Parse error, drop-down menu

HogDog · October 19, 2008

Hello all,

My php is giving me this error: Parse error:

parse error, unexpected T_STRING, expecting ',' or ';' in /home/doc_update.php on line 39

What I'm trying to do is populate a drop-down menu with a list of products from a database, then the admin could select the product, then a type of file(documentation, firmware, software), then enter the directory of the file. this info would then be inserted in to a certain table in the same database.

This is my php file:

<?php
session_start(); 
if(!isset($_SESSION['adminctrl'])){ 
header('Location: admin.php'); die('<a href="admin.php">Login first!</a>');
   }
   elseif(!isset($_SESSION['production'])){
   header('Location: production.php'); die('<a href="production.php">Login first!</a>');
   }
$query = mysql_connect("*********************", "****", "*******") or die(mysql_error());
mysql_select_db('*********', $query) or die(mysql_error());

$len = strlen($_POST['type']);
$len = strlen($_POST['product']);
$len = strlen($_POST['directory']);
@mysql_query('INSERT INTO `product_docs_support` (type, product, directory) VALUES (\''.mysql_real_escape_string($_POST['type']).'\', \''.mysql_real_escape_string($_POST['product']).'\',\''.mysql_real_escape_string($_POST['directory']).'\')');
echo"Update was successful.";
php?> 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Customer Update</title>
</head>
<html>
<body>
<form method="post" action="customer_info.php">
<?php
$query = mysql_connect("****************", "*****", "********") or die(mysql_error());
mysql_select_db('********', $query) or die(mysql_error());

         $conn ="SELECT product_id FROM `product`" ;
         $result = mysql_query($conn,$query);
  
         while($row=mysql_fetch_row($result))
          {
           $user[] = $row[0];
          }
         
         echo "<selectname="product">\n";
         foreach( $url as $y  )
                {
                 echo "<option value="$y">\n" $y."<option/>\n";
                }
         echo "</select>\n";

mysql_free_result($result);

php?> 

<br /> 
<selectname="type">
<option value="doc">Documentation<option/>
<option value="firmware">Firmware<option/>
<option value="software">Software<option/>
</select>

directory:
<input type="text" name="directory" />

<input type="submit" name="submit" value="Update" />
</form>
</body>
</html>

I'm a newbie but I'm guessing its somewhere in here:

    $conn ="SELECT product_id FROM `product`" ;
         $result = mysql_query($conn,$query);
  
         while($row=mysql_fetch_row($result))
          {
           $user[] = $row[0];
          }
         
         echo "<selectname="product">\n";
         foreach( $url as $y  )
                {
                 echo "<option value="$y">\n" $y."<option/>\n";
                }
         echo "</select>\n";

mysql_free_result($result);

I'm guessing what I did to try and populate a drop-down menu won't work that way.

thanks for any help.

JasonLewis · October 19, 2008

You have multiple errors, all of the same type. Look at your script, you do this a lot:

echo "<option value="something">Something</option>";

Although it's not always with options, its with other stuff as well. You cannot open a string with double quotes the have double quotes inside the string because PHP thinks you are closing it off. You can solve it by escape the quotes:

echo "<option value=\"something\">Something</option>";

Or you can use single quotes on the inside, or outside:

echo "<option value='something'>Something</option>";

philipolson · October 19, 2008

Also, it's ?> and not php?> ...

Also, 100% of the time a parse error is for invalid PHP syntax.

Also, most of the time when PHP says an error is on a certain line the problem exists on the line before it (because it becomes a problem after)... anyway, so here PHP said line 39 and I reckon line 38 had the problem... the first one at least.

Also, when posting a question, it's helpful to say which line is in the error... so in this case state which line is 39.

Sorry I wrote "Also," so much here, I think it's getting late.

HogDog · October 19, 2008

thanks a lot guys.

Now I'm getting this error on line 38:

Parse error: parse error, unexpected T_VARIABLE, expecting ',' or ';' in /home/doc_update.php on line 38

I've looked over my code but can't seem to find the problem. thanks for all your help.

<?php
session_start(); 
if(!isset($_SESSION['adminctrl'])){ 
header('Location: admin.php'); die('<a href="admin.php">Login first!</a>');
   }
$query = mysql_connect("**************", "******", "*************") or die(mysql_error());
mysql_select_db('*******', $query) or die(mysql_error());

$len = strlen($_POST['type']);
$len = strlen($_POST['product']);
$len = strlen($_POST['directory']);
@mysql_query('INSERT INTO `product_docs_support` (type, product, directory) VALUES (\''.mysql_real_escape_string($_POST['type']).'\', \''.mysql_real_escape_string($_POST['product']).'\',\''.mysql_real_escape_string($_POST['directory']).'\')');
echo"Update was successful.";
?> 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Customer Update</title>
</head>
<html>
<body>
<form method="post" action="customer_info.php">
<?php
$query = mysql_connect("*****************", "***********", "**********") or die(mysql_error());
mysql_select_db('***********', $query) or die(mysql_error());

         $conn ="SELECT product_id FROM `product`" ;
         $result = mysql_query($conn,$query);
  
         while($row=mysql_fetch_row($result))
          {
           $user[] = $row[0];
          }
         
         echo "<selectname='product'>\n";
         foreach($url as $y){
                 echo "<option value='$y'>\n" $y."<option/>\n";
                }
         echo "</select>\n";

mysql_free_result($result);

