[SOLVED] Going Insane Here - Help Please

[widget]

Using php and mySQL

 

Table consists of

 

user_name

amulet_id

 

What I am trying to do is display a table of 9 cells (3 by 3) kinda like a puzzle.

Labeled 1 to 9

 

If the user has a puzzle piece, it shows in the table.

 

example, if they have piece number 9 then in the database would be...

 

user_name: pinkkitty

amulet_id: 9

 

Here is my code

 

$amulet_piece = mysql_query("SELECT * FROM amulet WHERE user_name = $username");

if ($amulet_piece[amulet_id] = 1)
{
$amulet1 = "<img src=$base_url/images/real image here.gif>";
}
else
{
$amulet1 = "<a href=amulet.pro.php?amulet=1><img src=$base_url/faded image here.gif></a>";
}

if ($amulet_piece[amulet_id] = 2)
{
$amulet2 = "<img src=$base_url/images/real image here.gif>";
}
else
{
$amulet2 = "<a href=amulet.pro.php?amulet=2><img src=$base_url/faded image here.gif>";
}

 

I have tried all sorts of combination's and I cant get it to work correctly, any help is greatly appreciated.

 

I just know theres probably an easier way to accomplish what I am doing.

 

[mtoynbee]

You'll need to use "mysql_fetch_array" or "mysql_fetch_assoc" to assign a variable to the result set.

 

 

[widget]

erm example please

 

Im a total noob really

[mtoynbee]

Also I've just noticed you need to use a double equals on your if statement otherwise you are assigning that value not validating it. This would also return as true.

 

$amulet_piece = mysql_query("SELECT * FROM amulet WHERE user_name = $username");

if ($amulet_piece[amulet_id] == 1)
{
$amulet1 = "<img src=$base_url/images/real image here.gif>";
}
else
{
$amulet1 = "<a href=amulet.pro.php?amulet=1><img src=$base_url/faded image here.gif></a>";
}

[widget]

I had == at one point

 

its one of the combination's I tried

[mtoynbee]

$amuletQuery = mysql_query("SELECT * FROM amulet WHERE user_name = $username");

// slight edit added
while ($amulet_piece = mysql_fetch_array($amuletQuery, MYSQL_ASSOC)) 
{

// Your code here

}

[widget]

so would I still put the if(amuletpiece ==1) {do this} else {do that} etc etc

[mtoynbee]

Yes

[widget]

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource

[widget]

$amuletQuery = mysql_query("SELECT * FROM amulet WHERE user_name = $username");

while ($amulet_piece = mysql_fetch_array($amuletQuery))
{

if ($amulet_piece[amulet_id] == 1)
{
$amulet1 = "<img src=$base_url/images/user_images/opg_1/items/item_30893.gif>";
}
else
{
$amulet1 = "<a href=amulet.pro.php?amulet=1><img src=$base_url/farm/images/1.gif></a>";
}

}

[mtoynbee]

try this instead:

 

while ($amulet_piece = mysql_fetch_assoc($amuletQuery))
{

}

 

This will make $amulet_piece an array like $amulet_piece['id'] etc..

[widget]

same error

 

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

[mtoynbee]

Silly question, but I assume you have a line in your code to connect to the database?

[widget]

yes

 

the script does an include

 

all other database connections on the page work

[mtoynbee]

This is a textbook example of this type of query:

 

<?php
mysql_connect("localhost", "mysql_user", "mysql_password") or
    die("Could not connect: " . mysql_error());
mysql_select_db("mydb");

$result = mysql_query("SELECT id, name FROM mytable");

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    printf("ID: %s  Name: %s", $row["id"], $row["name"]);
}

mysql_free_result($result);
?>

 

You might want to try tweaking this script and see what outputs.

 

[mtoynbee]

Ah, it could be your query is erroring!

 

Try '".$username."' instead of just $username.

[widget]

tried that, it outputs the correct database information

[widget]

changed .$username.

 

nothing happened

[mtoynbee]

Did you try single quotes?

 

Also try this:

 

$link = mysql_connect("localhost", "mysql_user", "mysql_password");

mysql_select_db("db_name", $link);
$amuletQuery = mysql_query("SELECT id, name FROM mytable",$link);
echo mysql_errno($link) . ": " . mysql_error($link) . "\n";

 

 

 

[widget]

gave this result

 

0:

 

 

the database connection is fine

[widget]

All i need to do is, find out if the user has a number 1 2 3 4 5 6 7 8 and/or 9

 

then if so, display a certain image.

 

Man its sooo frustrating and I really appreciate your help here

 

 

[widget]

Theres something wrong with my if else

[mtoynbee]

 

 

Try == "1" instead of == 1

 

 

sometimes if the number format isn't correct this can sort the problem.

 

I expect this can be solved by int_val also

 

What does the code after this look like?

[widget]

I moved the info to another table and all is working