[[SOLVED] Stupid LEFT JOIN not working]

you could try this

$db->query("SELECT item.id, item.name, order.amount, order.item_id FROM `order` LEFT JOIN `item` ON order.item_id = item.id WHERE order.user_id='$user_id'") or die(mysql_error());

 

note the double quotes arround the query string and single quotes around the php variable

 

print_r($iteminf); should tell you whats exactly in the returned array

[[SOLVED] Mysql is rounding up my currency values problem]

if the column was float(6,2) that should do it for you

[open new window maximised]

I used this in one of my sites and the pop up window opened in the middle of the page

 

function PopupMe(num){
myleft=(screen.width)?(screen.width-800)/2:100;
mytop=(screen.height)?(screen.height-600)/2:100;
settings='top=' + mytop + ',left=' + myleft + ',width=500,height=400,location=no,directories=no,menubar=no,toolbar=no,status=no,scrollbars=yes,resizable=no,fullscreen=no, titlebar=no'
var address = "designmode.php"; //put this in for you
popupwin=window.open(address,'', settings);
popupwin.focus();
}

 

then call the function onClick

[open new window maximised]

thye only way you can do that is to stipulate the height and width of the window, work out the minimum size the window has to be ot show all the information then use those sizes when you create the window

onClick = "window.open('designmode.php', '', 'chrome=yes, resizable=yes,width=350,height=250')

 

[[SOLVED] get next id in mysql...]

when you wrote the function you stipulated that you would supply it with $menu

 

function get_menu_id($menu) {

 

and when you call it you dont give it that information

 

echo get_menu_id()

 

it fails because it wants to know what $menu is, you dont use it in the function anyway so I am confused as to why you want it

[Adding a date to mysql from php]

what is the datatype for the column you are adding the date too????

[Syntax Error]

it is telling you that the query cannot run because the variable $user does not exist

 

try echo out the query instead of running it and see what I mean

 

try

$update = "UPDATE `user` SET `weight` = 'weight' + $weight WHERE `username` = '$user'";
echo $update;

 

you will find that there is no information for $user

[[SOLVED] if statement in QUERY]

SELECT * FROM news WHERE title LIKE '%".$_GET['q']."%' OR content LIKE '%".$_GET['q']."%' AND pageid = 3496 ORDER BY pageid DESC LIMIT 10

[the .open function]

you could try and change the check line to read

 

if (XMLreq)

 

because the "if" statement returns either true or false, if the item exists the it is true (code above) if it does not exist then it will fail

[First Use]

when I first started ( and in fact I still do ) i installed and use <a href="http://www.apachefriends.org/en/xampp.html">XAMPP</a>. It installed everything I needed to get going, PHP, MYSql, and Apache server. it also installed PHPMyAdmin which is a graphical interface for creating and editing tables. it comes with installation instructions and everything to get started I believe

 

regards

 

paul

[SMTP Servers]

http://www.softstack.com/freesmtp.html

[[SOLVED] concatenating a variable to use for sql search]

you could change this line

$data.= "\")";

 

to

$data.= "\") or die ("error in query " . mysql_error))";

 

and see what errors come up, i think the main problem will be in the following code

 

$data = "mysql_query(\"SELECT * FROM music WHERE";
if ($composition != "") {
$data.= " composition LIKE '$composition'";
}
if ($origin != "") {
$data.= " AND origin LIKE '$origin'";

 

if you have a composition and an origin your query string will be

SELECT * FROM music WHERE composition LIKE blah AND origin LIKE blah "

 

whereas if you dont have a composition but have an origin your query string will be

 

SELECT * FROM music WHERE AND origin LIKE blah "

 

and will fail big style

 

HTH

[the .open function]

I kno wnothing about AJAX, you are wondering why I am answering here then, I will try and explain using a small amount of programming knowledge

I have been looking around on the web for information....

 

1.  xxx.open is a function call, that is fact

2.  for your function to fail it must be that your XMLreq is not an object that .open works with...

