Syntax error!

[BigX]

Hi guys,

 

I'm kinda new to PHP! Working on a site now but I'm getting the next parse/syntax error:

 

Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in [path removed] on line 95

 

The code I have is:

<?php

$query = "SELECT id_usr FROM ekpoule_users WHERE level_usr = User ORDER BY id_usr ASC";
$berekenen = mysql_query($query);
   
while($row = mysql_fetch_array($berekenen, MYSQL_ASSOC))
{

// Zwitserland - Tsjechie (uitslag)

$query = mysql_query("SELECT uitslag FROM echte_uitslagen WHERE wedstrijd = 'zwitserland_tsjechie'");
$echte_uitslagen = mysql_fetch_array($query);
$query = mysql_query("SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = $row['id_usr']");
$poule_a = mysql_fetch_array($query);
if( $poule_a['ZWI_TSJ_uitslag'] == $echte_uitslagen['uitslag'] )
{
    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '15' WHERE id = $row['id_usr']");
}
else
{
    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '0' WHERE id = $row['id_usr']");
}

}
?>

 

line 95 is this one:

 

$query = mysql_query("SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = $row['id_usr']");

 

Hopefully some of you guys know the error! I suspect it has someone to do with the user_id = $row['id_usr']");

 

I have tried '$row['id_usr']' "); but that didn't do the trick!

 

thanks in advance

 

edit by thorpe: Please use

tags to enable syntax highlighting.

[AndyB]

I suspect that the problem is caused by naming a table `poule a`.  The parser can't make sense of the space.  Better to rename that table without spaces, e.g. poule_a

[BigX]

I suspect that the problem is caused by naming a table `poule a`.  The parser can't make sense of the space.  Better to rename that table without spaces, e.g. poule_a

 

Yeah I feel you on that one! I figured this out also but I have used `poule a` a whole lot on the site before figuring that out but the strange thing is that when using `poule a` it never gave any problems before (using only poule a did but the ` ` solved the problem)! All my next poules I called poule_b, poule_c etc but I don't think `poule a` is the problem since this code for example is working fine and it also uses `poule a`:

 

<?php

$query = mysql_query("SELECT uitslag FROM echte_uitslagen WHERE wedstrijd = 'zwitserland_tsjechie'");

$echte_uitslagen = mysql_fetch_array($query);

$query = mysql_query("SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = '{$_SESSION['kt_login_id']}'");

$poule_a = mysql_fetch_array($query);

if( $poule_a['ZWI_TSJ_uitslag'] == $echte_uitslagen['uitslag'] )

{

    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '15' WHERE id = '{$_SESSION['kt_login_id']}'");

echo "15";

}

else

{

    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '0' WHERE id = '{$_SESSION['kt_login_id']}'");

echo "0";

}

    ?>

[BigX]

Also when I try in my_sql:

 

SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = 4

 

it works and it gives the results so I really think the problem lies in user_id = $row['id_usr']");

 

Anyone any other suggestions??

 

Greetz

[mikefrederick]

do this

 

user_id='$row[id_usr]'");

[nethnet]

Try this:

 

<?php

$sql = "SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = \"{$row['id_usr']}\"";
$query = mysql_query($sql) or die("Error in the query: " . mysql_error());

?>

 

If that does not work, try echoing $sql to ensure that the query is exactly what it should be.

[BigX]

do this

 

user_id='$row[id_usr]'");

 

Thanks this seemed to work! I replaced it and the syntax error was gone!  The problem now is that I get:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in [path removed] on line 95

 

This is the code now around line 95:

 

$query = "SELECT id_usr FROM ekpoule_users WHERE level_usr = User ORDER BY id_usr ASC";
$berekenen = mysql_query($query);
   
while($row = mysql_fetch_array($berekenen, MYSQL_ASSOC))
{

 

When inserting the above in My_sql I get:

#1054 - Unknown column 'User' in 'where clause'

 

In the WHERE part I have a column level_usr where there could be filled in 'Admin' or 'User' and I only want to select the Users form this column! Should I place User between ' ' or something?? Normally this worked for me...

 

Thanks in advance for putting time into this

 

Greetz

[nethnet]

It should be level_usr = 'User'.  Don't forget quotes.

[BigX]

It should be level_usr = 'User'.  Don't forget quotes.

 

Thanks when I tried level_usr = 'User' I did get a result when putting this into my_sql (2, 3, 4, 5 etc)

but I still get the:

 

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in [path removed] on line 95

 

Line 95 is the While loop:

 

while($row = mysql_fetch_array($berekenen, MYSQL_ASSOC))

{

 

I don't know what's wrong here! I know there are valid my_sql results but it still has a problem somewhere!