?> 

<br /> 
<selectname="type">
<option value="doc">Documentation<option/>
<option value="firmware">Firmware<option/>
<option value="software">Software<option/>
</select>

directory:
<input type="text" name="directory" />

<input type="submit" name="submit" value="Update" />
</form>
</body>
</html>

JasonLewis · October 19, 2008

Look at line 38, your not breaking out of the string properly.

echo "<option value='$y'>\n" $y."<option/>\n";

Should be this:

echo "<option value='$y'>\n".$y."</option>\n";

(I also fixed your HTML, you had <option/> not </option>)

HogDog · October 19, 2008

thanks a ton, I can't believe missed that and the html. Thanks

However I am presented now with these errors:

 Warning: mysql_fetch_row(): supplied argument is not a valid MySQL result resource in /home/doc_update.php on line 31

Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in /home/doc_update.php on line 42

<?php
session_start(); 
if(!isset($_SESSION['adminctrl'])){ 
header('Location: admin.php'); die('<a href="admin.php">Login first!</a>');
   }
$query = mysql_connect("***********", "********", "***********") or die(mysql_error());
mysql_select_db('********', $query) or die(mysql_error());

$len = strlen($_POST['type']);
$len = strlen($_POST['product']);
$len = strlen($_POST['directory']);
@mysql_query('INSERT INTO `product_docs_support` (type, product, directory) VALUES (\''.mysql_real_escape_string($_POST['type']).'\', \''.mysql_real_escape_string($_POST['product']).'\',\''.mysql_real_escape_string($_POST['directory']).'\')');
echo"Update was successful.";
?> 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Customer Update</title>
</head>
<html>
<body>
<form method="post" action="customer_info.php">
<?php
$query = mysql_connect("************", "********", "*********") or die(mysql_error());
mysql_select_db('*********', $query) or die(mysql_error());

         $conn ="SELECT product_id FROM `product`" ;
         $result = mysql_query($conn,$query);
  
         while($row=mysql_fetch_row($result))
          {
           $user[] = $row[0];
          }
         
         echo "<select name='product'>\n";
         foreach($url as $y){
                 echo "<option value='$y'>\n" .$y."</option>\n";
                }
         echo "</select>\n";

mysql_free_result($result);

?> 

<br /> 
<select name="type">
<option value="doc">Documentation</option>
<option value="firmware">Firmware</option>
<option value="software">Software</option>
</select>
<br/>
directory:
<input type="text" name="directory" />

<input type="submit" name="submit" value="Update" />
</form>
</body>
</html>

HogDog · October 19, 2008

sorry for the double post. I changed my code around a little bit and emilinated two errors. Now I'm getting this message:

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /home/content/J/a/p/JapII/html/reftek/doc_update.php on line 17

heres my code, thanks a ton.

<?php
session_start(); 
if(!isset($_SESSION['adminctrl'])){ 
header('Location: admin.php'); die('<a href="admin.php">Login first!</a>');
   }
$query = mysql_connect("***************", "************", "************") or die(mysql_error());
mysql_select_db('**********', $query) or die(mysql_error());

$len = strlen($_POST['type']);
$len = strlen($_POST['product']);
$len = strlen($_POST['directory']);
@mysql_query('INSERT INTO `product_docs_support` (type, product, directory) VALUES (\''.mysql_real_escape_string($_POST['type']).'\', \''.mysql_real_escape_string($_POST['product']).'\',\''.mysql_real_escape_string($_POST['directory']).'\')');
echo"Update was successful.";

         $conn ="SELECT product_id FROM `product`" ;
         $result = mysql_query($conn,$query);
         while($row=mysql_fetch_assoc($result))
          {
           $url[] = $row['product_id'];
          }
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Customer Update</title>
</head>
<html>
<body>
<form method="post" action="customer_info.php">
<br/>
Product:<br/>
<?php
         echo "<select name='product'>\n";
         foreach($url as $y){
                 echo "<option value='$y'>\n" .$y."</option>\n";
                }
         echo "</select>\n";
?>
<br /> <br/>
File Type:<br/>
<select name="type">
<option value="doc">Documentation</option>
<option value="firmware">Firmware</option>
<option value="software">Software</option>
</select>
<br/><br/>
Directory:<br/>
<input type="text" name="directory" />
<br/>
<input type="submit" name="submit" value="Update" />
</form>
</body>
</html>

AndyB · October 20, 2008

Trap errors while developing.

Change:

 $result = mysql_query($conn,$query);

to:

 $result = mysql_query($conn,$query) or die("Error: ". mysql_error(). " with query ". $conn;

HogDog · October 20, 2008

okay just did that, now I get this:

Parse error: parse error, unexpected ';' in /home/doc_update.php on line 16

philipolson · October 20, 2008

Still you're making helping you difficult by not saying what line is what... like, what is line 15 and 16?

HogDog · October 20, 2008

I don't know I'm guessing, line 16 :-\

HogDog · October 20, 2008

oops, I read your post incorrectly, philipolson, line 16 is:

$result = mysql_query($conn,$query) or die("Error: ". mysql_error(). " with query ". $conn;

-thanks

AndyB · October 20, 2008

my fault.

It should be

$result = mysql_query($conn,$query) or die("Error: ". mysql_error(). " with query ". $conn);

HogDog · October 21, 2008

okay thanks here whats displaying now.

Error: Table 'raytech.product' doesn't exist with query SELECT product_id FROM `product`

pugboy · October 21, 2008

The error means exactly what it says-- the table product does not exist in your Reytech database. That is why the product_id cannot be selected.

HogDog · October 21, 2008

dang, typo in my database.

Thanks a ton guys!

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Parse error, drop-down menu

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