[SOLVED] No Errors but Data not Submitting???

[cmaclennan]

Hey Guys,

 

I have a form that at one time was submitting to a database as it should and for some reason it is submitting ok and not giving any kind of errors yet the data isnt ending up in the table?? What could i be missing? when I echo out the statement to see if the data is being posted by the script everything shows up normal but it's not getting entered.

 

Thanks for your help

[.josh]

post code.

[cmaclennan]

Sorry one other thing i have just noticed the form submits to 2 tables and the data is going into the second table ok but not the first, below is the query.

 

// Make the query.		
	$query = "INSERT INTO rma (rma_id, date, store_company, address, address2, city, state_prov, zip_postal, country, phone, ext, contact) VALUES ('$ri', '$dt', '$co', '$ad' , '$ad2', '$ct', '$sp', '$zp', '$cou', '$ph', '$ex', '$cn')";
	$result = @mysql_query ($query); // Run the query.
	$rma_id = mysql_insert_id();

	$result = @mysql_query ($query); // Run the query.
	foreach($_POST as $key => $value)
{
	//looping through all posted fields
	if(strpos($key, 'item_') !== false)
	{
		//found a part field, need to find out the _x number, then insert the entire row
		$suffix = substr($key, -1); //this gets the _x number
		$item = $_POST['item_' . $suffix];
		$description = $_POST['description_' . $suffix];
		$qty = $_POST['qty_' . $suffix];
		$serial = $_POST['serial_' . $suffix];
		$mfg = $_POST['mfg_' . $suffix];
		$tariff = $_POST['tariff_' . $suffix];
		$value = $_POST['value_' . $suffix];

		$query = "INSERT INTO rma_parts (item, description, qty, serial, mfg, tariff, value, rma_id) VALUES ('$item', '$description', '$qty', '$serial', '$mfg', '$tariff', '$value', '$rma_id')";
		$result = mysql_query($query);
	}
}
	if ($result) { // If it ran OK.

		// Send an email, if desired.

		// Print a message.
		echo '<h1 id="mainhead">Thank you!</h1>
	<p>Your entry has now been registered.</p><p><br /></p>';	

		// Include the footer and quit the script (to not show the form).

		exit();

	} else { // If it did not run OK.
		echo '<h1 id="mainhead">System Error</h1>
		<p class="error">Your entry could not be registered due to a system error. We apologize for any inconvenience.</p>'; // Public message.
		echo '<p>' . mysql_error() . '<br /><br />Query: ' . $query . '</p>'; // Debugging message.

		exit();
	}

	mysql_close(); // Close the database connection.

} else { // Report the errors.

	echo '<h1>Error!</h1>
	<p class="error">The following error(s) occurred:<br />';
	foreach ($errors as $msg) { // Print each error.
		echo " - $msg<br />\n";
	}
	echo '</p><p>Please try again.</p><p><br /></p>';

} // End of if (empty($errors)) IF.

} // End of the main Submit conditional.

[.josh]

you sure you're looking in the right table/database?

[cmaclennan]

Yep all the tables im using are on the same database and its the only one the site connects too and there is only the one table with that name and it enters the parts (2nd part of query) but not the first.

[PFMaBiSmAd]

And remove the @ in front of the mysql_query() statements. If you are getting a php error generated on a mysql_query() statement, it means that your database connection is not present and you should never get to the point of executing a mysql_query() statement. What php errors where you getting before you put the @ in the code?

[cmaclennan]

This is the same query structure as i am using succesfully on a number of other forms, I have just noticed that when i try to enter the query directly in the database it seems to think it's creating a duplicate entry for the primary key and won't allow it to go in.

[Ken2k7]

So why would you run it twice? See the lines with your two comments up there? You ran the same query twice.

[cmaclennan]

Im not sure i understand or perhaps i explained wrong? what am i running twice? and I dont get any errors when i remove the @ from the mysql_query () statements

[Ken2k7]

You don't see the 2 lines you have comments on? The comments that says "// Run the query" Read the first 6 lines of the code you posted.

[cmaclennan]

Yes ok I do see that now, I understood from someone else that it should be that way given the mysql_insert_id tag? I assume that is not necessary then?

[cmaclennan]

Ok so I removed that part but am still stuck with the original issue of the data posting to the second table in the query bit not the first. Could there be something else I am missing?

[Ken2k7]

Gah... that's not what I meant.

