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Solved

results from db

Posted on 2000-04-04
6
168 Views
Last Modified: 2006-11-17
im trying to get results from database.
i want to write "good" if the $username
variable is show in the db,and "bad"
if it isn't.
here's the code:
$sql="select name,passwd from user where name=$username";
mysql_connect("haag","elad");
$result=mysql_db_query("appscan",$sql);
$row=mysql_fetch_object($result);
if ($row){
echo ("good");
}else{
echo ("bad");
}
....

it is not working...
0
Comment
Question by:eladr
  • 3
  • 3
6 Comments
 
LVL 8

Expert Comment

by:us111
ID: 2684268
mysql_db_query needs 3 parameters:

int mysql_db_query(string database, string query, int [link_identifier] );

try:

$row = mysql_query($query, $sql);
0
 
LVL 8

Expert Comment

by:us111
ID: 2684299
Oops other mistakes

$sql="select name,passwd from user where name=$username";
$conn = mysql_connect("database", "login", "password");
$result=mysql_query($sql, $conn);

$row=mysql_fetch_row($result);
if ($row){
     echo ("good");
}else{
     echo ("bad");
}
0
 
LVL 1

Author Comment

by:eladr
ID: 2684350
not working...
where im "telling" what is the database name?
sorry...completley new to php.
know ASP...
0
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LVL 1

Author Comment

by:eladr
ID: 2684360
that's the error:
Warning: Supplied argument is not a valid MySQL result resource in /home/elad/public_html/index.php on line 8

8 $row=mysql_fetch_row($result);
0
 
LVL 1

Author Comment

by:eladr
ID: 2684380
that's the error:
Warning: Supplied argument is not a valid MySQL result resource in /home/elad/public_html/index.php on line 8

8 $row=mysql_fetch_row($result);
0
 
LVL 8

Accepted Solution

by:
us111 earned 2 total points
ID: 2684467
Sorry mistake, try that:

<?
$sql="select * from table";
$conn = mysql_connect("localhost", "", "");

$result = mysql("yourdatabase",$sql,$conn);

$row=mysql_fetch_row($result);
if ($row){
        echo ("good");
}else{
        echo ("bad");
}
?>
0

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