[Resolved] check this code for errors? please

gat

<?php 

if(!($dbh=mysql_connect ("localhost", "*****", "*****"))) 
die("Cannot connect to the database"); 

if (!mysql_select_db("kriminalyouth",$dbh)) 
die("Cannot open the database"); 

$thumb = substr($thumb, 0, 4); 
$thumb = EscapeShellCmd($thumb); 

$query = "SELECT pic_path, pic_date FROM walls_berks 
WHERE pic_id >= $thumb 

if (!($result = @ mysql_query ($query, $dbh))) 
die("Cannot execute query against the database"); 


if ($row = @ mysql_fetch_array($result)) 
{ 
$filename = $row["pic_path"]; 
echo "<img SRC=\"$pic_path\"><br>"; 
} 
echo "$pic_path<br>"; 
$anno = $row["pic_date"]; 
echo "</center><p>$anno</p><center><table BORDER=0 WIDTH=\"100%\">"; 

$lastthumb = $thumb - 1; 
$nextthumb = $thumb + 1; 

echo "<tr>"; 

echo "<td WIDTH=\"50%\" valign=\"top\"><h4><a href=\"photoview.php?thumb=$lastthumb\">"; 
echo "View Previous Photograph</a></h4></td>"; 

echo "<td WIDTH=\"50%\" valign=\"top\"><h4><a href=\"photoview.php?thumb=$nextthumb\">"; 
echo "View Next Photograph</a></h4></td>"; 
?>

I get "Cannot execute query against the database"
and no errors????

please help.

scoppc

put this:

$query = "SELECT pic_path, pic_date FROM walls_berks WHERE pic_id >= $thumb LIMIT 1";

before this:

if (!($result = @ mysql_query ($query, $dbh))) 
die("Cannot execute query against the database");

gat

ok, i edited the last post.

thing is, i will have to take the code home with me to my home pc.
so i can test it then.

thanks for the idea...

does everything else look like it should work ??????

scoppc

I think u'll get the $thumb to be just BLANK coz it is being set after the query.
It wont be available. =)
So, again, just move it before the query.

$thumb = substr($thumb, 0, 4); 
$thumb = EscapeShellCmd($thumb);

BeastRider

Originally posted by gat
$query = "SELECT pic_path, pic_date FROM walls_berks 
WHERE pic_id >= $thumb 
LIMIT 1"; 
I get "Cannot execute query against the database"
and no errors????

please help. [/B]

I created the select statement that way because there are gaps in my key. If you know your key increases sequentially by 1, the you can change it to:

$query = "SELECT pic_path, pic_date FROM walls_berks 
WHERE pic_id = $thumb;

The only other reasons I can think of is that you have bad names for your table variables.

gat

ok, ill try that then....
(i edited the top post again)

BeastRider

Originally posted by scoppc
I think u'll get the $thumb to be just BLANK coz it is being set after the query.
It wont be available. =)
So, again, just move it before the query.
$thumb = substr($thumb, 0, 4); 
$thumb = EscapeShellCmd($thumb); 
[/B]

$thumb will not be blank if it contains a number. The substr command just takes the first four characters of the input. The second command strips all special characters from the input.

gat

thanks for your help beastrider....
ill get back tomorrow if it does not work at home.

once again, many thanks.
Also, you might know?

i would like a sentence "there are (***) images stored"

but i have many tables containing images, and i would like the total of all of them.

phpmyadmin, shows the 'sum' of all the records when you select the main DB, so it should be simple i guess??

can you help with that??

BeastRider

You have two approaches. I assume all the tables are in one database. The first is to do a select and then a count for each table, summing them up. The second is to merge the tables as described in 7.2 of the MySql manual, then do a count on the whole table. A third idea would be to do the counts once, then store the results in a special table created for image statistics.