[Resolved] mysql_fetch_array

[Anthony_john5]

• Before I start I'd like to point out that I have read the FAQ, it's just that I need a little more guidance. 🙂

I am trying to query a database for the picture id number of the latest entry to my photo album so that I can then echo it to show how many pictures have been submitted.

Currently I have the following script:

<?php

include("/home/www/teenonyM/scripts/database_functions.php");

# Function from the file above.
OpenConnection();

# Assign query to a value that can be passed to the custom error handler.
$query = "SELECT `pic_id` FROM `phpbb_album` WHERE 1 ORDER BY `pic_id` DESC  LIMIT 0 , 1 ";
@mysql_query($query, $connection)
	or ThrowError("error with query: " . mysql_error() . "\n\nQuery: " . $query);

?>

Which as I expected returns a blank page.

Q1) Is this code correct so far and from this can I use the mysql_fetch_array to set the value of variable to the pic_id value?

After reading the FAQ I tried chaning the above code to:

<?php

include("/home/www/teenonyM/scripts/database_functions.php");

# Function from the file above.
OpenConnection();

# Assign query to a value that can be passed to the custom error handler.
$query = "SELECT `pic_id` FROM `phpbb_album` WHERE 1 ORDER BY `pic_id` DESC  LIMIT 0 , 1 ";
@mysql_query($query, $connection)
	or ThrowError("error with query: " . mysql_error() . "\n\nQuery: " . $query);

$pics = mysql_fetch_array($query['pic_id']);

echo "$pics";

?>

However I receive the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/teenonyM/scripts/test.php on line 13

Q2 What does the error message mean and what do I need to change to the code?

• Anthony

[Wynder]

$result = mysql_query....

while( $row = mysql_fetch_array($result) )
{
   echo row['field'];
}

[Anthony_john5]

I added

while( $row = mysql_fetch_array($query) ) 
{ 
   echo row['pic_id']; 
}

Making the whole script to:

<?php

include("/home/www/teenonyM/scripts/database_functions.php");

# Function from the file above.
OpenConnection();

# Assign query to a value that can be passed to the custom error handler.
$query = "SELECT `pic_id` FROM `phpbb_album` WHERE 1 ORDER BY `pic_id` DESC  LIMIT 0 , 1 ";
@mysql_query($query, $connection)
	or ThrowError("error with query: " . mysql_error() . "\n\nQuery: " . $query);

while( $row = mysql_fetch_array($query) ) 
{ 
   echo row['pic_id']; 
}

?>

However when run I receive the error

Parse error: parse error, unexpected '[', expecting ',' or ';' in /home/www/teenonyM/scripts/test.php on line 15

Is there an error earlier on in the script? Any suggestions as what to try next?
- Anthony

[njm]

You're missing the '$' sign. Should be:

while( $row = mysql_fetch_array($query) )  

{  

   echo $row['pic_id'];  

}

hth, njm

[Wynder]

My bad.. sorry -- forgot the $. 🙂

[Anthony_john5]

Still not working! 😕

Currently my code reads:

<?php 

include("/home/www/teenonyM/scripts/database_functions.php"); 

# Function from the file above. 
OpenConnection(); 

# Assign query to a value that can be passed to the custom error handler. 
$query = "SELECT `pic_id` FROM `phpbb_album` WHERE 1 ORDER BY `pic_id` DESC  LIMIT 0 , 1 "; 
@mysql_query($query, $connection) 
    or ThrowError("error with query: " . mysql_error() . "\n\nQuery: " . $query); 

while( $row = mysql_fetch_array($query) )   

{   

   echo $row['pic_id'];   

}

?> 

And when run gives the error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/teenonyM/scripts/test.php on line 13

Line 13 is the line that starts; 'while( $row'

The only thing that I think may be wrong is that the variable $query has merely been assigned to the mysql_query not to it's result. Am I right in thinkning this? If so what should I change?

Thanks for both of you helping so far, hopfully once I've got this figured out this time I'll know it for life.

• Anthony

[Wynder]

Gotta read my post a little closer:

$result = mysql_query.... 

while( $row = mysql_fetch_array($result) ) 
{ 
   echo $row['field']; 
}

The result comes from the mysql_query call.. you fetch the array from that.

[njm]

Wynder - my bad too 😉 I copied the wrong code into my response.

However, I'm pretty sure we've got there now...

njm

[Anthony_john5]

Originally posted by Wynder
Gotta read my post a little closer:

$result = mysql_query.... 

while( $row = mysql_fetch_array($result) ) 
{ 
   echo $row['field']; 
}

The result comes from the mysql_query call.. you fetch the array from that. [/B]

My apologies, I read your post correctly, just misunderstood it. The part that I don't fully understand is the $resuly= mysql_query.... part. What should I be replacing with the dots. Sorry this is taking longer than it should 🙂

• Anthony

[Wynder]

The rest of what you had in your original statement -- just stick a "$result =" in front of the mysql_query statement.

[Anthony_john5]

As I was posting that I tried making that line to read:

$result = mysql_query($query);

Which worked, however after having read your post I tried changing it to:

$result = mysql_query("SELECT `pic_id` FROM `phpbb_album` WHERE 1 ORDER BY `pic_id` DESC  LIMIT 0 , 1 ");

which also worked. So that I learn things properly, which is the 'proper' way of doing it the first or second code snippet?

• Anthony

[Wynder]

That is a pure matter of choice...

Generally people assign the query to a variable and then use it, because sometimes you dynamically create it..

$query = "SELECT * FROM tblFoo ORDER BY name";

if( $sort == "ascending")
   $query .= " ASC";
else
   $query .= " DESC";

[Anthony_john5]

Thank-you 🙂