Problems with extracting values from a row...

[ack]

I am trying to extract field data from a specific row,

 

Here is my code:

 

    $query = "SELECT * FROM music WHERE title = '%$selected%'";

 

    $musicresult = mysql_query($query) or die ("Couldn't execute query.");

 

    $row = mysql_fetch_array($musicresult);

 

      /* extract the contents of the row */

    Line 98 extract($row);

   

echo $title." "."<BR>";

    echo $select."<BR>";

    /* set contents of selected to 1  */

$selectval = $select;

++$selectval ;

 

    echo $selectval."<BR>";

/* Put the data into the database */

//$query= "UPDATE music SET select=$selectval WHERE title=$selected";

//$result = mysql_query($query) or die ("could not save new data!");

 

 

 

 

Here is the error message:

 

 

Warning: extract() [function.extract]: First argument should be an array in /home/content/a/c/k/acksfaq/html/theackshow/THANKYOU.PHP on line 98

 

The two echo statements that follow line 98 do not generate an output

 

About the only thing that is different with is table compared with a table that this code works with is the fact that this table (music) is not indexed.

 

What the heck am I doing wrong??

 

Thanks,

 

Ack

 

 

 

 

[PFMaBiSmAd]

Just because a query executes without any errors, does not mean that it matched any rows. You should use mysql_num_rows to find out if there are any rows in the result set before you attempt to fetch the data from the result set.

 

If you echo the $query variable, you will likely be able to see why it is not matching anything in your table (I suspect that $selected is either empty or does not match any title.)

[wigwambam]

Have you tried changing your query to:-

 

$query = "SELECT * FROM music WHERE title LIKE '%$selected%'";

[ack]

Thanks,  PFMaBiSmAd and wigwambam!

 

I tried "LIKE" instead of "=" and was able to pull the field contents in the row.  I guess that somehow a leading or trailng space manged to get appended to the string that I was looking for because the contents of the title field was extracted from the same table then passed from a dynamic php page to this page using URL parameters...

 

So, I have the data, now I want to increment the content of a counter field (to track how many times this particular row of data has been selected) and store it back into the table...

 

The incrementing is working but the I can't seem to manage the simple task of storing the incremented variable back into to table...

 

 

$query= "UPDATE music SET select=$selectval, WHERE title=$title";

$selectresult = mysql_query($query) or die ("could not save new data!");

 

So just how moronic am I???

 

I've slept a few times since  I last monkeyed with php...

 

Thanks in advance!

[BlueSkyIS]

as $query is invalid SQL syntax but you don't get the die() message, I suspect either error reporting is turned off or the code below is never reached:

 

$query= "UPDATE music SET select=$selectval, WHERE title=$title";
$selectresult = mysql_query($query) or die ("could not save new data!");

[ack]

as $query is invalid SQL syntax but you don't get the die() message, I suspect either error reporting is turned off or the code below is never reached:

 

$query= "UPDATE music SET select=$selectval, WHERE title=$title";
$selectresult = mysql_query($query) or die ("could not save new data!");

 

 

I do get the die("could not save new data!" ) when the code executes.

 

I have discovered that sample code found at various PHP sites tend to lack a bit when it comes to accuracy.  Thus it is often hard to understand what the proper syntax is required for proper execution.  I end up playing for hours with the code until I somehow hit upon the solution.  Very frustrating...

 

Assuming that the query contains invalid SQL syntax, what is that syntax error?  Alternatively, is there a way to debug this problem? 

 

 

By the way,...

 

This project's purpose  is to create a music request page for an Internet radio show that I have.  A plug-in has been written for winamp, but it cannot be easily adapted to other similar applications.  A PHP-based clone would be useful for other "DJs" wanting to offer request funtionality without having to sacrifice the ability to use the audio mixing software of their choice.

 

Bonus:

 

It makes for a good introduction to PHP!

 

 

 

Humbly submitted,

 

Ack

[PFMaBiSmAd]

You can use mysql_error() to find out why your query is failing. You can also look at it.

 

You have an extra comma after $selectval, that wasn't present in the first post. You also need single-quotes around the string data values in the query.

[ack]

You can use mysql_error() to find out why your query is failing. You can also look at it.

 

You have an extra comma after $selectval, that wasn't present in the first post. You also need single-quotes around the string data values in the query.

 

Well, I removed the extra comma and I tried all the variations that i could think of with single quotes including "%" as seen in the original post's $query string an all I got was the die() message - could not save new data!

 

Also, I was unable to see any results with

 

echo mysql_errno($query) ;

 

 

Placed after the

 

$query= "UPDATE music SET select=$selectval, WHERE title=$title";

$selectresult = mysql_query($query) or die ("could not save new data!");

 

Very puzzling...

 

[kenrbnsn]

Your query is still wrong. Try:

<?php
$query= "UPDATE music SET select='$selectval' WHERE title='$title'";
$selectresult = mysql_query($query) or die ("Problem with the query: $query<br>" . mysql_error());
?>

 

Ken

[ack]

Ken!

 

Thanks!

 

Your snippet suggestion revealed - in an obtuse way - the problem with the query along with the fact that the variable content is correct:

 

Problem with the query: UPDATE music SET select='1' WHERE title='Free Xone Janet Jackson The Velvet Rope Soul and R&B.wma'

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near

 

'select='1' WHERE title='Free Xone Janet Jackson The Velvet Rope Soul and R&B.wma' at line 1

 

(Sorry about the choice of the song title.  For some reason I typed in "Jackson" and clicked on the first title that appeared    )

 

The at line 1 is puzzling...

 

<?php                                line 1

session_start();              line 2

?>                                      line 3

....

 

 

Gotta get to work.  I'll check with the hosting company and find out the version of PHP that is running.

 

Thanks!

[PFMaBiSmAd]

The 'at line 1' part of the sql error message refers to line 1 of the query (queries can be formed with multiple lines of sql.)

 

The part of the query that is called out in - the right syntax to use near 'select is where the error is occurring at.

 

Select is a reserved mysql keyword, as in SELECT * FROM your_table.... Either rename your column to something else or you must enclose it in back-ticks `` every time you use it in a query.

[ack]

kenrbnsn AND PFMaBiSmAd:

 

You are the BEST!

 

You have been patient with a guy who, in a matter of 8 months, had a great deal of "learnin'" leak out of his noggin!

 

Naming a field "select" is pretty stupid.

 

That was the biggest goof that i made with this project.  Many smaller ones preceeded it.

 

All of you helped me get on the right track - which I appreciate very much!

 

If any of you get the offroading bug and buy a Suzuki Samurai or Sidekick (know by other names incluiding Asuna, Tracker and Vitara), feel free to stop by http://www.acksfaq.com for help!

 

Most Sincerely,

 

Ack