Array Issue??

[ndog4ever]

I am trying to make a simple array based off what ID a user has for a login. I know the variable is getting passed correctly because I tested it with a print statement. However, I am stuck on why i cannot get my array to work. I have tried printing the $username variable and it's blank. I am not sure what else to do or try? I appreciate any help.


$db = @mysql_select_db($db_name, $connection) or die("Couldn't Select Database."

$sql = "select username, password from $table_name where username = \"$id\"
";

$result = @mysql_query($sql, $connection) or die("Couldn't execute query."

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

$username = $row["username"];
$password = $row['password'];

}

?>

 

[AnakinPt]

do an echo of the query and run it in mysql.
Your code is ok, so the problem can be a problem with the query (table_name) or about the connection. Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[ndog4ever]

i did an echo of the result and this is what i got....

Resource id #2


I know that the record "dan" has an id number of 2.

is that what it should look like?

 

[AnakinPt]

i said the query

but if you get and resource id in the result i don't get why you don't get the values

just try one thing

instead of the while,

list($username,$password)=mysql_fetch_row($result);

echo "USERNAME: $username/$password";

and check what you gen on the screen Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[ndog4ever]

i commented out the while loop and i get the following now

---->>> USERNAME: /Resource id #2

$result = @mysql_query($sql, $connection) or die("Couldn't execute query."

list($username,$password)=mysql_fetch_row($result);
echo "USERNAME: $username/$password";

It seems like to me that the array function is screwed up, I am new to php and mysql so I don't know. Do you have any other suggestions, thanks for all your help so far.

Nate

 

[hos2]

first you used singel quotes since doubles are normal I believe on
$password = $row['password'];

I check on the login if the username and the password give a result in the query like

$query="SELECT name, password FROM users where name like '$name' and password like '$password'";
$rs=mysql_query($query,$conn);
$aantal=mysql_num_rows($rs);

and then I use a bit of javascript since I'm not quite experienced with php how to do that part but I continue if the number of rows =1 and otherwise go back

<script language=&quot;javascript&quot;>
function move(url)
{
window.location.replace(url);
}
<?
if ($aantal <> 1){
print &quot;move('login.php3');&quot;;
}
else {
print &quot;move('cookie.php3?loginnaam=$naam&loginid=$lluid&sleutel=$sleutel');&quot;;
}

print (&quot;
</script>

 

[AnakinPt]

hos:
you can use single, double or none quotes while refering to an array index. $array['username'], $array[&quot;username&quot;], $array[username] referes to the same thing.


i think that $table_name may not be set, and so, to check it, do:

echo $sql;

and see if the sql query is properly well. Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[KarveR]

I'd change the sql query slightly:
$sql = &quot;select username, password from '$table_name' where username = '$id'&quot;; ***************************************
Party on, dudes!

 

[AnakinPt]

there will be an error ...

you cannot use the quotes in the table name.
Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[KarveR]

Anikin, you are right about the table name quotes, my mistake.
I still think the connect looks odd tho.
------------------------------------------
here's one of mine:
$db = mysql_connect(&quot;server_ip&quot;,&quot;user&quot;,&quot;password&quotor die ('cannot connect to server');

mysql_select_db(my_db,$db) or die ('cannot select database');

$result = mysql_query(&quot;select * from my_table&quot;,$db) or die ('Cannot select from that table.');

***************************************
Party on, dudes!

 

[AnakinPt]

i'm not user to mysql. I dont devellop in mysql for more than one year. I don't remember the order of functions, bu evetything seems ok in his code. It must be something strange. My odds goes to that $table_name. I don't see if he setted the value of the var before. he didn't show in the full code.

Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[ndog4ever]

this is my code from the beginning on the page where the person would select the name of the contact to update or look at.

<?
$db_name = &quot;test&quot;;
$table_name = &quot;contacts&quot;;
$connection = @mysql_connect(&quot;localhost&quot;, &quot;nate&quot;, &quot;nate&quot or die (&quot;Couldn't Connect.&quot;
$db = @mysql_select_db($db_name, $connection) or die(&quot;Couldn't Select Database.&quot;
$sql = &quot;select id, username, password from $table_name order by username&quot;;
$result = @mysql_query($sql, $connection) or die (&quot;Couldn't execute query.&quot;

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

$id = $row['id'];
$username = $row['username'];
$password = $row['password'];

$option_block .= &quot;<option value =\&quot;$id\&quot;>$username </option>&quot;;

}


this is the output i am getting when i do an echo $sql:

select username, password from contacts where username = &quot;2&quot;


I am guessing it's passing the variable right, here is my sample data for my contacts table:


mysql> select * from contacts;
+----+----------+------------------+
| id | username | password |
+----+----------+------------------+
| 1 | nate | 5e8b7ec24bbc800a |
| 2 | dan | 7b08d75b22acbf46 |
+----+----------+------------------+

Well thanks for all the help, this is driving me crazy. If you have any more ideas I would appreciate your posts.

thanks

 

[AnakinPt]

of course you are not getting nothing.

you are selecting where username is equal to 2 and not the id.

change the select to

where id=2

but the program you show, you are concatenating only options. For them appear, you must put them inside a select.

And in that code, you are not echoing nothing to the screen. Anikin
Hugo Alexandre Dias
Web-Programmer
anikin_jedi@hotmail.com

 

[ndog4ever]

the last code i posted was the code for the page that the user would select a username:

the option block code for that is :

$option_block .= &quot;<option value =\&quot;$id\&quot;>$username </option>&quot;;

the following php code is the array that i am trying to display back to the screen
based on the user's ID

$sql = &quot;select username, password from $table_name where username = \&quot;$id\&quot;
&quot;;
$result = @mysql_query($sql, $connection) or die(&quot;Couldn't execute query.&quot;
while ($row = mysql_fetch_array($result)) {
$username = $row[&quot;username&quot;];
$password = $row['password'];
}

when i echo the above the sql statement i get this:

select username, password from contacts where username = &quot;2&quot;

Contacts is the correct table. when i try to echo $username or $password, nothing shows up on the screen. sorry for the confusion earlier.

 

[ndog4ever]

ive figured it out, thanks for all your help, i feel like an idiot, but i guess staring at a computer screen all day can drive you crazy. I had to set my select statement to :

where id = \$id\;

thanks again