PHP - Posting From <table>

I have an issue with my script, the date for some reason stopped posting after I changed the format in the date() section, I wanted it to post so it shows month-day-year, and it seems for some reason all it accepts is Y-m-d...

Can someone help me out here please?

Code: [Select]
<?php
require_once('connectvars.php');

  if (isset($_POST['submit'])) {
    // Connect to the database
    $dbc = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
$PO_Date = date('MM "/" DD "/" YY',strtotime($_POST['PO_Date']));
$query = "INSERT INTO ncmr (PO_Date) VALUES ('$PO_Date')";
            mysqli_query($dbc, $query);



// Confirm success with the user
        echo '<p>The date have been successfully entered</p>';
mysqli_close($dbc);

// Clear the score data to clear the form
            $PO_Date = "";
}
echo'<form form id="all" method="post">';
echo '<div id="abd"><span class="b">On:&nbsp;&nbsp;</span><input type="text" name="Added_By_Date" value="" /></div>';
echo '<div id="button"><input type="submit" value="Submit Edits" name="submit" /></div>';
echo '</form>'
?>

TWO questions:

I haven't actually CONNECTED and POSTED.  My form has two input fields that combine in a TOTAL field. I notice that the address bar is carrying TOTAL=input1+input2 when it TRIES to connect.  Do I need to include
the TOTAL field in the database if the info is NOT relavant data? the strg and chick TOTALS? 

Will the database accept PARTIAL data from a form that has 25 field with only 6 being populated for testing?

might have it nevermind.

can anyone help me as to whay I get this error: Problem with the query: INSERT INTO options (item_id, option ) VALUES('1', 'hi' )
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 'option ) VALUES('1', 'hi' )' at line 1

with this code: Code: [Select]
foreach($_POST["user"] AS $key => $val) {
$user = $val;
$id1 = $_POST['id1'][$key];


$sql = "INSERT INTO options (item_id, option )





VALUES('$id1', '$user' )";
$rs = mysql_query($sql) or die ("Problem with the query: $sql <br />" . mysql_error());


} Code: [Select]
<input type="checkbox" name="yes" id="checkme" /><div id = "extra"> <div id="user"> Enter User Names:<br> <input type="text" name="user[]"/><br></div>
<div id="add_user" style="display: none;"><input type="text" name="user[]"/>

Hi guys,

I have a list box.

say the code is
Code: [Select]
<select name="drop1" id="Select1" size="4" multiple="multiple">

    <option value="1">item 1</option>

    <option value="2">item 2</option>

    <option value="3">item 3</option>

    <option value="4">item 4</option>

    <option value="0">All</option>

</select>

The user will select few items.
How do i capture the data through the "POST" method.

Code: [Select]
$data = $_POST['drop'];
would not work coz there is an array if data.

Can any one help or point me in the right direction?

Thanks in advance

How can I post this variable $healspell when I already have an echo within a php?

Here's what I have:



echo '<p>You cast Nature\'s Cure (Recovered <?=$healspell?>)</p>';


It's not printing anything out at the moment, obviously, and if I just put $healspell, it just echos out: $healspell.

Thanks!

i am basically trying to retrieve the response of a chatbot from mysql database with query below. As i run script i don't get any error from the server but it seems the post doesn't take $data2 which is an array that includes the keys "user_id" and "text" that are normally accepted (eg: $dmessage->post('direct_messages/new/wrap_links=true', 'text'->'example', '12345');

As result of the above i can't post to Twitter the chatobot response as the recipient (my other twitter account) doesn't get the answer...what do you think?

Code: [Select]

include_once ('connect.php');

connect_to_database();

$query2 = mysql_query ("SELECT response FROM conversation_log WHERE id = (SELECT MAX(id) FROM conversation_log)") or die (mysql_error());

$row = mysql_fetch_row($query2);

$response = $row[0];

//this is the array with the parameters to be used to send bot responses via Twitter API. It normally takes
$data2 = array("text"=>"$response" , "user_id"=> "$userid");

//this modifies the array values with response e userid fields
$data2["text"] = "$response";
$data2["user_id"] = "$userid";

require_once('config_oauth.php'); // includes the applications Oauth keys

require_once($_SERVER['DOCUMENT_ROOT'].'Program-O/gui/xml/oauth/twitteroauth.php'); //Full path to twitteroauth.php library

$dmessage = new TwitterOAuth($consumer_key,$consumer_secret, $oAuthToken, $oAuthSecret); // create new instance with the credentials

//sends the bot response via Twitter API by going through the array $data2
$dmessage->post('direct_messages/new/wrap_links=true', $data2);

Is there anyway to do this:

my form is posting to self, then checking login creds. if they are incorrect it spits out "invaild bla bla" but if they are correct send them to a different page?

here is the short version of what I have.

if(isset($_POST['submit'])) {

//check inputs vs db

if ($check) {
header("Location: http://". $_SERVER['SERVER_NAME']."/include/process_renew.php");
} else { 
echo "<font color='red'>Username or Password did not match.</font> <br /><br />";
   }
}


}

I tested it using echo first and it worked fine then replaced the echo with header and now when I submit it just refreshes the page.

is this not possible?

