what's wrong with this code?

seanpaulcail

mysql_select_db('a0017855');
$query ="select * from Eli2 where $page_name = etruria";
$result=mysql_query($query);

?>
<body background="images/<?php echo  &row[bckgrnd_image]?>" leftmargin="0" topmargin="0" style="margin:0px">

I get an error which follows:

Parse error: parse error in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria.php on line 3

thanks

Tracer

try to echo out $page_name and see if this var is actually set....

It appears that your SQL query is failing...

seanpaulcail

I've also tried the following:

<?
include ('header_eli.php')
$connection= mysql_connect("localhost", "a0017855', "snuren34");

mysql_select_db("a0017855", $connection);
$query ="select * from Eli2 where $page_name = etruria";
$result=mysql_query($query);

?>
<body background="images/<?php echo  &row[bckgrnd_image]?>" leftmargin="0" topmargin="0" style="margin:0px">

thanks

LordShryku

<?php echo &row[bckgrnd_image]?>

Shouldn't that be
<?php echo $row['bckgrnd_image']; ?>

greeno76

Is the variable $row specified?

Plus can you specify $page_name as a field?

I would have thought it was something more like this:

mysql_select_db('a0017855');

$query ="select * from Eli2where page_name = "\etruria\" ";

$result=mysql_query($query)
or die ("Couldn't Execute Query");

while ($row = mysql_fetch_array($result))

{ echo $row["bckgrnd_image"]; }

seanpaulcail

here's the error message

Parse error: parse error in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria.php on line 4

here's the code

<?

mysql_select_db('****');
$connection= mysql_connect('localhost", '*****', '*****');
mysql_select_db("****", $connection);


$result=mysql_query("select * from `Eli2`where `page_name` = "\etruria\"");
or die ("Couldn't Execute Query");

$info = mysql_fetch_field($result);


echo "$page_name";
print(info->pagename);
print(info->bckgrnd_image);
?>

what am I doing wrong?

thanks

LordShryku

$connection= mysql_connect('localhost", 'a0017855', 'snuren34');

Mixing single quotes and double quotes on localhost

seanpaulcail

here's the code for line 8

$result=mysql_query("select * from `Eli2`where `page_name` = "\etruria\"");

here's the error message

zWarning: Unexpected character in input: '\' (ASCII=92) state=1 in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria.php on line 8

Parse error: parse error in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria.php on line 8/PHP] 

thanks again

greeno76

I think you need to escape "etruria" to \"etruria\"

seanpaulcail

it just didn't come across when I cut and pasted

thanks

LordShryku

There's a lot wrong with this code. There was only one slash. There wasn't a space between Eli2 and where. There's no $'s on your print variables. You selected the DB twice(?), once before the connection, once after. Try this instead:

<?php
$connection= mysql_connect("localhost","a0017855","snuren34");
mysql_select_db("a0017855",$connection);

$result = mysql_query("select * from Eli2 where page_name = 'etruria'");
or die ("Couldn't Execute Query");

$info = mysql_fetch_field($result);

echo $page_name;
print($info->pagename);
print($info->bckgrnd_image);
?>

seanpaulcail

I cut and pasted your code but I get the following error:

Parse error: parse error in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria1.php on line 6

thanks for the help

LordShryku

Ugh, yeah, just remove the ; from the them of this line:

$result = mysql_query("select * from Eli2 where page_name = 'etruria'");

seanpaulcail

<?

$connection= mysql_connect('localhost', '****', '****');
mysql_select_db("****", $connection);

$query = ('select * from Eli2 where page_name = etruria');
$result = mysql_query($query);


$db_result = mysql_fetch_array($result);

echo $db_result['page_name'];
echo $db_result['bckgrnd_image'];

?>

but I get the following error message

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /usr/home/web/users/a0017855/html/phpMyEdit2_bu/etruria.php on line 10

I thought Mysql_fetch_array was a valid function.

thanks for the help

LordShryku

It is, but your SQL's bad.
Use
$query = ("select * from Eli2 where page_name = 'etruria'");

Notice the quoting...

seanpaulcail

I solved it, with your help of course

here is the working code.

I used page_number instead of page_name in the query

<?

$connection= mysql_connect('localhost', '****', '*****');
mysql_select_db("*****", $connection);

$query = ('select * from Eli2 where page_number = "5"');
$result = mysql_query($query);


$db_result = mysql_fetch_array($result);

echo $db_result['page_name'];
echo $db_result['bckgrnd_image'];

?>

thanks for the help