 

Whole code:

 

<?php

$query = "SELECT id_usr FROM ekpoule_users WHERE level_usr = 'User' ORDER BY id_usr ASC";
$berekenen = mysql_query($query);
   
while($row = mysql_fetch_array($berekenen, MYSQL_ASSOC))
{

// Zwitserland - Tsjechie (uitslag)

$query = mysql_query("SELECT uitslag FROM echte_uitslagen WHERE wedstrijd = 'zwitserland_tsjechie'");
$echte_uitslagen = mysql_fetch_array($query);
$query = mysql_query("SELECT ZWI_TSJ_uitslag FROM `poule a` WHERE user_id = '$row[id_usr]'");
$poule_a = mysql_fetch_array($query);
if( $poule_a['ZWI_TSJ_uitslag'] == $echte_uitslagen['uitslag'] )
{
    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '15' WHERE id = '$row[id_usr]'");
}
else
{
    mysql_query("UPDATE puntentelling SET zwitserland_tsjechie_uitslag = '0' WHERE id = '$row[id_usr]'");
}
}
?>

[nethnet]

That warning is telling you that $berekenen is not of type RESOURCE like it should be in order to be used in the mysql_fetch_array() function.  Most likely, it is of type BOOLEAN, more specifically FALSE because the query failed.  Add this for debugging purposes and see if a mysql error is returned:

 

<?php

$berekenen = mysql_query($query) or die("Query error: " . mysql_error());

?>

[BigX]

That warning is telling you that $berekenen is not of type RESOURCE like it should be in order to be used in the mysql_fetch_array() function.  Most likely, it is of type BOOLEAN, more specifically FALSE because the query failed.  Add this for debugging purposes and see if a mysql error is returned:

 

<?php

$berekenen = mysql_query($query) or die("Query error: " . mysql_error());

?>

 

When I added what you told me I got:

 

Query error: No database selected

 

I kinda understand what you were saying about the RESOURCE not being valid but is there a way to change that?? Maybe I should explain a bit more what I'm trying to accomplish in the first place! I'm making a poule for the upcoming European Championships Soccer (we love that in Europe  )! I have a table with all the people that join the poule and everyone has a unique id (id_usr) using the $row I need to select this unique id to run  a script for all the users to give them points (or not) for a specific match result! So the script need to select id=2 (id=1 is the Admin) and give that person points for the matches (all the matches are between the { } of the While loop)/ After this id=3, id=4 etc! A person really good at PHP told me this was the way to go, but he's away for 6 weeks starting yesterday so I can't ask him for help! Is this While loop idea the right thing to use??

[nethnet]

It is saying you haven't selected a database.  You need these two lines of code to initiate a database session:

 

<?php

$link = mysql_connect("server","username","password");
mysql_select_db("database",$link);

?>

 

It's best to put that in a file called db.php and include it on every page.  You'll have to change all the values in quotes to your actual values.

[BigX]

It is saying you haven't selected a database.  You need these two lines of code to initiate a database session:

 

<?php

$link = mysql_connect("server","username","password");
mysql_select_db("database",$link);

?>

 

It's best to put that in a file called db.php and include it on every page.  You'll have to change all the values in quotes to your actual values.

 

Well on top of my page I have:

 

<?php require_once('Connections/ekpoule08.php'); ?>
<?php
//initialize the session
if (!isset($_SESSION)) {
  session_start();
}
?>

 

And in ekpoule08.php it says:

 

<?php

# FileName="Connection_php_mysql.htm"

# Type="MYSQL"

# HTTP="true"

$hostname_ekpoule08 = "#####";

$database_ekpoule08 = "#######";

$username_ekpoule08 = "######";

$password_ekpoule08 = "########";

$ekpoule08 = mysql_pconnect($hostname_ekpoule08, $username_ekpoule08, $password_ekpoule08) or trigger_error(mysql_error(),E_USER_ERROR);

?>

 

Is this what you mean?? I have the same thing on all the other pages of the site and the connection with the database works fine there:

 

joost.dvdcollectie.com/ekpoule2008

 

on the first page index there is allready connection being made with the database...

[nethnet]

Okay but you are still NOT selecting a database.  You have opened a connection to the MySQL server, but until you select a database, your script is not going to work.

 

Back to what I said before, use the function mysql_select_db() to select the database.

[BigX]

Ahhh it finally did work  Thanks very much!!!!!!! You're a life saver! Well maybe that's a bit too  but I really appreciate you putting your time into helping me out here! THANX

[nethnet]

No problem, good luck with the rest of your project.

 

Regards,

Theo