PHP - Problems With Conditionals In Php

Hello,

Alright I am having a few issues here. Let me explain what I am doing...this is a script that allows a customer to split an order..he can split certain items out of one order and it will create another order with the items he split added to this order, and then subtracting the items from the old order. The products are stored in one field called products, they are stored in a delimited list like this qty, product name, price each, product id | qty, product name, price each, product id |etc.

Now, before you say why are you storing your products like that?? The reason I have to store my products like so is because the customer does not want the products stored in the order table to be dependent off the products table like they were before. In that case if he modified a product it would modify all of the orders with that product stored in them. Make sense? I am aware a delimited list probably wasn't the best way to do this but it was the best way I could think of.

Ok here is my list of issues.
  1) Everytime you split an order, it is not creating a new order to store the items that were split into. It created an order the first time I did it, and now all it does is it keeps updating the products field in that order, rather than creating a new order.
  2) Next I do not know how I can update the old products (stored in the $prod array). It needs to update the products qtys after the order was split, and then if all of the qty for that product were moved to the new order it just needs to remove that product from the $prod array rather than updating the quantity.

The code is below. Please let me know if I can provide any more information to help! Thanks everyone!


// Get Old Order
$get_order = @mysql_query("SELECT * FROM orders WHERE order_id = {$_POST['order_id']}");
$order = @mysql_fetch_assoc($get_order);

// Get Old Order Items
$products = $order['products'];
//breaking products text down for display
$prod = array();

$_products = explode('|', $products);
foreach ($_products AS $p) 
$prod[] = explode(',', $p);

/*
$get_items = @mysql_query("SELECT product_id, qty FROM order_items WHERE order_id = {$order['order_id']}");
$items = array();
while(($row = @mysql_fetch_assoc($get_items)) !== false) {
  $items[] = $row;
}
*/

if(empty($prod)) {
  header("Location: tracking.php");
  die();
}


// Create New Order
mysql_query("INSERT INTO orders SET customer_id = {$order['customer_id']}, order_status = {$order['order_status']}, order_date = '{$order['order_date']}', order_date_paid = '{$order['order_date_paid']}', order_shipping = '{$order['order_shipping']}', order_shipping_fee = '{$order['order_shipping_fee']}', order_insurance = '{$order['order_insurance']}', order_insurance_fee = '{$order['order_insurance_fee']}', order_insurance_total = '{$order['order_insurance_total']}', order_grand_total = '{$order['order_grand_total']}', order_date = '{$order['order_date']}', order_filled = '{$order['order_filled']}', order_ship_date = '{$order['ship_date']}'");
$get_new_order = @mysql_query("SELECT MAX(order_id) AS order_id FROM orders") or die(mysql_error());
$new_order_id = @mysql_result($get_new_order, 'order_id', 0);


// Add Items to New Order & Remove Items from Old Order
$new_items = array();
$_new_items = '';
foreach($prod as $p2) {
  for($i = 0; $i < $p2[0]; $i++) { 
    if(!empty($_POST[trim($p2[3]).'_'.$i])) {
      $new_items[trim($p2[3])]++;
    }
  }
}

//construct new static products list
foreach($new_items as $id=>$qty) {
$get_product = mysql_query("SELECT name, price FROM products WHERE product_id = '{$id}'");
$got_product = mysql_fetch_assoc($get_product);
$_new_items .= $qty.','.$got_product['name'].','.$got_product['price'].','.$id.'|';
//echo $id.' - '.$qty.'<br>';
}
//remove last character in products text before going into DB
$_new_items = substr_replace($_new_items ,"",-1);

//update products field in new order
mysql_query("UPDATE orders SET products = '{$_new_items}' WHERE order_id = '{$new_order_id}'");

Hi, I hope someone can help me.  I am trying to generate a script for our Intranet that tells me which birthdays = today.  My field in my table = d/m/y, eg 09/05/2011 (9 May 2011).  How do I get the script to just look at the day & month, not the year?  My script looks like this:

Code: [Select]
<?php

echo date("d/m/Y") . "<br />";
$myDate = date('d/m/Y');
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("dbname", $con);

$sql = "SELECT * FROM detail WHERE dob='$mydate'";
$result=mysql_query($sql);
echo mysql_num_rows($result);

?>

the image in my comments box wont show up it gives me an error

Code: [Select]
Notice: Undefined index: avatar in /home/ecabrera/public_html/profile.php on line 277
and i dont whats wrong with it

Code: [Select]
// display comments
$perpage = 10;
$start=0;
if(@$_GET['s'])
$start = $_GET['s'];


$query = mysql_query("SELECT * FROM profile_comments WHERE profile_id='$getid' ORDER BY id DESC LIMIT $start, $perpage");
$numrows = mysql_num_rows($query);
if ($numrows > 0){

