need help moving scripts from local to remote

[Vizonz]

Fixed the include problem

now this is a problem

 

Also ken helped me with some scripts and when i try to run them on the server now i get

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in  on line 25
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /

 

these work fine locally but not on the server

[ocpaul20]

and the error message is......?

[Vizonz]

$query = "SELECT DISTINCT
  requestlist.songID, songlist.ID, requestlist.ID, songlist.artist, songlist.title, songlist.count_requested, songlist.composer, requestlist.status
FROM
  radio.requestlist
  INNER JOIN radio.songlist ON requestlist.songID = songlist.ID
WHERE
  requestlist.status = 'played'
GROUP BY
  requestlist.status, songlist.title, songlist.artist
ORDER BY
  songlist.count_requested DESC, songlist.artist DESC LIMIT 30;
";
$result = mysql_query ($query)
or die("Query error: ". mysql_error());
$num=mysql_num_rows($result);
echo "<ol>\n";
while ($row = mysql_fetch_assoc($result)) {
   echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
}
echo "</ol>\n";
?><br />

 

and i am getting

 

Query error: SELECT command denied to user

 

i am still learning but i am taking a guess  that i dont have privledges set up on that database for that user.

 

what do i run through phpmyadmin to grant that

[trq]

Those errors simply mean your query failed and you have done nothing to catch the error. Take  look at mysql_error to see what the actual error is then try and fix it.

[Vizonz]

Those errors simply mean your query failed and you have done nothing to catch the error. Take  look at mysql_error to see what the actual error is then try and fix it.

Kinda lost me  what exactly i do ? i am learning so when you say that it kinda means nothing to me LOL sorry to be blunt but i am clueless about this

[Vizonz]

like i am completely lost  i have the exact same code running locally and it works without a problem

[trq]

Actually, your calling....

 

or die("Query error: ". mysql_error());

 

which should show you the data you need. I'm not sure if the code you have posted is actually the code causing the problem now.

[Vizonz]

Actually, your calling....

 

or die("Query error: ". mysql_error());

 

which should show you the data you need. I'm not sure if the code you have posted is actually the code causing the problem now.

 

okay lets start from here

 

include("dbinfo.inc.php");
mysql_connect($host,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = "SELECT DISTINCT
  requestlist.songID, songlist.ID, requestlist.ID, songlist.artist, songlist.title, songlist.count_requested, songlist.composer, requestlist.status
FROM
  radio.requestlist
  INNER JOIN radio.songlist ON requestlist.songID = songlist.ID
WHERE
  requestlist.status = 'played'
GROUP BY
  requestlist.status, songlist.title, songlist.artist
ORDER BY
  songlist.count_requested DESC, songlist.artist DESC LIMIT 30;
";
$result=mysql_query($query);
$num=mysql_num_rows($result);
echo "<ol>\n";
while ($row = mysql_fetch_assoc($result)) {
   echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
}
echo "</ol>\n";
?><br />

 

thats the code in the page.

 

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource  on line 24


      Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource  on line 26

 

line 24

$num=mysql_num_rows($result);

 

Line 26

while ($row = mysql_fetch_assoc($result)) {

 

thats where i am rright now

 

 

[trq]

You need to check your queries succeed before using there results. At minimum....

 

if ($result = mysql_query($query)) {
  if (mysql_num_rows($result)) {
    echo "<ol>\n";
    while ($row = mysql_fetch_assoc($result)) {
      echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
    }
    echo "</ol>\n";
  } else {
    // no results found, handle error.
} else {
  // query failed, handle error.
  trigger_error(mysql_error());
}
?>

 

always.

[Vizonz]

parse error  Parse error: syntax error, unexpected T_ELSE

 

} else {  -on that line

 

sorry i am trying to learn but its a slow process

[trq]

I don't think there are any parse errors in the code I posted. Post your current code.

