Help inserting into DB via form

[cmccomas]

I need help with inserting something into a DB field based on which page I am on.

The page address would be like:

../comment.php?blogID=19

I use this query to insert into the database, but I also want to include the blogID # as a field.

$query = "INSERT INTO content(page, content, linkname) VALUES('$page', '$content', '$linkname')";

Where/how do I do the #?

Mucho Thankso

[kingbear]

Add a field into the table 'content' called blogID and do this:

$blogID = $_GET["blogID"];

$query = "INSERT INTO content(page, content, linkname, blogID) VALUES('$page', '$content', '$linkname', $blogID)";

[cmccomas]

kingbear,

Thanks. Actually the table is called 'contents' and there in an integer field called 'blogID' already in there. I tried the code you posted and here's what is happening.

If I do it exactly like you have it without the ' around $blog towards the end of the query I get this error.

Error in query: INSERT INTO comments(commentUSER, commentENTRY, commentTITLE, blogID) VALUES('Test', 'TEst', 'Test', ). 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 ')' at line 1

If I put a ' around the $blog it inserts the info to the database, but sets the blogID field to 0.

Here's my full code:

<?
	// form not yet submitted
	// display initial form
	if (!$submit)
		{
?>
	<table cellspacing="5" cellpadding="5">
		<form action="<? echo $PHP_SELF; ?>" method="POST" name="theform" id="theform">
		<tr> 
			<td width="100" align="right"><font class="text-blog"><b>Name: </b></font></td>
			<td align="left"><input size="40" maxlength="250" type="text" name="commentUSER"></td>
		</tr>
		<tr> 
			<td width="100" align="right"><font class="text-blog"><b>Title: </b></font></td>
			<td align="left"><input size="40" maxlength="250" type="text" name="commentTITLE"></td>
		</tr>
		<tr> 
			<td width="100" align="right" valign="top"><font class="text-blog"><b>Comment: </b></font></td>
			<td align="left"><textarea name="commentENTRY" cols="40" rows="10"></textarea></td>
		</tr>
		<tr> 
			<td colspan=2 align="center"><input type="Submit" name="submit" value="Add"></td>
		</tr>
		</form>
	 </table>
<?
	}
		else
			{
				// includes
				include("dbconnex.php");

			// set up error list array
			$errorList = array();
			$count = 0;

			// validate text input fields
			if (!$commentTITLE) { $errorList[$count] = "Invalid entry: Please enter your name"; $count++; }

			// check for errors
			// if none found...
			if (sizeof($errorList) == 0)
				{
					// open database connection
					$connection = mysql_connect($host, $user, $pass) or die ("Unable to connect!");

					// select database
					mysql_select_db($db) or die ("Unable to select database!");

					// generate and execute query
					$blogID = $_GET["blogID"]; 

					$query = "INSERT INTO comments(commentUSER, commentENTRY, commentTITLE, blogID) VALUES('$commentUSER', '$commentENTRY', '$commentTITLE', $blogID)";
					$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());

					// print result
					echo "<font class='text-blog'>Your comment was succesfully added. </font><a href='../index.php'>Go back</a>.</font>";

					// close database connection
					mysql_close($connection);
				}
			else
				{
					// errors found
					// print as list
					echo "<font size=-1>The following errors were encountered: <br>";
					echo "<ul>";
					for ($x=0; $x<sizeof($errorList); $x++)
						{
							echo "<li>$errorList[$x]";
						}
							echo "</ul></font>";
						}
					}
?>

[Kudose]

The SQL will be correct once you change $blogID in the VALUES portion of your SQL to '$blogID'

[cmccomas]

Originally posted by Kudose
The SQL will be correct once you change $blogID in the VALUES portion of your SQL to '$blogID'

Kudose,

When I do that, it inserts 0 into the DB in the blogID field.

[Kudose]

Then that means that $blogID is either set at 0 or FALSE.

What happens when you run this:

echo "blogID: ".$_GET['blogID'];

Also, what is the exact URL you are using to execute this script?

[cmccomas]

I get:

blogID: 12

Here's how the table is setup:
commentID - INT - pk
blogID - INT - fk
commentUSER - TEXT
commentTITLE - TEXT
commentENTRY - TEXT

The page that this code is on is:

../comment.php?blogID=12

[cmccomas]

http://www.christophermccomas.com/entry.php?blogID=12

There is one comment there, but I entered that into the db via myPHPadmin.

[qadeer_ahmad]

Try this query string this solve your problem.

                    $query = "INSERT INTO comments(commentUSER, commentENTRY, commentTITLE, blogID) VALUES('".$commentUSER."', '".$commentENTRY."', '".$commentTITLE."', '".$blogID."')";