$next = $start + $perpage;
$prev = $start - $perpage;

while($row = mysql_fetch_assoc($query)){
$user_id = $row['user_id'];
$user_name = $row['user_name'];
$comment = nl2br($row['comment']);
$date = $row['date'];
                $avatar = $row['avatar'];

echo "<img src='avatars/$avatar'></img><a href='$site/profile?id=$getid'></a>

<b> on $date</b><br />";
echo "<div style='margin-left: 10px;'>$comment</div><hr>";

}
}
else
echo "This user has no profile comments.<br />";
// end diplay comment area

I'm trying to either update my database if some data matches the if statement, or insert the data in a new row if it doesn't match (if possible).
The code will update the database just fine if the data matches, but if not, it won't insert the new data. (When I leave off the if statement it will insert just fine). I fear I'm not using the if...else correctly.

Thanks for any help.

<?php

header("Location: admin_schedule.php");

include("opendatabase.php");

$date=("$_POST[gamedate]");
$week=("$_POST[Week]");
$game=("$_POST[Game]");
$hometeam=("$_POST[team_name_1]");
$awayteam=("$_POST[team_name_2]");

$check = mysql_query("SELECT week_id,game_id FROM schedule");
$row[] = mysql_fetch_array($check);

if ($row[week_id]='$week' && $row[game_id]='$game')
{
mysql_query("UPDATE schedule 
SET week_id = '$week', game_id = '$game', date = '$date', H_team  = '$hometeam', A_team = '$awayteam'
WHERE week_id = '$week'
AND game_id = '$game'");
}

else
{
mysql_query("INSERT INTO schedule (week_id,game_id,date,H_team,A_team)
VALUES('$week','$game','$date','$hometeam','$awayteam')");
}

mysql_close($con);

I want to gettwitter feeds,
this gives me a xml
http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3

based on this:
http://www.w3schools.com/PHP/php_xml_simplexml.asp

i try to get the xml with:
<?php$xml = simplexml_load_file("http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3");?>()

i get this error:
Quote

Warning: simplexml_load_file(http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3): failed to open stream: HTTP request failed! HTTP/1.1 400 Bad Request in /home/vhosting/x/vhost0033377/domains/doekewartena.nl/htdocs/www/temp/twitter03.php on line 2 Warning: simplexml_load_file(): I/O warning : failed to load external entity "http://twitter.com/statuses/user_timeline.rss?screen_name=fleuragemapvv&count=3" in /home/vhosting/x/vhost0033377/domains/doekewartena.nl/htdocs/www/temp/twitter03.php on line 2


i have no clue, i'm very new to php
can i even use simplexml_load_file cause it has no .xml file type extention.

First
Since I first started learning PHP, I've been writing code for forms like this:

$name = $_POST["name"];

if ($name == "") {
echo "You didn't type anything";
}

I've written scripts like this and tested them on a GOdaddy server, and they always worked fine.

Now, after testing scripts like that on my own server (WAMP2 [PHP5, Mysql, Apache]), I get an error saying "Unidentified index: name", and so I have to nest that if statement AS WELL AS the variable assignment inside another if statement like this:

if (isset ($_POST["name"])) {
$name = $_POST["name"];
if ($name == "") {
echo "you didn't type anything.
}
}

The error doesn't actually stop the script from running, but a medium sized box appears in the browser with the error message contained that's after the element in the document of which is getting an invalid error.

Which way is truly the correct way? Because now that I have to use the second way in order for my scripts to run correctly on my own server, I have to write more code and it all gets disorganized and more complicated to read.

So...is this a configuration thing that I can tweak in WAMP2 to stop messages like that from popping up? Or is the second way the way you're supposed to do it?

Also, I was wondering why ereg got deprecated and what the replacement of it is in PHP5? I've asked before and even googled it but I simply get taken to a "Pearl" function on PHP.net. I'm not sure what it is, or how to use it, and haven't got any hints from either PHP.net or anywhere else.

Any suggestions?

I have a config file that lists a file like so..

Code: [Select]
$file1 = 'storagefolder/mydatabase.txt';
How can I delete the file "mydatabase.txt" that is in another directory 'storagefolder' ?

Code: [Select]
unlink($file1);gives an error that says not such file or directory

I have built a simple website to share photos with people at work.  And it is based on code - known to work - from a larger website that I built a few years ago.

When you land on the site, you have a login screen, and if the username/password match what is hardcoded, then I set $_SESSION['loggedIn'] = TRUE; and I redirect to the menu.php page.

If your credentials do not match, then I redirect the user to an access-denied.php (403) page.