You guys helped me create a function to gather information for my database. Now I need some help with posting that information.

Function code and Form:
<?php include("opendatabase.php"); ?>

<?php
function createMatchup($week, $game)
{
$query1 = "SELECT team_name 
FROM schedule, teams 
WHERE schedule.week_id = $week 
AND schedule.game_id = $game 
AND schedule.A_team = teams.team_id";

$query2 = "SELECT team_name 
FROM schedule, teams 
WHERE schedule.week_id = $week 
AND schedule.game_id = $game 
AND schedule.H_team = teams.team_id";

$result1 = mysql_fetch_assoc(mysql_query($query1));
$result2 = mysql_fetch_assoc(mysql_query($query2));

$output = "
<input type=\"text\" size = \"3\" name=\"w$weekg$gameA\"> {$result1['team_name']} vs. {$result2['team_name']} 
<input type=\"text\" size = \"3\" name=\"w$weekg$gameH\">";

return $output;

}

$game1 = createMatchup(1, 1);
$game2 = createMatchup(1, 2);
$game3 = createMatchup(1, 3);
$game4 = createMatchup(1, 4);
$game5 = createMatchup(1, 5);
$game6 = createMatchup(1, 6);
$game7 = createMatchup(1, 7);
$game8 = createMatchup(1, 8);
$game9 = createMatchup(1, 9);
$game10 = createMatchup(1, 10);
$game11 = createMatchup(1, 11);
$game12 = createMatchup(1, 12);
$game13 = createMatchup(1, 13);
$game14 = createMatchup(1, 14);
$game15 = createMatchup(1, 15);
$game16 = createMatchup(1, 16);

mysql_close($con);

?>

<form action="insert_spreads.php" method="post">
<?php echo $game1; ?>
<br>
<?php echo $game2; ?>
<br>
<?php echo $game3; ?>
<br>
<?php echo $game4; ?>
<br>
<?php echo $game5; ?>
<br>
<?php echo $game6; ?>
<br>
<?php echo $game7; ?>
<br>
<?php echo $game8; ?>
<br>
<?php echo $game9; ?>
<br>
<?php echo $game10; ?>
<br>
<?php echo $game11; ?>
<br>
<?php echo $game12; ?>
<br>
<?php echo $game13; ?>
<br>
<?php echo $game14; ?>
<br>
<?php echo $game15; ?>
<br>
<?php echo $game16; ?>
<br /><br />
<input type="Submit" value="Submit Spreads">
</form>
I'm wondering if I can use this code for posting to the database (insert_spreads.php):
<?php

header("Location: admin_schedule.php");

include("opendatabase.php");

$w$weekg$gameA=("$_POST['w$weekg$gameA']");
$w$weekg$gameH=("$_POST['w$weekg$gameH']");

$sql="
UPDATE schedule 
SET A_pt_spread= '$w$weekg$gameA',H_pt_spread= '$w$weekg$gameH'
WHERE week_id = '$week'
AND game_id = '$game';

$result = mysql_query($sql) or die(mysql_error().$sql);

mysql_close($con)

?>



Thanks for helping.

hello, i have a while loop that displays multiple tables of data depending on how many are in the database... i have a link if you click it will redirect to another page that prompts you to save an xls file with the table data... this was all working for me before when i had the while loop display rows of data in one table...well i decided that i would like the while loop to display a separate table each time... when i changed it... the xls file only shows one table of data... how can i set up so it will export all tables of data... is there some kind of loop i can set up for it to append the tables??

i have the table data all in a variable $table which i put into a hidden form field that redirects to the xls export page.... any help is appreciated...   

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>

<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>Admin</title>



</head>

<body>

<?php





$sql= mysql_query("SELECT * FROM programs")
           or die (mysql_error());

while($row = mysql_fetch_array($sql)){

$table =

"<table border='1' width='600' style='border-collapse:collapse;'>" .
"<th colspan='4'>Programs</th>" .
"<tr>" .
"<td>ID</td>" . 
"<td>Views</td>" .
"<td>Enabled</td>" .

