AW: [PHP-DB] selecting and updating fields

[Lars Jedinski]

You could do:

mysql_query("update url set nrviews=nrviews+1 where url='$url'");
if(mysql_affected_rows()==0)
  mysql_query("insert into url (urn, nrviews) values ('$url',1)");

Lars

> -Ursprüngliche Nachricht-
> Von: Claudiu Bandac [mailto:[EMAIL PROTECTED] 
> Gesendet: Montag, 13. Oktober 2003 05:40
> An: [EMAIL PROTECTED]
> Betreff: [PHP-DB] selecting and updating fields
> 
> 
> Hi !
> 
> I've got a form that passes a variable to a script.
> The script connects to a database and I need to check if a 
> field containing that variable  already exists in the table, 
> and if it does, to select the "nr_of_views" field, increment 
> it, and update the table and if it doesn't, to create a new 
> field with that variable and set nr_of_views to 1
> 
> the table in my database that looks like this :
> url   nrviews
> www.google.com1
> www.yahoo.com 3
> 
> 
> And I've tried something like this:
> 
> 
> $query = mysql_query("select * from url") or die (mysql_error());
> while ($row = mysql_fetch_row($query)) 
>   {
> if ($url === $row["url"]) 
>   {
> $nrviews = $row["nrviews"];
>   $nrviews++
> mysql_query("update url set 
> nrviews='$nrviews' where url='$url'") or
> die (mysql_error());
>}
>else 
> {
> $nrviews=1;
>   mysql_query("insert into url 
> (url, nrviews) values ('$url','$nrviews')") or
>   die (mysql_error());
>  }
>  } 
> 
> 
> AND STILL DOESN'T WORK !!! 
> When I get a variable that is not in the database, the whole 
> thing goes crazy ! (it inserts lots of fields with that 
> variable and nrviews=1
> 
> 
> Please HELP !
> 
> Thank You !
> 
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[PHP-DB] selecting and updating fields

[Claudiu Bandac]

Hi !

I've got a form that passes a variable to a script.
The script connects to a database and I need to check if a field containing that 
variable  already exists in the table, and if it does, to select the "nr_of_views" 
field, increment it, and update the table and if it doesn't, to create a new field 
with that variable and set nr_of_views to 1

the table in my database that looks like this :
url nrviews
www.google.com  1
www.yahoo.com   3


And I've tried something like this:


$query = mysql_query("select * from url") or die (mysql_error());
while ($row = mysql_fetch_row($query))
{
if ($url === $row["url"])
{
$nrviews = $row["nrviews"];
$nrviews++
mysql_query("update url set nrviews='$nrviews' where url='$url'") or
die (mysql_error());
 }
 else
  {
$nrviews=1;
mysql_query("insert into url (url, nrviews) values 
('$url','$nrviews')") or
die (mysql_error());
   }
 }


AND STILL DOESN'T WORK !!!
When I get a variable that is not in the database, the whole thing goes crazy ! (it 
inserts lots of fields with that variable and nrviews=1


Please HELP !

Thank You !

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