can anyone please help me on this?? query details

alternative

i have some problem when i try to query a detail of my news.

can anyone please help me on this??

index.php

<?php

include ('db_fns.php');

function find_data($newsid){
	if($newsid=""){
		echo "bad";
	}
	else{
		echo "good";
	$conn = db_connect();
	$campusnews_sql = "select * from campusnews where campusnews_id = '$newsid'";
	$campusnews_result = mysql_query($campusnews_sql, $conn);
	$result = @mysql_query($query);
	$result = @mysql_fetch_array($result);
	//echo "$result";
	return $result;
	}
}

function output_data($news){
    if(is_array($news)){
	echo "output good";
	print $news["campusnews_index"];
	print "<br>";
	}
	else{echo "<br>output bad";}
}

$news = find_data($newsid);
output_data($news);

?>

db_fns.php

function db_connect()
{
   $result = @mysql_pconnect("localhost", "content", "password"); 
   if (!$result)
      return false;
   if (!@mysql_select_db("content"))
      return false;

   return $result;
}

attributes

campusnews_id

campusnews_topic

campusnews_index

campusnews_date

bubblenut

Originally posted by alternative
i have some problem when i try to query a detail of my news.

Not a very descriptive explination, can you clarify?

alternative

for an example

index.php?newsid=1

then it will display the details of campusnews_id=1

display the topic, index, and the date of the record..

bubblenut

Try using

$news = find_data($_GET['newsid']);

😕

What is the actual problem (appart from "it's not working"). Are you getting any output or none? Are you getting errors? Please try to be a little more explanitory.

alternative

now i am trying the query the record ( only 1 record, which selected by the newsid)

the problem is my codes cant work from the function

$campusnews_sql = "select * from campusnews where campusnews_id = '$newsid'";
$campusnews_result = mysql_query($campusnews_sql, $conn);
$result = @($query);
$result = @mysql_fetch_array($result);
//echo "$result";
return $result;

there is no record output.

those echo "good"; or echo "bad"; is use for me to check the function is it run properly.

and then output is :

good<br>output bad

eg:

index.php?newsid=80
--------------------------------------output that i want --------
title : bla bla bla
index : bla bla bla long text

date : xxxx-xx-xx

query the record 80 from the database..
like that....

by the way, do you have the actual codes for the detailed query?

bubblenut

Firstlty, have you made the change I suggested?

Secondly, I've spotted the error I think (if not the error an error which would cause the script to fail), let's see if you can spot it too. Try putting a little debuging code in

...
$campusnews_sql = "select * from campusnews where campusnews_id = '$newsid'";
echo($campusnews_sql."<br />\n");
$campusnews_result = mysql_query($campusnews_sql, $conn) or die(mysql_error()."<br />\n");
$result = mysql_query($query) or die(mysql_error()."<br />\n");
$result = mysql_fetch_array($result); 
...

Just a note, putting the @ before a function call supresses the errors, this is very unwise unless you're absolutely certain you're never going to make a mistake.
<voice name="Rolf Harris">Can ya' see what it is yet?</voice>

alternative

by the way, do you have any completed application for the displaying detailed data from database records?

alternative

i found out the problem.

my php file named as index.php

then for records queries is index.php?id=3

but i find out that the id=3 is not pass to my codes


<html>
<body>
<?php

$id=3;

$db = mysql_connect("localhost", "root");
mysql_select_db("mydb",$db);
// display individual record
if ($id) {
$result = mysql_query("SELECT * FROM employees WHERE id='$id'",$db);
$myrow = mysql_fetch_array($result);
printf("First name: %s\n<br>", $myrow["first"]);
printf("Last name: %s\n<br>", $myrow["last"]);
}
?>
</body>
</html>

i try to create $id=3 inside my code, and then record can be query out properly. but if i use index.php?id=3
then id value not pass to my codes.

for ur information, i am using phpdev.
thanx in advance.

alternative

anyone can help me?

thanx in advance...

bubblenut

Originally posted by bubblenut
Try using
$newsid = find_data($_GET['newsid']);
[/B]

If you make me say it again I'll have to slap you!

alternative

ok, i switch everything to become like this..

this is the error msg that i get.

Fatal error: Call to undefined function: find_data() in c:\phpdev2\www\test4detail2.php on line 5

my code is


<html>
<body>
<?php
$id = find_data($_GET['id']);
//$id=3;

$db = mysql_connect("localhost", "root");
mysql_select_db("mydb",$db);
// display individual record
if ($id) {
$result = mysql_query("SELECT * FROM employees WHERE id='$id'",$db);
$myrow = mysql_fetch_array($result);
printf("First name: %s\n<br>", $myrow["first"]);
printf("Last name: %s\n<br>", $myrow["last"]);
}
?>
</body>
</html>