mysql_fetch_array help

Spartaman

Ok, heres the script:

<?php

$per_page = 25;       // number of links listed pr page
$category  = 2;          // 1 is the code for 'clean humor

$server     = "localhost";       // domæne
$brugernavn = "********";     // username
$password   = "********";      // password
$db_navn    = "*********";         // name of database
$dbt_navn   = "*********";      // navn of table in the database

mysql_connect($server, $brugernavn, $password)
   or die( "Unable to connect\n". mysql_error() );
mysql_select_db("$db_navn")
  or die("Unable to select db ".mysql_error()."\n");

 $category = $_POST['category'];
 if ( isset( $_POST[ 'startnr' ] ) ) {
      $startnr = $_POST[ 'startnr' ];
} else {
      $startnr = 0;     // default
}

$list = mysql_select_db( "
    SELECT *
    FROM  $dbt_navn
    WHERE category = $category
    ORDER BY id DESC
    LIMIT  $startnr, $per_page
" );
while ( $row = mysql_fetch_array( $list ) ) {
       echo "<a href='" .$row['url'] ."'>" .$row['title'] ."</a><br>\n";
}
?>
<form method='post' action='content.php?var=flashmovies'>
<input type='hidden' name='startnr' value="<?php echo $startnr; ?>">
<input name='submit' type="submit" value="show previous page" onclick="<?PHP echo "this.form.startnr=" .(($startnr>=$per_page)?($startnr-$per_page):0); ?>">
<input name='submit' type="submit" value="show next page" onclick="<?PHP echo "this.form.startnr=" .($startnr+$per_page); ?>">
</form>


</body>
</html>

I get this error:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/httpd/vhosts/sparta.cn/httpdocs/flashgames.php on line 31

Line 31 being the "while" statement. Can anyone help?!

Thanks in advance.

laserlight

Well, that error is quite common, you probably should have searched the forums.

The reason is usually because the corresponding query is incorrect.

In your case, you used
mysql_select_db()
instead of
mysql_query()

Spartaman

Thank you so much.

Now I have 1 more problem.
Heres my MySQL table code:

CREATE TABLE yourtablename (

id int(10) AUTO_INCREMENT PRIMARY KEY,

title varchar(128) not null,

url varchar(256) not null,

category int not null

)

It is saying:
Unable to select db You have an error in your SQL syntax near 'fileindex' at line 1

Thank you for your time.

tsinka

Most likely the variable $db_navn contains some invalid database but it is hard to imageine.

Can you post the database name ?

Spartaman

$server = "localhost"; // domæne
$brugernavn = "files"; // username
$password = "*********"; // password
$db_navn = "fileindex"; // name of database
$dbt_navn = "yourtablename"; // navn of table in the database

tsinka

Hi,

can you post the current code again. I think the problem is not the mysql_select_db but the mysql_query because the error message means that you have a syntax error in your SQL query. Did you copy the die (..) part from the mysql_select_db call to the mysql_query so you had this ?

mysql_query("$query")
or die("Unable to select db ".mysql_error()."\n");

I was a bit confused about that error showing up with a mysql_select_db.

The error migh be

WHERE category = $category which should probably be
WHERE category = '$category'

Just a suggestion 🙂

FraXTC

Originally posted by Spartaman
Thank you so much.

Now I have 1 more problem.
Heres my MySQL table code:

It is saying:
Unable to select db You have an error in your SQL syntax near 'fileindex' at line 1

Thank you for your time.

Always check the most logic solution: did you use the

use yourdatabasename;

before trying to create the table??

Norman_Graham

Listen, Spartaman, no time to check the manual right this second, but I'm pretty sure varchar(256) is illegal - you can only have varchar up to varchar(255). Anything above that has to be one of the TEXT column types.

Check it out

Norm

laserlight

good call, I didnt notice that.

The length of a CHAR column is fixed to the length that you declare when you create the table. The length can be any value between 1 and 255. (As of MySQL Version 3.23, the length of CHAR may be 0 to 255.)
...
Values in VARCHAR columns are variable-length strings. You can declare a VARCHAR column to be any length between 1 and 255, just as for CHAR columns.

