A straight forward script that DOES NOT WORK!!!

[mysql_query]

Hi. I have a script which is really straight forward, but for the past 4 days I couldn't figure out why this stup*d script won't work. Here it is:

the form...

<form method="POST" action="pratingsystem.php">
 <?php
echo '<input type="hidden" name="jokeId" value="$id">';
?>
 <input type="radio" value="5" checked name="rating">Excellent&nbsp;
  <input type="radio" name="rating" value="4">Good&nbsp;
  <input type="radio" name="rating" value="3">OK&nbsp;
  <input type="radio" name="rating" value="2">Not-so-good&nbsp;
  <input type="radio" name="rating" value="1">Sucks</p>
  <p><input type="submit" value="Submit" name="B1"></p>
</form>

$id is gottten like this from the URL (something.php?id=1)

Proccessing script:

<?php
$db = mysql_connect("localhost","","");
mysql_select_db("dbthawowi_2",$db);;
$query = 'INSERT INTO evaluation (id, rating) VALUES ('.$_POST['jokeId'].','.$_POST['rating'].')';
$result=mysql_query($query, $db);
echo "operation successful";
?>

The script returns an "operation successful" (my comment), but nothing gets entered into the db. WHY???

Thanks.

[levi_501_dehaan]

you may want a little error checking stuff

<?php
if(isset($jokeId) && isset($rating))
{
$db = mysql_connect("localhost","","");
mysql_select_db("dbthawowi_2",$db);;
$query = 'INSERT INTO evaluation id='.$jokeid.', rating='.$rating.'
$result=mysql_query($query, $db) or die (mysql_error());
echo "operation successful";
}
?>

try that

[mysql_query]

<?php 
if(isset($jokeId) && isset($rating)) 
{ 
$db = mysql_connect("localhost","",""); 
mysql_select_db("dbthawowi_2",$db);; 
$query = 'INSERT INTO evaluation id='.$jokeid.', rating='.$rating.' 
$result=mysql_query($query, $db) or die (mysql_error()); 
echo "operation successful"; 
} 
else{
print("something went wrong...");
}
?> 

The above gives me an error on the last line (?>)

Any idea why? Yes, I put in my username and password.

Another thing.

$query = 'INSERT INTO evaluation id='.$jokeid.', rating='.$rating.' 

I have never done this before. Does this require POST or GET? I have POST on the form. Thanks.

[sarahk]

lets start with the first echo
<?php
echo '<input type="hidden" name="jokeId" value="$id">';
?>

which should be

<?php
echo "<input type='hidden' name='jokeId' value='$id'>";
?>

for some reason if you use single quotes $id won't be substituted and your hidden field won't have a value. You can check this by viewing source.

then after
$query = 'INSERT INTO evaluation (id, rating) VALUES ('.$POST['jokeId'].','.$POST['rating'].')';
I'd have

echo $query . "<br>";

Visually check it and if it seems ok then
I'd copy the output into your mysql database (I use phpMyAdmin) and see if the query runs from there.

If not, then it's a connection problem.

good luck

[sarahk]

if you get an error on the last line it means some control character isn't matched up right eg { without a }

have a look at

mysql_select_db("dbthawowi_2",$db);;
$query = 'INSERT INTO evaluation id='.$jokeid.', rating='.$rating.'

which should be

mysql_select_db("dbthawowi_2",$db);
$query = 'INSERT INTO evaluation id='.$jokeid.', rating='.$rating.' ;

the semi colon's were mucked about.

[levi_501_dehaan]

<?php
if(isset($jokeId) && isset($rating))
{
$db = mysql_connect("localhost","","");
mysql_select_db("dbthawowi_2",$db);
$query = "INSERT INTO evaluation id=".$jokeid.", rating=".$rating.";
$result=mysql_query($query, $db) or die (mysql_error());
echo "operation successful";
}
else{
print("something went wrong...");
}
?>

also on the first part
this may work even better 🙂
<?php
echo "<input type=\"hidden\" name=\"jokeId\" value=".$id.">";
?>

sorry i didnt fix some of the errors, but when ppl post code you should look for errors, cuz sometimes we're just giving you an answer not the code 🙂. so check for parse errors and such.
also to check and make sure that variables are being passed right you may want to look at variables that you think may be sketchy with
echo var_dump($variable); and see what it outputs, i use it alot, it helps with buggy code.

[mysql_query]

[breath] OK, I am finished this portion of the script [/breath]. Here's what i ended up with:

the form...

<form method="POST" action="pratingsystem.php">
 <?php
echo "<input type=\"hidden\" name=\"jokeId\" value=".$id.">"; 
?>
 <input type="radio" value="5" checked name="rating">Excellent&nbsp;
  <input type="radio" name="rating" value="4">Good&nbsp;
  <input type="radio" name="rating" value="3">OK&nbsp;
  <input type="radio" name="rating" value="2">Not-so-good&nbsp;
  <input type="radio" name="rating" value="1">Sucks</p>
  <p><input type="submit" value="Submit" name="B1"></p>
</form>

the proccessing script:

<?php 
if(isset($jokeId) && isset($rating)) 
{ 
$db = mysql_connect("localhost","",""); 
mysql_select_db("dbthawowi_2",$db); 
$query = "INSERT INTO evaluation (id, rating) VALUES ('$jokeId','$rating')";
$result=mysql_query($query, $db) or die (mysql_error()); 
echo "operation successful"; 
} 
else{
print("something went wrong...");
}
?> 

Soon, I will need to do something with it. I will try to figure it out on my own, but expect a question within a few days 😃

Thanks everyone.