3.  this tells me that it was not created properly in the first place.

4.  something has to be there or the "!= null" would have failed

5.  I cant seem to find a function call named "new AJAX()", I have found a xmlReq = new XMLHttpRequest(); ..

 

I might be way off but thats what I thought ....

[Cut off text at designated character amount.]

$newstring = substr($oldstring,0, 200);

[[SOLVED] Question about str_replace]

<?php
function Encrypt($text) {
$length=strlen($text);
for ($i = 0; $i < $length; $i++) {
$char = ord($text[$i]); //gets the ascii character number of the letter
$char1 = chr($char +1); //adds 1 onto the ascii charachter a=b f=g etc
$output .= $char1; //remakes the string
}
return $output;
}

$text = "hello world";
echo Encrypt($text); //prints out ifmmp!xpsme
?>

 

rewrite it slightly to decrypt the line by making it minus1

[quick function help]

echo $err;

$multivar = func("blah");

function func($func) {

$returnArray[0] = "bar";
$returnArray[1] = $func;

return $returnArray;

}

echo $multivar[0]; //prints out "bar"
echo $multivar[1]; //prints out "blah"

 

HTH

[[SOLVED] if statement]

= assigns a value to a variable...

== checks to see if two values are equal

=== checks to see if two things are exactly the same value and type

 

in an if statement you are checking for equality and not assignation - use == not =

 

HTH

[[SOLVED] Checking if the query is empty?]

$query = "Blah blah blah"; 
$result = mysql_query($query) or die ("Error in query " . mysql_error());
$rows = mysql_num_rows($result);
if ($rows < 1)
{
echo "Oops no rows returned";
}
else
{
blah blah blah echoing out your data
}

 

hope that helps

[Parse error]

try this

<?php  
	  $sql = "SELECT * FROM articles";
	  $run = mysql_query($sql) or die ("error in query " . mysql_error()); //just in case you query is duff
	  while($data = mysql_fetch_assoc($run))
	  {
	  echo "<a href=\"index.php?p=articles&name=".$data['name']."\">".$data['name']."</a>";
	  }
	   ?>

 

if you are using fetch_array you have to use the numbered key of the data you wish to get (i.e $data[1]), using fetch_assoc allows you to use the name of the column for the data(i.e $data['name']), but either way unless you use $data it wont understand you are calling for information from an array object

 

hope that helps

[[SOLVED] Using multidimensional array]

ok here we go,  will try for you

 

array('First Name'=>$row['FirstName'],'Last Name'=>$row['LastName']);

 

means

 

1. create an array object

2. lets say that $row['FirstName'] is "Fred"

3. create a key called 'First Name' and make the value of that key equal Fred (that's what => means)

 

[[SOLVED] how? after form proccessed, replace echo with a redirect to thankyou page?]

if it works how you want it can you click the "SOLVED" button?

[Is it bad design to join 5 tables?]

the whole idea is to reduce the size of data that is stored, if you have 5 tables that are normalised and dont produce any duplicated information then it is quite all right to use a query that joins the five tables to get the dynamic information

[[SOLVED] how? after form proccessed, replace echo with a redirect to thankyou page?]

if (mail($target_address, $subject, $message, $headers)) {
header('Location: thankyou.html');
exit;
}

[foreach repeating the data twice from mysql]

I have just been playing with it as it was printing the data twice for me too until i changed this line

 

while($question = mysql_fetch_array($questions_datasource))

 

to

 

while($question = mysql_fetch_assoc($questions_datasource))

 

might be worth a go

[inserting date error]

it could be if you have set your due_date column as a datetime then it will be expecting it to come at it in the format yyyy-mm-dd, you could try

 

$sql = "INSERT INTO invoices SET client_id='$_POST[c_id]', service_name='$_POST[s_name]',
total_price='$_POST[t_price]', current_date='$date' , due_date=DATE_fORMAT('$_POST[d_date]' , '%Y-%m-%d')";