Spartaman

I got the part about the varchar, It wouldnt let me create a table doing 256, so I chaneged it to 255.

I did name the table yourtablename.

And I will try that tsinka.

tsinka

Hi,

if it doesn't work then post the most recent version of your code again 🙂

Spartaman

<?php

$per_page = 25;       // number of links listed pr page
$category  = 2;          // 1 is the code for 'clean humor

$server     = "localhost";       // domæne
$brugernavn = "files";     // username
$password   = "********";      // password
$db_navn    = "fileindex";         // name of database
$dbt_navn   = "yourtablename";      // navn of table in the database

mysql_connect($server, $brugernavn, $password)
   or die( "Unable to connect\n". mysql_error() );
mysql_query("$db_navn")
  or die("Unable to select db ".mysql_error()."\n");

 $category = $_POST['category'];
 if ( isset( $_POST[ 'startnr' ] ) ) {
      $startnr = $_POST[ 'startnr' ];
} else {
      $startnr = 0;     // default
}

$list = mysql_query( "
    SELECT *
    FROM  $dbt_navn
    WHERE category = '$category'
    ORDER BY id DESC
    LIMIT  $startnr, $per_page
" );
while ( $row = mysql_fetch_array( $list ) ) {
       echo "<a href='" .$row['url'] ."'>" .$row['title'] ."</a><br>\n";
}
?>
<form method='post' action='content.php?var=flashmovies'>
<input type='hidden' name='startnr' value="<?php echo $startnr; ?>">
<input name='submit' type="submit" value="show previous page" onclick="<?PHP echo "this.form.startnr=" .(($startnr>=$per_page)?($startnr-$per_page):0); ?>">
<input name='submit' type="submit" value="show next page" onclick="<?PHP echo "this.form.startnr=" .($startnr+$per_page); ?>">
</form>


</body>
</html>

CREATE TABLE yourtablename (

id int(10) AUTO_INCREMENT PRIMARY KEY,

title varchar(128) not null,

url varchar(255) not null,

category int not null

)

tsinka

Hi,

replace

mysql_query("$db_navn")

with

mysql_select_db("$db_navn")

The problem was only the second mysql_select_db in your first posted code snipplet 🙂

Spartaman

WEEEEEEEEEEEEEEEEEEE!

THANK YOU!!!

Spartaman

Ran into some more trouble

<?php 

if ( isset( $_POST['submit'] )) {

   $server     = "localhost";       // domæne
   $brugernavn = "files";     // username
   $password   = "**********";      // password
   $db_navn    = "fileindex";         // name of database
   $dbt_navn   = "yourtablename";      // navn of table in the database

   mysql_connect($server, $brugernavn, $password)
     or die( "Unable to connect\n". mysql_error() ); 

mysql_select_db("$db_navn") 
  or die("Unable to select db ".mysql_error()."\n"); 

$title = $_POST['title'];
   $url  = $_POST['url'];
   $category = $_POST['category'];

   $res = mysql_query( "
         SELECT COUNT( * )
         FROM $dbt_navn
         WHERE url = '$url'
   ");
   if ( mysql_result( $res, 0, 0 )) {
        // The given url is already in the database. shoould we proceed and insert it again ?
   }

   mysql_query( "
        INSERT INTO $dbt_navn
        VALUES (0, '$title', '$url', $category)
   ");
   if ( mysql_affected_rows() == 0 ) {
        // could not insert, one of the fields had not been filled.
   }

} // end of the IF statement

?>
<html>
<head></head>
<body>
<form method='post' action="content.php?var=indexscript">
<input name='title' type='text' value="Title">
<input name='url' type='text' value="URL">
<SELECT name='category'>
   <option value='1'>Flash Movies</option>
   <option value='2'>Flash Games</option>
   <option value='3'>Videos</option>
</select>
<input name='submit' type="submit" value="Send it">
</form>
</body>
</html>

This is the script I use to add files onto the db, which is then pulled by the other script.

I ran into a couple mysql errors with this one as well, but I was able to fix them, however, now whenever I submit something, it goes back to indexscript (this one) which it should, but it doesnt show up on any of the pages.

Spartaman

Can anyone help?