Here is the problem...

Occasionally, when you try to log in you will get routed to the access-denied page.  But then if you try a second time you end on on the menu page.

I uploaded my otherwise working code to my hosted webserver, and now I can never seem to log in.

It seems to me that something is getting screwed up with the session variable?

Any ideas what could be causing this strange behavior?

Hello,   Can anyone tell me what I'm doing wrong here?  I am trying to run a query against a table for scenes from movies.  Each movie would have up to 8 scenes.  For the life of me I can't get this loop to work.  The movie id is being passed in the url, tested to make sure it's being passed and it is.  The code is pasted below. 

require_once ('../mysqli_connect.php');

$movie_id = $_GET['movie_id'];

$q = "SELECT * FROM scenes WHERE movie_id = $movie_id";

$r = @mysqli_fetch_array($dbc, $q);

$row = @mysqli_fetch_array($r, MYSQLI_ASSOC);

while ($row = @mysqli_fetch_array($r, MYSQLI_ASSOC));{
 
echo '<a href=" '.$row['scene_id'].'.mp4 ">'.$row['scene_name'].'</a>';}

In advance thanks a million!!
Edited by Zane, 28 May 2014 - 10:01 AM.

I am having a problem inserting information into my database.  It will not create a record, no matter what I do.  I have used the exact same template for two other pages, and it works fine, but for this one, it simply will not work.  If I echo each variable, it will print out, but even if I reduce the insert to one field, it still will not work.  Please help...  Here is my code.  

<?php
session_start();
?>

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

<?php
require('connect.php');

$ID = $_POST['ID'];
$CustomerFirstName = $_POST['Customer_B'];
$CustomerLastName = $_POST['Customer_A'];
$CustomerAddress = $_POST['Customer_C'];
$CustomerCity = $_POST['Customer_D'];
$CustomerState = $_POST['Customer_E'];
$CustomerZipCode = $_POST['Customer_F'];
$CustomerPhone = $_POST['Customer_G'];
$CustomerEmail = $_POST['Customer_H'];
$SCLastName = $_POST['Service_A'];
$SCFirstName = $_POST['Service_B'];
$SCAddress = $_POST['Service_C'];
$SCCity = $_POST['Service_D'];
$SCState = $_POST['Service_E'];
$SCZip = $_POST['Service_F'];
$SCPhone = $_POST['Service_G'];
$SCEmail = $_POST['Service_H'];
$DateRequested =  date('Y-m-d', strtotime($_POST['Date_Requested']));
$DateScheduled =  date('Y-m-d', strtotime($_POST['Date_Scheduled']));
$AssignedTo = $_POST['Assigned_To'];
$ServiceBranch = $_POST['Service_Branch'];
$Warranty =  isset($_POST['Warranty'])?$_POST['Warranty']:"No";
$WorkRequested = $_POST['Work_Requested'];

$sql = "INSERT INTO WorkOrder (ID, CustomerLastName, CustomerFirstName, CustomerAddress, CustomerCity, CustomerState, CustomerZip, CustomerPhone, CustomerEmail, SCLastName, SCFirstName, SCAddress, SCCity, SCState, SCZip, SCPhone, SCEmail, DateRequested, DateScheduled, Warranty, WorkRequested, ServiceBranch, AssignedTo) 
VALUES ('$ID', '$CustomerLastName', '$CustomerFirstName', '$CustomerAddress', '$CustomerCity', '$CustomerState', '$CustomerZipCode', '$CustomerPhone', '$CustomerEmail', '$SCLastName', '$SCFirstName', '$SCAddress', '$SCCity', '$SCState', '$SCZip', '$SCPhone', '$SCEmail', '$DateRequested', '$DateScheduled', '$Warranty', '$WorkRequested', '$ServiceBranch', '$AssignedTo')";

if ($conn->query($sql) == TRUE) {
   echo "<script type='text/javascript'> alert('Work Order Created'); </script>";
   echo "<script type='text/javascript'> document.location = 'Main.php'; </script>";
} else {
   echo "<script type='text/javascript'> alert('Record Not Inserted'); </script>";
   echo "<script type='text/javascript'> document.location = 'Work_Order.php'; </script>;
}

?>

  </body>
</html>

Hi. I have 2 problems which I'm hopeing someone can help me with.