[Vizonz]

<?php
include("dbinfo.inc.php");
mysql_connect($host,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = "SELECT DISTINCT
  requestlist.songID, songlist.ID, requestlist.ID, songlist.artist, songlist.title, songlist.count_requested, songlist.composer, requestlist.status
FROM
  radio.requestlist
  INNER JOIN radio.songlist ON requestlist.songID = songlist.ID
WHERE
  requestlist.status = 'played'
GROUP BY
  requestlist.status, songlist.title, songlist.artist
ORDER BY
  songlist.count_requested DESC, songlist.artist DESC LIMIT 30;
";
$result=mysql_query($query);
$num=mysql_num_rows($result);
if ($result = mysql_query($query)) {
  if (mysql_num_rows($result)) {
    echo "<ol>\n";
    while ($row = mysql_fetch_assoc($result)) {
      echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
    }
    echo "</ol>\n";
  } else {
    // no results found, handle error.
} else {
  // query failed, handle error.
  trigger_error(mysql_error());
}
?>

 

and its a parse error

Parse error: syntax error, unexpected T_ELSE in   on line 34

 

i took out some of the html above it  so yeah

 

    // no results found, handle error.
} else {

line 34 is the else

[trq]

Sorry, there is a missing }

 

if ($result = mysql_query($query)) {
  if (mysql_num_rows($result)) {
    echo "<ol>\n";
    while ($row = mysql_fetch_assoc($result)) {
      echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
    }
    echo "</ol>\n";
  } else {
    // no results found, handle error.
  }
} else {
  // query failed, handle error.
  trigger_error(mysql_error());
}
?>

 

Also, you can remove those extra calls to mysql_query() and mysql_num_rows() that you have.

[Vizonz]

Notice: SELECT command denied to user  thats what i get with that now

[trq]

Whatever user your using to login to mysql with doesn't have sufficient permissions to execute a SELECT on that db.

 

See http://dev.mysql.com/doc/refman/5.1/en/privilege-system.html

[Vizonz]

thats what i thought it had to do with. its a remote server not dedicated and they dont have the user permissions in phpmyadmin  is there a query to add privledges or is it something the server has to add

[trq]

Do you have shell access? You cannot execute a query (even to fix the problem) if you don't have sufficient permissions.

[Vizonz]

yeah i had to contact them for them to open a ticket to  srt permissions on the database LOL  dedicated servers so much better . but thanks i will let you know  if it works after that

 

by the way love your blog  nice design

[trq]

by the way love your blog  nice design

 

Thanks, its just some template.

[Vizonz]

i like it   stupid server tells me  we opened a ticket for are tier 3 team  and you will get a responce within 48 hours.  48 hours to check a few things in phpmyadmin.  aplus hosting great  support team LOL

[Vizonz]

Alright they telling me i have access to the database that something is wrong with the coding anyone help me ?

[trq]

Post your current code and error.

[Vizonz]

sorry wasnt paying attencion its where it was last nothing changed  still same thing same error

 

From the server

Regarding your issue with database connection, has returned from our investigation. It has been determined that database connection is working properly. We have tested and we were able to access database and run sql query for it (please see connection logs below). Please check code in top30.php. We have now closed the ticket with regards to this issue. If you do require any further assistance, please do not hesitate to contact us and we will be glad continue working with you.

[Vizonz]

<?php
include("dbinfo.inc.php");
mysql_connect($host,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");
$query = "SELECT DISTINCT
  requestlist.songID, songlist.ID, requestlist.ID, songlist.artist, songlist.title, songlist.count_requested, songlist.composer, requestlist.status
FROM
  radio.requestlist
  INNER JOIN radio.songlist ON requestlist.songID = songlist.ID
WHERE
  requestlist.status = 'played'
GROUP BY
  requestlist.status, songlist.title, songlist.artist
ORDER BY
  songlist.count_requested DESC, songlist.artist DESC LIMIT 30;
";

$result=mysql_query($query);
if ($result = mysql_query($query)) {
  if (mysql_num_rows($result)) {
    echo "<ol>\n";
    while ($row = mysql_fetch_assoc($result)) {
      echo "<li> {$row['artist']}  - <a href='{$row['songID']}'>{$row['title']}</a>        ({$row['count_requested']} ) </li>\n";
    }
    echo "</ol>\n";
  } else {
    // no results found, handle error.
  }
} else {
  // query failed, handle error.
  trigger_error(mysql_error());
}
?>

theres the code  though

[Vizonz]

i am an idiot i know why LOL its a new database name  i forgot to change it in the query \

 

where it had radio. in the query was the localhost database i just removed that and it now works.  thanks for the help with the actual php though the rest was my stupidity