whats wrong with this sql statement

[icymint3]

$num_result = mysql_affected_rows( /*$link*/ ); // $link is optional, and since you only use one database dont pass it 

$link would be the result of a call to

/* connect to database */
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password'); 

which is probably nestled safely in conn.php

[leocrawf]

"$num_result = mysql_affected_rows()"
i tried that and still got the same error.

[icymint3]

$num_result = mysql_affected_rows( $dblink ); 

[leocrawf]

tried that but this is the result:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'WHERE id = 2' at line 1

[icymint3]

post the code youre using, i'm pretty sure you put in the comma again.

[leocrawf]

Ran the code agian and this is what i am getting:

"Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/someurl/update.php on line 18
success= 1udate=3"

However the dtatbase field is updated. Wierd enough. Still would rather it work perfectly though.
this is the code:

<?php
  include_once("conn.php");
  
  $udate = rawurlencode($_GET["udate"]);
  $id = rawurlencode($_GET["id"]);

// This is the SQL statement that will update the appropriate table in the database
$query = "UPDATE ipcinfo SET udate = $udate WHERE id = $id";
if (!$result = mysql_query($query)) {
  die(mysql_error());
}
$num_result = mysql_affected_rows($dblink);
/* This just gets the number of rows in the Query - It's just a check to see if the id exists - If not it echos out an error statement. */

// If the number of rows is not equal to one then it echos out an error statement 

  if ($num_result == 1) {
  if($row = mysql_fetch_array($result));
   echo"success= 1";
   echo"udate=$udate";
   }
  else
  {
    echo( "success=0" );
  }

  // Clean up
  mysql_close($dblink);
?>

line 18 is this : if($row = mysql_fetch_array($result));

thanks for any help.

[recursion]

if($row = mysql_fetch_array($result));
   echo"success= 1";
   echo"udate=$udate";
   }
  else
  {
    echo( "success=0" );
    //add print statement below
    print (mysql_error());
  } 

Try adding a print statement for the last error it should give a more descriptive message as to what the error cause is.

[icymint3]

thats because your query was an UPDATE and updates do not return results (records) as SELECT does, mysql_fetch_array has no meaning in this context just remove it.

if ($num_result == 1) 
{
   if($row = mysql_fetch_array($result)); // <-- dont know if you noticed this semicolon here

   echo"success= 1";
   echo"udate=$udate";
}
else
{
   echo( "success=0" );
} 

corrected code

<?php
  include_once("conn.php");
  
  $udate = rawurlencode($_GET["udate"]);
  $id = rawurlencode($_GET["id"]);

  // This is the SQL statement that will update the appropriate table in the database
  $query = "UPDATE ipcinfo SET udate = $udate WHERE id = $id";
  $result = mysql_query($query) or die(mysql_error());

  /* This just gets the number of rows in the Query - It's just a check to see if the id exists - If not it echos out an error statement. */
  $num_result = mysql_affected_rows($dblink);

  echo "success = $num_result"; // show success code ... comment this out
  if ($num_result == 1) echo "udate = $udate"; // if its a success then  show $udate

  // Clean up
  mysql_close($dblink);
?>

[death_knight]

yeh.. madness that.. update edits records based on ids.. not return as icymint3 pointed out. you cant fetch on a update resource.

[leocrawf]

Thanks you all it worked perfectly. boy that was a difficult trip. Glad to know that closure is here.