[RESOLVED] scratching my head: why would simple mysql_connect line NOT work?

marcnyc

Hello,
I was wondering if any changes were introduced in version 5.1.4 of PHP that would not make a simple line of code like this work anymore:

$link = mysql_connect('localhost', 'user', 'pw');

I had my friend try this on his 5.1.2 PHP local machine and it worked for him but I keep getting the error:

Parse error: syntax error, unexpected T_VARIABLE in /home/chaindlk/public_html/reviews/last.php on line 9

where 9 is that line...

part of this function:

function show_randomId() {
$link = mysql_connect('localhost', 'user', 'pw');
$db_stream = mysql_select_db(SERVER_DB_NAME, $link);
$result = mysql_query("SELECT reviews2_reviews.id AS N_REVIEWS FROM reviews2_reviews ORDER BY RAND( ) LIMIT 1", $mysql_stream);
$r = mysql_fetch_object( $result );
echo get_byId($r->N_REVIEWS);
}

Thanks...

NogDog

The error message seem to indicate that there's something wrong earlier in the script, but it's not until it gets to the $link variable on line 9 that it realizes there is a problem. We may need to see the script from line 1 through that function to see if there's something missing or wrong.

marcnyc

actually I tried making a standalone file and putting this code at the beginning of it (replacing the database constant with the db's name) and it still outputs the same error... very strange indeed!

Kudose

Shouldn't $mysql_stream be $db_stream?

marcnyc

yes, you are right... I had already caught that, but the problem is not that... i had corrected it after posting and the problem remains

Kudose

Thanks, but I already knew that wasn't the problem.

Can I see some fresh code?

edit
Also, it's best practice to post up your most recent code so you don't get comments like my previous. Not trying to be a <insert explict word>, it's just easier to help people that way.
/edit

Weedpacket

I detect no syntax errors in that code; might be the same problem as this

NogDog

If it's the case of a stray, non-printing character screwing things up, you may need to open the file in a text editor and do a "save as", making sure you specify the file type as a plain ASCII text file (not UTF8 or some proprietary format).

marcnyc

I've been doing all my test remotely editing the file. I open the file in my FTP software which opens it in BBEdit (I am on Mac). I have always worked like this and never had issues.
I don't know exactly what a stray character is and how it got there, but I downloaded the file to my hd, opened it in BBEdit and Saved it As with Unix line breaks, and Western Mac OS roman encoding (which should be ASCII I guess) as opposed to one of the unicode UTF formats and other formats.
I did notice that my FTP sofware uses UTF8 as default encoding, so, although this has never given me any problems, since you are advising against this, I switched the default encoding on the software to Western Mac OS Roman as well.
I re-uploaded the file replacing the old one but I still get the same error:

Parse error: syntax error, unexpected T_VARIABLE in /home/chaindlk/public_html/reviews/last.php on line 4

which you can see for yourself here:
http://www.chaindlk.com/reviews/last.php

Hope somebody has some other ideas or troubleshooting processes for me to go through. Thanks

NogDog

Can you copy-and-paste the code here within [ php ]...[ /php ] tags (without the spaces I typed here)? If not everything, at least from the very beginning through the line where the error is occurring.

marcnyc

Sure!

function show_randomId() {
   $link = mysql_connect('localhost', 'user', 'pw');
   $db_stream = mysql_select_db(SERVER_DB_NAME, $link);
   $result = mysql_query("SELECT reviews2_reviews.id AS N_REVIEWS FROM reviews2_reviews ORDER BY RAND( ) LIMIT 1", $db_stream);
   $r = mysql_fetch_object( $result );
   echo get_byId($r->N_REVIEWS);
}
show_randomId();

NogDog

No such error when I ran it on my machine. 🙁

marcnyc

the same is happening to my friend on his machine... weird...
can you attach a php file with that code in it and I'll upload your code and give it a try? that way we'll know if the problem is encoding or my server...

NogDog

File attached...

bradgrafelman

If that doesn't work, you could try standard MySQL debugging methods... using [man]mysql_error/man and all...

marcnyc

Ok, so your file actually did not produce the same error, so I guess the mistery is partially solved...
Now I get different errors but at least at first sight they seem errors that can be addressed:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in /home/chaindlk/public_html/reviews/debug.php on line 7

Warning: mysql_fetch_object(): supplied argument is not a valid MySQL result resource in /home/chaindlk/public_html/reviews/debug.php on line 8

Notice: Trying to get property of non-object in /home/chaindlk/public_html/reviews/debug.php on line 9

NogDog

MySQL is not able to process your query for some reason. Try changing your query line to something like this to get some debug info:

$sql = "SELECT reviews2_reviews.id AS N_REVIEWS FROM reviews2_reviews ORDER BY RAND( ) LIMIT 1";
$result = mysql_query($sql, $db_stream) or die("Query failed: $sql - ".mysql_error());

bradgrafelman

The error has nothing to do with the query itself.

I would make sure that you can connect to the server as well as select the database (add the same debug statements NogDog did for the mysql_query() line in his code snippet above).

NogDog

bradgrafelman wrote:
The error has nothing to do with the query itself.

I would make sure that you can connect to the server as well as select the database (add the same debug statements NogDog did for the mysql_query() line in his code snippet above).

Oops! You're right: my eyes thought they saw the "not a valid result resource id" message (or whatever it is when your query fails), but it's a "not a valid link resource", indicating the mysql_connect() failed.

bradgrafelman

Oops! Now it's my turn to have bad eyes.

You're expecting to get a resource object from [man]mysql_select_db[/man], except this function doesn't return a resource! Just use this:

<?php
function show_randomId() {
   $link = mysql_connect('localhost', 'user', 'pw') or die('Couldn\'t connect: ' . mysql_error());
   mysql_select_db(SERVER_DB_NAME, $link) or die('Couldn\'t select DB: ' . mysql_error());
   $result = mysql_query("SELECT reviews2_reviews.id AS N_REVIEWS FROM reviews2_reviews ORDER BY RAND( ) LIMIT 1", $link) or die('Couldn\'t run query: ' . mysql_error());
   $r = mysql_fetch_object( $result );
   echo get_byId($r->N_REVIEWS);
}
show_randomId(); 
?>

Also, note that unless you have two different connections to MySQL servers, giving the link reference is superfluous.