please help me here !!! rate code

[webtuto]

why this code douesnt work !!!!!!!!

<form method="post" action="">
<b>Rate this tutorial</b> <select value="rate">
<option name=1>1</option>
<option name=2>2</option>
<option name=3>3</option>
<option name=4>4</option>
<option name=5>5</option>
<input type="submit" value="rate" name="value">
</form>
<?php
if($_POST['value']){
$sql="insert into `tutorialz`.`tuto` SET `rate`='$_POST[rate]' WHERE `id`='$get'";
$res = mysql_query($sql) or die('Database error: ' . mysql_error());
if($res) {
  echo"Thanks for rating<br><br>";
}else{echo"it was a problem in the rate please try again!<br><br>";}
}

I use it to rate the tutorialz in my script

 

and it gives this error

 

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 'WHERE `id`='5'' at line 1

[nikefido]

if($_POST['rate'])

 

'{$_POST['rate']}'

[webtuto]

ok i did it but still the problem

[nikefido]

you still didn't fix the $_POST[rate] in the query according to what you are showing me.

[adam291086]

There are a few problems

 

a)

`id`='$get'"; - Where is $get defined are you trying to get $id from the url

 

b)

`tutorialz`.`tuto` - This should be table name. What are you trying to do?

[duclet]

Please echo out the SQL and post it here.

[DyslexicDog]

what is the datatype assigned to your rate column? If it is integer your trying to assign a string and it will kick back an error!

[webtuto]

here is the new code

 

<form method="post" action="">
<b>Rate this tutorial</b> <select value="rate">
<option name=1>1</option>
<option name=2>2</option>
<option name=3>3</option>
<option name=4>4</option>
<option name=5>5</option>
<input type="submit" value="rate" name="value">
</form>
<?php
if($_POST['value']){
$sql="INSERT INTO `rate` VALUES ('',$_POST['rate'],$get)";
$res = mysql_query($sql) or die('Database error: ' . mysql_error());
if($res) {
  echo"Thanks for rating<br><br>";
}else{echo"it was a problem in the rate plz try again!<br><br>";}
}
?>

 

but gives this error =>

 

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 '3)' at line 1

[duclet]

Again, post your SQL statement here after it has been processed.