[Query duplicate results]

* still returns game* results. First name etc is unique to rounds. Also none of them have the same number of results as game except game. Why game* results. It should only be looking a 1 game ID, which has say 8 people and  entry ID's, hence 8 names.

[Query duplicate results]

$PlayerQuery = "SELECT FirstName, SurName FROM players p, rounds m, entrants e, games t WHERE p.PlayerID = e.PlayerID AND m.GameID = '$ID' AND e.EntrantID = m.Player";
	$PlayerResult = mysql_query($PlayerQuery);
	$PlayerRow = mysql_num_rows($PlayerResult);

 

Can someone suggest why this returns game*the number of results. For each new game it returns an extra duplicate result. For example for 2 games:

j Smith

j Smith

t John

t John etc

 

There are no duplicate entry's! and results should be unique.  Not in a loop (positive)

[Ubuntu 11.04, blank screen when parsing PHP, code used before and fine]

Blank screen. Also I have test using older files I use fine on windows and ubuntu 10.10 before I updated to 11.04. Nothing will run, so I assume it is not a parse error.

[Get next result without array?]

I have a list of ID's returned from a table (even number). I always used a loop to get the next result, but I want to put these straight into a new table

 

INSERT INTO table VALUES ($ID1, $ID2);

 

This is recursive till there are no more ID's. How do I get the result then the next result without loops. ID1 and ID2 cannot be null. What is the best way?

[$row[1] undefined, mysql_num_rows = 2 ?]

loops, sorry i am very over tried, I cant believe the mistakes I have made this evening. Thanks.

[$row[1] undefined, mysql_num_rows = 2 ?]

I know im being blind today but can someone point out why:

$query = "SELECT peopleID FROM numpeople WHERE ID = '3';";
$result = mysql_query($query)
or die("failed.");
$row = mysql_fetch_row($result);
$numpeople = mysql_num_rows($result);
echo "<br /> row $row[1]";
echo "<br />numpeople $numpeople";	

 

$numpeople returns 2, but $row[1] is undefined? $row[0] is fine. (The value of  $row[0] is 7 and  $row[1] should be .

 

This is the second error that I really cant see....

[undefined constant?!]

Sorry, been coding for days to finish this - I think im going a bit insane! Im sure they had $ before.... The universe removed them!

 

Thanks anyway for sorting my stupidity.

[undefined constant?!]

No im not, its a weird font but they all start with $?

 

$entrantQuery etc

[undefined constant?!]

$entrantQuery = "SELECT entrantID FROM entrants WHERE Game = $idT";
$entrantResult = mysql_query(entrantQuery);
$entrantRow = mysql_fetch_row(entrantResult);

 

I have these at the start of my php (after getting $idT), but I get the error

Notice: Use of undefined constant entrantQuery - assumed 'entrantQuery' in C:\Entrants.php on line ..

 

Notice: Use of undefined constant entrantResult - assumed 'entrantResult' in C:\www\Entrants.php on line..

 

Ive used this code lots of times before, what am I missing? They are variables that I am defining!

[GET_REQUEST/POST in a loop var Entry$i]

I have forum options in a loop (ie value = Entry$i)

 

I need to call GET_POST('Entry$i') and loop though the INSERT query $i times. If I put GET_POST in a loop, however, it does not work. It does not give an INSERT failed error so I assume it is not reading it. I have also tried putting if GET_POST('Entry1') outside the loop, but it does not like it. I dont know how many options are needed so I need it to generate the correct number. Then you can select $i options from duplicated option lists.  (Ie) participant 1,  2 and 3 from a list of possibles. 

 

I have also use GET_POST, I use a function. Other sumbits (I not use what they are called, please set me know), work fine on this page.

 

echo "Entry $Number: <select name = \"Entry$j\">";

for ($i = 1; $i<$EntNum+1; $i++)
{ etc etc

then 
for ($i = 0; $i<$NumberOfEntrys; $i++) {
if (isset($_POST['AddEntry']) && isset($_POST["Entry$i"]))

//rest of this bit not executed -just insert

Ideas? Thanks

[Show Next Result in dropdown box]

I dont think the above is explained very well, also If I could add ORDER BY to topic...

