Scaler Values

evan18h

I got a php warning: "cannot use a scalar value as an array".... how can i fix this?

Also how do i query just to show all values begining in a character...like show all enties beginning with just A... or just B.

TheDefender

Regarding the first question, we need more details and some code...

Regarding the second question:

SELECT * FROM tablename WHERE upper(left(last_name, 1)) = 'A'

This query selects all rows in tablename where the last_name field starts with an A. I added the upper to ensure integrity in case sensitivity, because things may be in there with upper or lower, and this way, will always upper the first letter, and compare it to a cap letter... No muss no fuss.

You could write this to loop through letters, or sub a variable in for the letter, so you can make it flexible. I personally use this type of query in a Corporate Directory I have written.

nemonoman

In MySQL

SELECT * FROM tablename WHERE last_name like 'a%'

Finds every last name that starts with a or A. MySQL is case insensitive in string comparison.

evan18h

ok... i have included some code...now the only problem I have with this code is that it is giving back an warning on this line:

$link[$xcount] = $row["link"];

the warning is: "cannot use a scalar value as an array".
What is a scalar value, and how is this any differnt from the other data I have put to arrays? How can I fix it?

$query = "SELECT `title`, `artist`, `date`, `link` FROM `dl_db` ORDER BY '".$_GET['type']."' ".$_GET['order']."";
	$result = mysql_query($query) or die("Could not successfully run query ($query) from DB: " . mysql_error());
	while ($row = mysql_fetch_assoc($result)) { 
		$title[$xcount] = $row["title"];
		$artist[$xcount] = $row["artist"];
		$date[$xcount] = $row["date"];
		$link[$xcount] = $row["link"]; 
		$xcount++; }
	mysql_free_result($result);

nemonoman

Show the results after adding this:

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

print_r($row); //<<<<<<ADD THIS
$title[$xcount] = $row["title"];

nemonoman

Wait a minute: I think the problem is with $link, not with the row array.

A scalar value as an array...means you're trying to recast a string or number as an array.

My guess is you say someplace:
$link='some value';

and when you try to add an array element:
$link['x']

You get the error code.

It might also be caused by $date, $title, $artist, etc.

Roger_Ramjet

Don't mean to intrude but I'm just wondering what mysql_fetch_assoc does. It's not in the online manual and I've never come across it before.

evan18h

I have tried the print_r function to display the data... and I can see it all. anyway, the rest of the arrays work....I have a table that lists them all and the only one that doesnt come up is the $link[] array
becuase of that scaler array warning. The data that goes into the $link variable is really no differnt than the other data variables.

What can I do to fix this? Would it matter if the links were to have spaces?

Roger this might answer your question:
http://us3.php.net/manual/en/function.mysql-fetch-assoc.php

TheDefender

Roger, it's in the online manual right here

nemonoman

<?
    $link=1;
    $link[2]=2;
?>

Warning: Cannot use a scalar value as an array in c:\www\sql\zzz.php on line 3

evan18h

how would I fix it! And how is it differnet from the other arrays?

nemonoman

You're trying to recast a variable previously assigned a numeric value by using the same variable name for one of your arrays:

Find the offending variable

$title
$artist
$date
$link (probably $link)

that you have previously defined as a number. Give it a different name.

OR
before your while loop,
[man]unset[/man]
the variables named above.

evan18h

Thanks a bunch.... it finally works now.
The whole problem was becuase I called my sql connection with $link...found that and changed the array to $link2[], now it works!