IF Condition

[ddtpmyra]

Hi,

I want to display a button that depends on my query result. But my PHP scripts look like didn't understand what I wanted to do (smile).

#check if the file is already exist
      $query3  = "SELECT *   FROM cmr_reviewers 
                        WHERE FileID = {$id}
                        AND reviewer_name = '$user_name'
                       AND reviewed='Y'";
 
      $result3 = mysql_query($query3);
       if($result3==1)
      {
      echo"image 1";
      } 
      else 
       {
      echo "image2";
       }

[Markus]

You're going about your mysql queries all wrong.

When you do mysql_query($qu ery), that function returns a 'resource id'. You have to then use any of the available functions (mysql_fetch_ar ray, mysql_fetch_ass oc, etc) to use the results of the query.

So:

while($rows = mysql_fetch_array($result3))
{
    if($rows['FileID'] == 1)
      {
        echo"image 1";
      } 
    else 
      {
        echo "image2";
      }
}

$rows becomes an array of the data returned by the query. You give your column name as the array key.

Hope this helps.

[Atli]

A minor nitpick.
As the mysql_fetch_* functions would only return a single row at a time, naming the variable that holds it's return value $row rather than $rows might be a bit less confusing. (I said minor! :])

[Markus]

A minor nitpick.
As the mysql_fetch_* functions would only return a single row at a time, naming the variable that holds it's return value $row rather than $rows might be a bit less confusing. (I said minor! :])

I'm giving you an infraction for purposely embarrasing me! :P

[Atli]

I'm giving you an infraction for purposely embarrasing me! :P

Hehe. Luckily, moderators are immune to infractions! :)
I had just spent most of the day debugging a Java class, only to find that the entire problem was caused by a single misspelled variable name. So I was a bit touchy about that at the time :P
(How was I supposed to spot the difference between "fields" and "files" in a single line amongst like 2000 lines!)