 

echo "<Option value = \"$i\">$PersonRows[0] $PersonRows[1] and $Person2Rows[0] $Person2Rows[1] </Option>";

 

I need to iterate though the results of $PersonRows[0] $PersonRows[1] and $Person2Rows[0] $Person2Rows[1]  for each $i

 

At the moment I get

Andy Smith
Andy Smith
Andy Smith

 

 

But I need

 

Andy Smith
Tim Carbrook
Tom Steptoe

 

etc,

 

Also I have

 

$Choicequery ="SELECT PersonA, PersonB FROM modules WHERE GroupID = '$UpdateNumber'"; or
$Choicequery ="SELECT PersonA, PersonB,StartDate FROM modules WHERE GroupID = '$UpdateNumber'";

 

But if I add ORDER BY StartDate, it returns nothing? I really need them ordered....

 

 

 

Any ideas?, thanks

[Why does this not print anything?]

I have got it now, i did need to define them as well, Thanks everyone!

[Show Next Result in dropdown box]

for ($i = 1; $i<$rows+1; $i++)
{
	$Choicequery ="SELECT PersonA, PersonB FROM modules WHERE GroupID = '$UpdateNumber'";
	$Choiceresult = mysql_query($Choicequery);
	$ChoiceRows = mysql_fetch_array($Choiceresult);

	$PersonQuery = "SELECT FirstName, LastName FROM Persons p, modules m, users e, group t WHERE m.modID = $ChoiceRows[0] AND e.UserID = m.PersonA AND p.PersonID = e.PersonID";
	$PersonResult = mysql_query($PersonQuery);
	$PersonRows = mysql_fetch_array($PersonResult);

	$Person2Query = "SELECT FirstName, LastName FROM Persons p, modules m, users e, group t WHERE m.modID = $ChoiceRows[0] AND e.UserID = m.PersonB AND p.PersonID = e.PersonID";
	$Person2Result = mysql_query($Person2Query);
	$Person2Rows = mysql_fetch_array($Person2Result);

	echo "<Option value = \"$i\">$PersonRows[0] $PersonRows[1] and $Person2Rows[0] $Person2Rows[1] </Option>";
}

 

How do I get it to show the next result when $i increases? All the options are the first people A and B $i number of times. Not sure how to use a loop here to go though the results. Whats the best way?

 

ie $PersonRows[0] $PersonRows[1] are the same for every option

[Why does this not print anything?]

Unknown column 'table.PersonID' in 'on clause'

 

?? It is correct i can view the table

[Why does this not print anything?]

SELECT oneID FROM table WHERE table.PersonID = game.PlayerA

 

returns unknown column game.PlayerA

 

but it does exist.

 

Not a blank page, just this value missing, but it may have been the double quotes. Im fairly new to php and have never used table.anything (its a good idea)

[Why does this not print anything?]

no errors

[Why does this not print anything?]

yes, although the names are not the same as the ones im using, but as this is part of an assignment I remained the tables to post here. Player A is a column name in game.

[Why does this not print anything?]

nothing

[Why does this not print anything?]

$query ="SELECT oneID FROM table WHERE table.PersonID = 'game.PlayerA'" ;
$result = mysql_query($query);
$row = mysql_fetch_array($result);
$oneID = $row[0];

[code]

If I then echo "$oneID" why does it not print anything?
$result echos resource7

[Ubuntu 11.04, blank screen when parsing PHP, code used before and fine]

Chromium browser has this:

 

The website encountered an error while retrieving http://localhost/admin.php. It may be down for maintenance or configured incorrectly.

Here are some suggestions:

Reload this web page later.

HTTP Error 500 (Internal Server Error): An unexpected condition was encountered while the server was attempting to fulfil the request.

[Ubuntu 11.04, blank screen when parsing PHP, code used before and fine]

html pages work fine

[Ubuntu 11.04, blank screen when parsing PHP, code used before and fine]

Hiya,

 

I have set up apache, php5 etc on my computers before, but since I upgraded to 11.04 I get a blank screen when running any PHP file.

 

I have used these PHP files before, so the code is not incorrect. I have check that all packages are installed correctly and restarted apache. This is on two ubuntu computers.

 

I normally use firefox.

 

apache2 php5-mysql libapache2-mod-php5 mysql-server