PHP - How To Echo And Insert At The Same Time

Hello,

I want to display a weekly events day and time. This event will always be on Saturday @ 10:45 am however, I want to output:
2011-09-17 10:45
then when that date is passed - I want it to change to the next Saturday:
2011-09-21 10:45
etc.

Any ideas?

hi im trying to insert the username and time into the online table the trouble is nothing is getting inserted into the table no errors
have echoed both variables to check they contain the right information and they do.
Code: [Select]
$time=time();
echo $time;
echo $myusername;

$insert_user=mysql_query("INSERT INTO cvisinko_db.online (username,time,) VALUES ('$myusername','$time')") ;

then the plan is display a basic online list
haven't been able to test this as nothing is getting inserted into the online table
Code: [Select]
$data = mysql_query("SELECT count(*) from cvisinko_db.online WHERE `time` >= DATE_SUB(NOW(), INTERVAL 30 minute);") or die(mysql_error());

Hi i need to count the number items in an album now on the album i have create a field called count which counts the number items in that album

but now i dnt knw how to update those as soon insert a item in the table item

so i thought woul be possible to create a script that when insert data into item tables also update album table row count by adding one like update count with A now A equal row count ++1  any help on start of it if possible???/

I have a form right now that Updates a table on my database.

I was wondering if I could possibly still have that working but also Insert the same information into another table on the database when the submit button is pressed?

I insert a time value using the following code     $ti= ( !empty($_POST['time']) ) ? "'{$_POST['time']}'" : 'NULL';    I want to insert a second time value into a differen column which would be the same time minus 1 hour   something along the lines of :   $ti2= ( !empty($_POST['time']) ) ? "'{$_POST['time'] -1 hour}'" : 'NULL';    What would be the correct way to do it

What needs to be changed to make this work...
Code: [Select]
// Build query.
$q2 = "UPDATE member
SET logged_in=?,
last_activity=time(),
updated_on=NOW()
WHERE id=?
LIMIT 1";

Apparently PHP doesn't like time()   


Debbie

Code: [Select]
$date = date('m-d-y');
$ip = $_SERVER['REMOTE_ADDR'];

mysql_query("INSERT INTO users VALUES ($username, $password, 0, $ip, $date)") or die(mysql_error());

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 '.60.116, 03-06-11)' at line 1


I'm not sure why I get this error. :/

I'm not looking for code, mostly just some direction, but I'm sure I'll be back as I try to shape the code.  Also, I know very little about coding jQuery, Ajax, etc.  Here is what I'm looking for:

Basketball coach enters a Page that only he (or admin) can see.  (I have that part figured out.)
Once there, he has a form to fill out with a player's name and other information.

I'd rather there not be 12 rows of empty cells.  (Coaches will view that as work)
I'd rather not have the Page reload on each "Add", but that would be the lesser of two evils.

Is there a way to keep a persistent connection so when a coach hits Add, the player gets added to the database, a new empty form shows up, and the player's information shows up below on a roster?

OK, have no idea what's going on... I've done this a million times... why wont this output!??
I must have a major brain meltdown and dont know it yet!!!

Code: [Select]
<?php

// this echoes just fine:
echo $_POST['testfield'];


// but this wont echo:
echo 
if (isset($_POST['testfield']))
{
$_POST['testfield'] = $test;
}
echo $test;


/// or even this DOESNT echo either!:
$_POST['testfield'] = $test;
echo $test;

?>

Hi All,

I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result.

Please help?

<?php   
$mysqli = mysqli_connect("removed", "removed", "removed", "removed");  
$sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'";  
$result = mysqli_query($mysqli, $sql);   
echo {$result['message']}; 
?>

Can anyone tell me why this is not INSERTing? My array data is coming out just fine.. I've tried everything I can think of and cannot get anything to insert.. Ahhhh!