Hi everyone, I can’t understand what happens…  When I try my site in WAMP, I have the follow errors:

Deprecated: mysql_query(): The mysql extension is deprecated and will be removed in the futu use mysqli or PDO (…)   Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in (…)   The same happened with the connect to mysql, but I solved using mysqli extension. But in this case is totally diferente. When I use the mysqli_query or mysqli_num_rows that are the alternatives presented I receive again other error, in PHP Manual says: "function.mysqli-query doesn't exist. Closest matches:"   Someone know solve this problem???     When I use MySQL

<?php
session_start();
if (isset($_SESSION["manager"])){
	header("location:index.php");
	exit();
}
?>
<?php
if(isset($_POST["username"])&&isset($_POST["password"])){
	
	$manager=preg_replace('#[^A-Za-z0-9]#i','',$_POST["username"]);
	$password=preg_replace('#[^A-Za-z0-9]#i','',$_POST["password"]);
	
	
	$cnn= include "../lojascript/connect_mysql.php";
	$sql=mysql_query($cnn, "SELECT id FROM admin WHERE username='$manager' AND password='$password' LIMIT 1");
	
	$existCount=mysql_num_rows($sql);
	if ($existCount==1){
		while ($row = mysqli_fetch_array($sql)){
		$id=$row["id"];
	}
	$_SESSION["id"]=$id;
	$_SESSION["manager"]=$manager;
	$_SESSION["password"]=$password;
	header("location:index.php");
	exit();
  }else {
	echo 'Informação incorrecta <a href="index.php"> Click here</a>';
	exit();
	}
}
?>

When I use MySQLi

 

<?php
session_start();
if (isset($_SESSION["manager"])){
	header("location:index.php");
	exit();
}
?>
<?php
if(isset($_POST["username"])&&isset($_POST["password"])){
	
	$manager=preg_replace('#[^A-Za-z0-9]#i','',$_POST["username"]);
	$password=preg_replace('#[^A-Za-z0-9]#i','',$_POST["password"]);
	
	
	$cnn = include "../lojascript/connect_mysql.php";
	$query= "SELECT id FROM admin WHERE username='$manager' AND password='$password' LIMIT 1";
	$result= mysqli_query($cnn,$query) or die(mysqli_error());
	$num_row = mysqli_num_rows($result);
	 
	if ($num_row==1)
	{
	while ($row = mysqli_fetch_array($result)){
		$_SESSION["id"]=$row["id"];
	}
	
	
	$_SESSION["id"]=$id;
	$_SESSION["manager"]=$manager;
	$_SESSION["password"]=$password;
	header("location:index.php");
	exit();
  }else {
	echo 'Informação incorrecta <a href="index.php"> Click here</a>';
	exit();
	}
}
?>

Hello,   I have problem with JS slider. Here is demo website http://goo.gl/bPzpAo I can't make the slider to have similar size items (seems cutted). I can't get, what's wrong there? Someone could help me with this?? I tried to resize images, also li elements modifications, but no solution. If someone could help me,  will be much appreciated!   Regards,   Zox

I wrote a simple mysql record update. mysqli_query($link, "UPDATE photo01 SET PhotoText='abcd' WHERE id= '1' "); Work fine until I use a variable: mysqli_query($link, "UPDATE photo01 SET PhotoText= $ZZ  WHERE id= $x "); If $ZZ contains only number it updated the cell. ANY letters does not. The cell is text and standard US ASCII I tried: mysqli_query($link, "UPDATE photo01 SET PhotoText={$ZZ} WHERE id={$x}"); same problem   Here is my loop: <?php
for ($x = 0; $x < 131 ;$x++) {
$ZZ = $_POST["z".$x];
if ($ZZ != '') { mysqli_query($link, "UPDATE photo01 SET PhotoText={$ZZ} WHERE id={$x}");
}
}

?>

Hello Im just new to coding into php and im getting this error? i cant seems to see database into my php ? what wrong need help

 

CREATE TABLE IF NOT EXISTS `payment_details` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `idnumber` varchar(30) NOT NULL,
  `date` varchar(30) NOT NULL,
  `amount` varchar(30) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=1 ;

Here my php code i cant seems to understand what wrong i dont get any error from mysql or from my php  please i dont know what wrong 

 

<label for="filter">Filter</label> <input type="text" name="filter" value="" id="filter" />
								<table cellpadding="1" cellspacing="1" id="resultTable">
									<thead>
										<tr>
											<th  style="border-left: 1px solid #C1DAD7"> Date </th>
											<th>Amount</th>
										</tr>
									</thead>
									<tbody>
								<?php
										include('../connect.php');
									$id=$_SESSION['SESS_MEMBER_ID'];	
										$result = mysql_query("SELECT * FROM student WHERE id='$id'");
										while($row = mysql_fetch_array($result))
											{
												echo '<tr class="record">';
												echo '<td style="border-left: 1px solid #C1DAD7"><div align="left">'.$row['date'].'</div></td>';
												echo '<td><div align="left">'.$row['amount'].'</div></td>';
												echo '</tr>';
											}
										?> 
									</tbody>
								</table>