adding MYSQL query info to php

[abda53]

im trying to add a query string into php..

<?
$query1 = "select vote1 from user where id=xxx";
echo/print($query1);
?>

i dont think you can add that straight into php.. whats the code to do that?
the "select vote1 from user where id=xxx" is the mysql query string im using..
the xxx is the code that actually shows which user id is logged in, which i think is going to have to be a php code.. but im not sure.. anyone have any ideas?

ok.. thats kinda confusing i guess..
how can i do a print on like $query1.. where $query1 is a mysql query string?

and can a mysql query string contain php code in it? kinda like...
select vote1 from user where id=$id

thanks for any comments/suggestions

[bernouli]

after you have the variable

$query1 = "select vote1 from user where id='xxx'";
you must run it with

$data = mysql_query($query1);

and can a mysql query string contain php code in it? kinda like...
select vote1 from user where id=$id

yes you can

[abda53]

do you mean like this??

<?
$query1 = "select vote1 from user where id=1";
$data1 = mysql_query($query1);
print($query1);
?>

when i try
print($data1);
i get a:
"Resource id #9"

when i try
print($query1);
i get a:
"select vote1 from user where uid=1"

and when i do a
<?
$query1 = "select vote1 from user where id=$uid";
$data1 = mysql_query($query1);
print($query1);
?>
i get a:
"select vote1 from user where id=abda53"

what i meant i guess.. is can you add php to a mysql query, where it will query the php $code like a id#?

still lost on this one

[bernouli]

all you do is correct,
you will just get the text, becase you print it instead run the query

<?
$query1 = "select vote1 from user where id=$uid";
$data1 = mysql_query($query1);
$datarows = mysql_num_rows($data1);
if($datarows > 0) {
//what you'll do with the result

}

?>

[abda53]

<?
$query1 = "select vote1 from user where id=$uid";
$data1 = mysql_query($query1);
$datarows = mysql_num_rows($data1);
if($datarows > 0) {
//what you'll do with the result
}
?>

now if i use a mysql_num_rows wouldnt that just give me a result showing how many rows i got? i want to show the actual data in a row.. like the actual result of query1

if i do like a..
<?
$query1 = "select vote1 from user where id=1";
$data1 = mysql_query($query1);
$datarows = mysql_result($data1,0);
if($datarows > 0) {
//what you'll do with the result
echo $datarows;
}
?>
i will get the results for id 1..

but if i change it and do like a
<?
$query1 = "select vote1 from user where id='id'";
?>

i end up getting Warning: Unable to jump to row 0 on MySQL result index 9 in /home2/www/abda53/vote/user/user.php3 on line 80 *line 80=$datarows = mysql_result($data1,0);

so how would i do a query for a result on vote1, but for the id make it like $id?

[mikek]

Ok, if you want to retrieve data from a database heres the code, I will use a database named "Test" and table named Users the rows will be called "Name", "Age", "Nickname"

Now if you want to retrieve the Name, Age and Nickname from the database Test, you would do as the following:
<?
mysql_connect('localhost','root,'password') or die ("Problem connecting to DataBase");
$query = "SELECT * FROM Users;";
$result = mysql_db_query("Test", $query);
while($i = mysql_fetch_row($result)) {

$Name= $i[0]; //This retrieves the Name
$Age= $i[1]; //This retrieves the Age
$Nickname= $i[2]; //This retrieves the Nickname
}
//then if you wanted to print out the name or age or nickname you would do
echo "The name of person 1 is $Name. The age of person 1 is $Age. The nickname of person 1 is $Age.
?>

[abda53]

i figured out what you did.. works now. thanks.