">
<table>
<?php
$tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error());

function odgovor1($string)
{
$string1 = explode("/", $string);
echo $string1[0];
}
function odgovor2($string)
{
$string1 = explode("/", $string);
echo $string1[1];
}


while ($row=mysql_fetch_array($tema))
{
$id=$row['prasanje_id'];
$prasanje=$row['prasanje_tekst'];
$tekst=$row['odgovor'];
?>
    <tr>
    <td>
    </td>
    </tr>
    <tr>
    <td class="formaP">
 <?php echo $prasanje?>
   </td>
   </tr>
   <tr>
   <td class="formaO">
      <p>
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>"  />
          <?php odgovor1($tekst) ?></label>
        <br />
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" />
          <?php odgovor2($tekst) ?></label>
        <br />
      </p></td>
</tr>
<tr>
<td>
<br />

</td>
</tr>
<?php 




}
?>
</table>
<input align="left"type="submit" name="submit"
value="Внеси" >
</form>
</td>
</tr>
</table>
prasanje = question
tekst/odgovor = answer

The answer table:
id - primary
question_id - the questions ID whose answer is selected in the radio group
user_id - cookie takes care of this
answer - the value from radio group
date - automatic

I don't understand where the empty value is. I've substituted the variables for text and still have the same problem.

Code:
Code: [Select]
$sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')";
Error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

Hi guys

I have a registration form working fine, my database is as below:

userid
username
email
password
repeatpassword


I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column

this is my code below:

<?php

include ("global.php");


//username session
$_SESSION['username']=='$username';
$username=$_SESSION['username'];

//welcome messaage
echo "Welcome, " .$_SESSION['username']."!<p>";

  if ($_POST['register'])
    {
     //get form data
     $name = addslashes(strip_tags($_POST['name']));

$update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'");
   
}

?>

    <form action='updateprofile.php' method='POST'>
    Company Name:<br />
    <input type='text' name='name'><p />
 
    <input type='submit' name='register' value='Register'>
    </form>


can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes.

thanks in advance
 

I have this code: <?php
$con = mysql_connect("localhost","hhh","hhh");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hhh", $con);

// --------------------
// Avatar insert check
// --------------------

session_start();

$name = $_POST[name];
$group = $_POST[group];
$age = $_POST[age];
$usernameid = $_SESSION[id];

$result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'");

$num = mysql_numrows($result);

if ($num == 0) {
mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp)
VALUES ('', '$usernameid', '$name', '$group', '$age', '0')");

header( 'Location: me/' ) ;
}
else echo 'Sorry, please pick a new name';
?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help?

well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.

INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())

  Unknown column 'timestamp' in 'field list'   i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.   the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.   whats going on please?  

Im trying to insert some values automatically into a table once the form loads, but Im getting an error. Here is the code

Code: [Select]
<?php
$aid = $_GET['aid'];
$sd = $_GET['sd'];


?>


<style>
#message {margin:20px; padding:20px; display:block; background:#cccccc; color:#cc0000;}
</style>

<div id="message">Your notification has been submitted.</div>

<div style="text-align:center ">
<?php
$connection = mysql_connect("localhost",
        "username",
        "password");

    mysql_select_db("articles", $connection); 

$query="INSERT INTO broken_links (articleid, article) VALUES ('$aid', '$sp')";
if (!mysql_query($sql,$con))
  {
  die('Error: ' . mysql_error());
  }
echo "Submitted";
mysql_close($con)
?>

<table border="0" cellpadding="3" cellspacing="3" style="margin:0 auto;" >
 
<input type="submit" id="Login" value="&nbsp;&nbsp;Thank you. Please press to close&nbsp;&nbsp;" onclick="tb_remove()"></td>
</tr>
</table>
</div>

Any help will be appreciated

I have an old site written for PHP 5.4 and under and trying (very trying) to get it to work with PHP 7x without much luck. Due to all the changes in 7 my code is one big error message, but one thing at a time. I cannot get the follow code to work at all, even though it worked in PHP 5.

Error:

QUERY ERROR: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'viewuser.php?u=666' id='member'>THE PREDATOR [666] was added to the hit' at line 1
The Query was INSERT INTO gangevents VALUES('','20', UNIX_TIMESTAMP(),'THE  PREDATOR [666] was added to the hit list.')

I have tried at least 20+ different ways of doing this but just can't get the right syntax to get it inserted into MySQL, the code below is just the latest version.  If I echo the a href line out, it works perfect. 

I am sure it is something ridiculously simple, but I have been 4 hours and counting on this now.

Thanks

gangevent_add_2($gangdata['gangID'], "<a href='viewuser.php?u=".$r['userid']."' ".$csscode[$r['userlevel']-1].">".$r['username']."</a> [".$r['userid']."] was added to your hitlist");

function gangevent_add_2($gang, $text) { 
	global $db; $csscode; 
	$db->query("UPDATE users SET gangevent = gangevent + 1 WHERE gang={$gang}");
	$db->query("INSERT INTO gangevents VALUES('','$gang', UNIX_TIMESTAMP(),'$text')");
}