[SOLVED] Error

[EagleAmerican]

Hi and much thanks to whomever can help me with my problem. I've tried it for hours to get it myself but just can't. So here goes:

 

Error Message when trying to go to page:

 

Parse error: parse error, unexpected T_VARIABLE, expecting ',' or ';' in /home/www/blah.blah/add_parasite.php on line 66

 

PHP lines causing problem:

 

 if (isset($_POST['parasitename'])) {
   $parasitename = $_POST['parasitename'];
   $parasitefrom = $_POST['parasitefrom'];
   $parasitegenre = $_POST['parasitegenre'];
   $parasitedescription = $_POST['parasitedescription'];
   $parasitetechinfo = $_POST['parasitetechinfo'];
   $sql = "INSERT INTO parasite SET
       parasitename='$parasitename',
       parasitefrom='$parasitefrom',
       parasitegenre='$parasitegenre',
       parasitedescription='$parasitedescription',
       parasitetechinfo='$parasitetechinfo',
       parasitedateadded=CURDATE()";
   if (@mysql_query($sql)) {
     echo '<p>The parasite '$parasitename' has been added.</p>';
   } else {
     echo '<p>Error adding submitted parasite: ' .
         mysql_error() . '</p>';
   }
}

 

Thanks,

Adam

[Hypnos]

You were using single quotes inside of single quotes. Next time please point out what line is 66.

 

<?php
if (isset($_POST['parasitename'])) {
   $parasitename = $_POST['parasitename'];
   $parasitefrom = $_POST['parasitefrom'];
   $parasitegenre = $_POST['parasitegenre'];
   $parasitedescription = $_POST['parasitedescription'];
   $parasitetechinfo = $_POST['parasitetechinfo'];
   $sql = "INSERT INTO parasite SET
       parasitename='$parasitename',
       parasitefrom='$parasitefrom',
       parasitegenre='$parasitegenre',
       parasitedescription='$parasitedescription',
       parasitetechinfo='$parasitetechinfo',
       parasitedateadded=CURDATE()";
   if (@mysql_query($sql)) {
     echo "<p>The parasite '$parasitename' has been added.</p>";
   } else {
     echo '<p>Error adding submitted parasite: ' .
         mysql_error() . '</p>';
   }
}

[EagleAmerican]

Hi and thanks so much for your help. That part works now. However now I can't get it to list all of the parasites in the database. The message I get is:

 

Parasite list:

 

Error performing query1:

 

Here's the coding where the problem occurs:

 

echo '<p>Parasite list:</p>';

$name = @mysql_query('SELECT parasitename FROM parasite');
if (!$result) {
exit('<p>Error performing query1: ' . mysql_error() . '</p>');
}
$from = @mysql_query('SELECT parasitefrom FROM parasite');
if (!$result) {
exit('<p>Error performing query2: ' . mysql_error() . '</p>');
}
$genre = @mysql_query('SELECT parasitegenre FROM parasite');
if (!$result) {
exit('<p>Error performing query3: ' . mysql_error() . '</p>');
}
$description = @mysql_query('SELECT parasitedescription FROM parasite');
if (!$result) {
exit('<p>Error performing query4: ' . mysql_error() . '</p>');
}
$techinfo = @mysql_query('SELECT parasitetechinfo FROM parasite');
if (!$result) {
exit('<p>Error performing query5: ' . mysql_error() . '</p>');
}

 

And the MySQL is structured like this:

 

 Full Texts  	 id  	 parasitename  	 parasitefrom  	 parasitegenre  	 parasitedescription  	 parasitetechinfo  	 parasiteadddate
Edit 	Delete 	1 	Test.Blah 	Unknown 	Unknown 	This is a Test Parasite. 	Test Parasite. 	0000-00-00

[Hypnos]

Your result variables aren't consistant.

 

<?php
$name = @mysql_query('SELECT parasitename FROM parasite');
if (!$result) {
exit('<p>Error performing query1: ' . mysql_error() . '</p>');
}

 

Should be:

<?php
$name = @mysql_query('SELECT parasitename FROM parasite');
if (!$name) {
exit('<p>Error performing query1: ' . mysql_error() . '</p>');
}

 

Notice how the var used in the IF statement is now the same as the one used in mysql_query? That's how it's suppose to be.

 

[EagleAmerican]

Ohh.. thank you so much. This is very deeply appreciated and I am so glad I found this wonderful place. This is my first PHP site and I'm doing it for a friend and this has help so much.

 

Thanks,

Adam