"</tr>";




$prog_titles = $row['titles'];


$prog_titles = explode(',' , $prog_titles);








$table .=

 "<tr>" .
 "<td>".$row['id']. "</td>" .
 "<td>".$row['views']. "</td>" .
 "<td>".$row['enabled']. "</td>" .  
 "<td></td>" .
 "</tr>" .







 "<tr>" .
 "<td>Title ID</td>" .
 "<td>Name</td>" .
 "<td>Description</td>" .
 "<td>Image</td>" .
 "</tr>" ;
 






foreach ( $prog_titles as $title ) {

$sql2= mysql_query("SELECT * FROM titles WHERE title = '$title' ");
$row2 = mysql_fetch_array($sql2);


$table .=

 "<tr>" .
 "<td>" . $row2['id'] . "</td>" .
 "<td>" . $row2['title'] . "</td>" .
 "<td>" . $row2['description'] . "</td>" .
 "<td>". "<img width='50px' src='images/" . $row2['img'] . "'/></td>" .
 "</tr>" ;



} // close foreach

$table .=
 "</table>";
 
 
 
 echo $table;
 echo "<br /><br /><br />";
 
  
 
 
} // close while


 
?>

<br />
<br />

<FORM action="export.php">
<INPUT type="submit" value="Export to XLS"> 
<INPUT type="hidden" value="<?php print $table;?>" name="export">
</FORM>


</body>
</html>

Hi, i have written a form that uses 3 dropdown boxes, one of them being populated from a database.

The form code is as follows;

Code: [Select]
<form id="pointsForm" name="pointsForm" method="post" action="points-engine.php">

<select name="car" id="car">
<?php
        // result of car name collection
        $result = mysql_query ($query);
        
        //Array stored in $cars
        while($cars=mysql_fetch_array($result)){
            
        //option values added by looping through array
        echo "<option value=$cars[id]>$cars[carName]</option>";
        }
?>
</select>

<select name="season" id="season">
<option value="low">Low season</option>
<option value="mid">Mid season</option>
<option value="high">High season</option>
    <option value="peak">Peak season</option>
</select>
   
    <select name="period" id="period">
<option value="day">One day</option>
<option value="weekend">Weekend</option>
<option value="week">One week</option>
</select>

<input type="submit" name="Submit" value="Submit" /> 

</form>
which posts to;

Code: [Select]
<?php

$car=$_POST['car'];
echo $car;

$season=$_POST['season']; 
echo $season;

$period=$_POST['period'];
echo $period;

?>
The initial form code is working fine, the car field is being populated from the database. My problem is the post value for car isn't working...

points-engine.php is echoing $season & $period fine but not $car

can anyone tell me why?

many thanks
Richard

I have a simple guestbook set up. Even when I set the max post size to 6000000 (if you're wondering why Id allow that its cause I'm just using it as a private message board with one other person), after a certain amount of characters, it splits the post up into several messages and in place of the name of the poster, it's an IP. Any ideas why this is happening?

Thanks.

I have a database in which i store articles that contain all type of special characters including slashes, single quotes, double quote etc.

Previously it was working fine, but now after the updating the xampp, things are not going as usual.

I insert are articles using following code

$query = "INSERT INTO `articles` (`source`, `heading`, `description`, `catagory`, `detail`, `date`, `timestamp`)
          values ('$source', '$heading', '$desc', '$catagory', '$detail', '$date', '$timestamp')";

Error I am getting is

Error Running query 1You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 's economy and emphasized that international community should provide its complet' at line 2

Can anybody tell me whats wrong with my script?

Hi all

I have a form which takes details from drop down menus. But what I need is a wild card if someone leaves the selected option value as "all" it adjusts the SQL query to not search that cell so it brings all entries without filtering.

My form looks like this:

Code: [Select]
<select name="area" id="area">
       <option value="all">--All areas--</option>
      <?php 
   $getareas = mysql_query (" SELECT * FROM `areas` ORDER by name ASC");
   while ($showareas = mysql_fetch_array($getareas)) { ?>
      <option value="<?php echo $showareas['id']; ?>"><?php echo $showareas['name']; ?></option>
      <?php } ?>
    </select>

My SQL query looks lie this:

Code: [Select]
$getads = mysql_query(" SELECT * FROM adverts WHERE categoryid = 1 AND areaid = '".$searchedarea."' AND makeid = '".$searchedmake."' AND modelid = '".$searchedmodel."' AND berth = '".$searchedberth."'

AND live = 1 AND approved = 1 AND paid = 1 AND dateexpired >= '".$todaysdate."' ORDER BY seller ASC , type ASC, id ASC") or die (mysql_error());

So if they select 'all" in the drop down field, the SQL query doesn't have 'AND areaid = '".$searchedarea."'' in it?

Is this possible, or can I add a wildcard into the SQL query?

Many thanks

Pete