<?php
  $query = "SELECT RegionID, City FROM geo_cities WHERE RegionID='135'";
  $results = mysqli_query($cxn, $query);
  $row_cnt = mysqli_num_rows($results);
  echo $row_cnt . " Total Records in Query.<br /><br />";
  if (mysqli_num_rows($results)) {
      while ($row = mysqli_fetch_array($results)) {
          $insert_city_query = "INSERT INTO all_illinois SET state_id=$row[RegionID], city_name=$row[City] WHERE id = null" or mysqli_error();
          $insert = mysqli_query($cxn, $insert_city_query);
          if (!$insert) {
              echo "INSERT is NOT working!";
              exit();
          }
          
          echo $row['City'] . "<br />";
          
          echo "<pre>";
          echo print_r($row);
          echo "</pre>";
      }
      //while ($rows = mysqli_fetch_array($results))
      
      } //if (mysqli_num_rows($results))
      
      else {
          echo "No results to get!";
      }
?>

Here is my all_illinois INSERT table structu

CREATE TABLE IF NOT EXISTS `all_illinois` (
  `state_id` varchar(255) NOT NULL,
  `city_name` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

Here is my source table geo_cities structu

CREATE TABLE IF NOT EXISTS `1` (
  `CityId` varchar(255) NOT NULL,
  `CountryID` varchar(255) NOT NULL,
  `RegionID` varchar(255) NOT NULL,
  `City` varchar(255) NOT NULL,
  `Latitude` varchar(255) NOT NULL,
  `Longitude` varchar(255) NOT NULL,
  `TimeZone` varchar(255) NOT NULL,
  `DmaId` varchar(255) NOT NULL,
  `Code` varchar(255) NOT NULL
) ENGINE=MyISAM DEFAULT CHARSET=latin1;

I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: 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 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1

query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '')


Code: [Select]
if (empty($errors)) {

require_once ('dbconnectionfile.php');

$query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company)
 VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8',
'$description9', '$description10', '$description11')";

$result = @mysql_query ($query);


if ($result) {

$who_donated=mysql_insert_id();

$query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why)
 VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')";
$result2 = @mysql_query ($query2);

if ($result2) {echo "Info was added to both tables! yay!";}

echo "table one filled. Table two was not.";
echo $who_donated;

//header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id");

exit();
} else {
echo 'system error. No donation added';

PHP date and time function is not showing correct time on my local system

I have the following php code

date_default_timezone_set("Africa/Lagos");

$date = date('d-m-y h:i:s');

echo "Server Time ".$date ."<br>";

echo "The time is " . date("h:i:sa")."<br>";

$current_datetime = date("Y-m-d") . ' ' . date("H:i:s", STRTOTIME(date('h:i:sa')));

echo "Current time1: ".$current_datetime . "<br>";

 

Output

Server Time 21-05-21 09:55:39
The time is 09:55:39am
Current time1: 2021-05-21 09:55:39

 

Expected Output

Server Time 21-05-21 10:55:39
The time is 10:55:39am
Current time1: 2021-05-21 10:55:3

 

Any help would be appreciated.

 

Edited May 21 by Ponel

I am trying to simulate an ad expiration and carry out an action if the ad is expired. And I cannot get the if/else to work properly... I've tried many variations and I cannot see what I am doing wrong here. Any tips please 3 hours and counting of no solution!

$ad_start = time()-14 . "<br />"; // 14 days from today in the past (negative)
echo $ad_start;

$current_time = time() . "<br />"; // current epoch time
echo $current_time;

$days_past = $ad_start - $current_time; // days past

echo "<br />$days_past days have past since the ad started!<br />";

if($days_past <= 14)
{
echo "<br />Ad is less than 14 days.  Not expired.";
}
else
{
echo "<br />Ad is over 14 days. Expired.";
}

Basically I have recently been playing around with parsing a csv file. What I am looking to do at this point is simply take the date/timestamp (part of the csv file), which is in the following format:DD/MM HH:MM:SS.100th/s
For the sake of argument, lets say I have this in an array string called $csv[0] and the file has several lines that span the course of a couple hours. I wouldn't mind having to use explode() to breakup/remove the date or 100th/s IF that would make things a lot simpler.

So where would I start in trying to achieve this?. The result I am looking for will simply return "X Seconds". Storing this in a string variable would be a bonus, as I plan to use this to divide a separate piece of information.

Any examples or ideas would be great. Thank you.

ps:
Here is an example time from the csv file itself:
Code: [Select]
11/19 22:23:18.143