debugging says something is wrong with this

But I don't know what it is

[Sarah Atkinson]

$addnomsql="INSERT INTO $thislist (name,reason) VALUES ('$nomname','$nomreason')";

[dreamcatcher]

Sarah,

Only problem here would be with the $thislist variable. Is it populated? Have you echoed it before your query to see what value it holds?

Have you tried using mysql_error()?

$addnomsql="INSERT INTO $thislist (name,reason) VALUES ('$nomname','$nomreason') or die(mysql_error())";

dc

[coopster]

You may also want to addslashes [php.net] or use mysql_real_escape_string() [php.net] on those variables.

[Sarah Atkinson]

HEre is the errror message i get back

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\Inetpub\wwwroot\thegreatest\nominate.php on line 35

not added due to database error: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 '\" (name,reason) VALUES ('testing ','testing one two')' at line

[jatar_k]

that means the query died

2 things

1. echo the full constructed query before sending it to mysql_query
echo '<p>',$addnomsql;

2. add an or die to your query statement
$query = mysql_query($addnomsql) or die ('<p>' . mysql_error());

though that may be the error from mysql, I hate those ones, they don't tell you anything. My guess is that the $thislist var is not a proper tablename or there is something extra in there

[Sarah Atkinson]

now i'm getting
Nominee was not added due to database error:Unknown column 'testing' in 'field list'

.

This is driving me nuts. I cazn't figure it out and my debugger isn't helping my at all. OF course it doesn't help that the only time I get to work on it is aboult 30 min before bed when everything is blurry and my eyes are stinging.

[jatar_k]

what output do you get when you echo your full query.

my guess is it is missing the tablename so it thinks your testing values are the column names

[Sarah Atkinson]

INSERT INTO tg_american (name,reason) VALUES (Testin,123)

Nominee was not added due to database error:Unknown column 'Testin' in 'field list'

[Sarah Atkinson]

I changed it to a different format here is the code and here is the error

if(isset($_POST['name'])){  
 $sql="SELECT name FROM $thislist";
 echo " sql2= $sql<br>";
 $results=mysql_query($sql);
 if(!$results){
 exit('<p>erros retrieving data!<br />' . 'Error:' . mysql_error() . '</p>');}
 $validnominecheck='yes';
 $nomname=$_POST['name'];
 $nomreason=$_POST['reason'];
 echo " <br>Nominee name= $nomname";
 while($stuff=mysql_fetch_array($results)){
  $dbname=$stuff['name'];
  if($dbname==$nomname)
  {$validnominecheck='no'; }
 }
 if($validnominecheck=='yes'){
  $addnomsql="INSERT INTO $thislist name=$nomname, reason=$nomreason";
  echo "$addnomsql <br>";
  if(@mysql_query($addnomsql)){
   echo'<p>Your nominee has been added</p>';
  }else{
   echo'<p>Nominee was not added due to database error:' . mysql_error().'</p>';
  }
 }else{
  echo '<p>Nominee not added because nominee has already been nominated</p>';
 }
 exit();

Nominee name= testing
INSERT INTO tg_american name=testing, reason=123

Nominee was not added due to database error: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 'name=testing, reason=123' at line 1

[dreamcatcher]

Sarah,

The syntax you are using for the insert is actually the update syntax. Try:

INSERT INTO $thislist SET (name,reason) VALUES ('$nomname','$nomreason');

dc

[Sarah Atkinson]

that was the first way I was doing it

this is the error i get

Nominee name= testing
INSERT INTO tg_american SET (name,reason) VALUES ('testing','1234')

Nominee was not added due to database error: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 '(name,reason) VALUES ('testing','1234')' at line 1

[dreamcatcher]

Oops, SET shouldnt be used in the INSERT query, sorry about that.

INSERT INTO $thislist (name,reason) VALUES ('$nomname','$nomreason');

I can`t really see any reason for that failing unless there is a problem with the $thislist variable, which there doesn`t seem to be as the name of the table is in the error query.