?
Solved

results from db

Posted on 2000-04-04
6
Medium Priority
?
190 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
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 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
What does it mean to be "Always On"?

Is your cloud always on? With an Always On cloud you won't have to worry about downtime for maintenance or software application code updates, ensuring that your bottom line isn't affected.

 
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 8 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

Featured Post

Independent Software Vendors: We Want Your Opinion

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Introduction This article is intended for those who are new to PHP error handling (https://www.experts-exchange.com/articles/11769/And-by-the-way-I-am-New-to-PHP.html).  It addresses one of the most common problems that plague beginning PHP develop…
This article discusses how to implement server side field validation and display customized error messages to the client.
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.
The viewer will learn how to create a basic form using some HTML5 and PHP for later processing. Set up your basic HTML file. Open your form tag and set the method and action attributes.: (CODE) Set up your first few inputs one for the name and …
Suggested Courses

719 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question