 

First of all, mysql_insert_id isn't a tag, whatever that means. It's a function.

 

Anyways, the point is you ran the same query twice. Forget about mysql_insert_id(). If you looked at the code, you will see you ran the same query twice. Why would you run it twice? That's the error you described. You're inserting rma_id twice, which caused a duplicate conflict because I'm guessing it's a PK. Though, you never really need to insert anything into a PK column.

 

And don't remove the mysql_insert_id() line either because you'll need it in your second query.

[cmaclennan]

Ok I removed the second submit of the query as you described not the mysql_insert_id () so it is not submitting twice however I am still having the same problem, and the only reason I have anything in the rma_id field is a function to generate a unique id number as opposed to just auto incrementing a basic number.

[Ken2k7]

Are you getting any errors? Also, please post an updated version of your code.

[cmaclennan]

Still not getting any error as at all and here is the code for the queries again, if you would like me to post the page just let me know.

 

// Make the query.		
	$query = "INSERT INTO rma (rma_id, date, store_company, address, address2, city, state_prov, zip_postal, country, phone, ext, contact) VALUES ('$ri', '$dt', '$co', '$ad' , '$ad2', '$ct', '$sp', '$zp', '$cou', '$ph', '$ex', '$cn')";
	$result = mysql_query ($query); // Run the query.
	$rma_id = mysql_insert_id();

	foreach($_POST as $key => $value)
{
	//looping through all posted fields
	if(strpos($key, 'item_') !== false)
	{
		//found a part field, need to find out the _x number, then insert the entire row
		$suffix = substr($key, -1); //this gets the _x number
		$item = $_POST['item_' . $suffix];
		$description = $_POST['description_' . $suffix];
		$qty = $_POST['qty_' . $suffix];
		$serial = $_POST['serial_' . $suffix];
		$mfg = $_POST['mfg_' . $suffix];
		$tariff = $_POST['tariff_' . $suffix];
		$value = $_POST['value_' . $suffix];

		$query = "INSERT INTO rma_parts (item, description, qty, serial, mfg, tariff, value, rma_id) VALUES ('$item', '$description', '$qty', '$serial', '$mfg', '$tariff', '$value', '$rma_id')";
		$result = mysql_query($query);
	}
}
	if ($result) { // If it ran OK.

		// Send an email, if desired.

		// Print a message.
		echo '<h1 id="mainhead">Thank you!</h1>
	<p>Your entry has now been registered.</p><p><br /></p>';	

		// Include the footer and quit the script (to not show the form).

		exit();

	} else { // If it did not run OK.
		echo '<h1 id="mainhead">System Error</h1>
		<p class="error">Your entry could not be registered due to a system error. We apologize for any inconvenience.</p>'; // Public message.
		echo '<p>' . mysql_error() . '<br /><br />Query: ' . $query . '</p>'; // Debugging message.

		exit();
	}

	mysql_close(); // Close the database connection.

} else { // Report the errors.

	echo '<h1>Error!</h1>
	<p class="error">The following error(s) occurred:<br />';
	foreach ($errors as $msg) { // Print each error.
		echo " - $msg<br />\n";
	}
	echo '</p><p>Please try again.</p><p><br /></p>';

} // End of if (empty($errors)) IF.

} // End of the main Submit conditional.

[cmaclennan]

it would appear as though when i submit the form it's not creating a unique id for rma_id as i intended and it's trying to enter the same one over and over again which is not allowing it to submit to the first table but i have no idea why its doing that?

[Ken2k7]

Where are you defining $ri?

[cmaclennan]

At the top of the page along with a number of trim functions and checks for required fields. Its the first entry as follows:

 

// Generate Unique RMA ID
{
	$ri = uniqid (rand (), false);
}

[cmaclennan]

I figured it out, thanks for your help everyone.

[cmaclennan]

It would appear as though I spoke to soon, back to having the same problem, it's only submitting to the second table and keeps trying to submit the same primary key to the first but i don't know why??

[Ken2k7]

Are you running INSERT inside a loop?

 

What is rma_id? Is it a PK?

[cmaclennan]

No not that im aware of at least, sorry im still a noob as im sure you've noticed.

[cmaclennan]

Ok so i've now even removed anything that would enter anything into the rma_id field in hopes of just allowing it to auto incremement and it still will not go into that table, im at a loss as to what's going on here and what i should do as the next step?