Need Help! Record Update Problem.

justinlyman

I am creating a script that modifys a record in my database.

This script is suppost to take updated data from a form and update my database. This script also uploads a photo if a change is made. The name of this photo is stored in the database. The problem I am having is that when The photo is not being updated, it replaces the name in the database with a blank entry, rather than leaving it unchanged. Please look at my script and see what I am doing wrong.

Thanks,
Justin

<?
//check for required form variables
if ((!$POST[addr]) || (!$POST[city])) {
header("Location:http://www.lymantech.net/");
exit;
}

//set up table and database names
$db_name = "🙂";
$table_name = "main";

//connect to server and select database
$connection = @mysql_connect("localhost", "🙂", "🙂") or die(mysql_error());
$db = @mysql_select_db($db_name, $connection) or die(mysql_error());

// Uploading Files
if ($FILES[pic1] != "") {
$pic1name = $FILES[pic1][name];
@copy($FILES[pic1][tmp_name], "/var/www/html/secure/pics/" .$FILES[pic1][name]);

}

//build and issue query
$pic1_sql = "UPDATE $table_name SET
pic1 ='$pic1name'
WHERE id ='$_POST[id]'";

if ($_FILES[pic1] != "") {
$pic1_result = @($pic1_sql,$connection) or die(mysql_error());
}

//build and issue query
$sql = "UPDATE $table_name SET
addr ='$POST[addr]',
city ='$POST[city]',
state ='$POST[state]',
zip ='$POST[zip]',
cnty ='$POST[cnty]',
price ='$POST[price]',
bed ='$POST[bed]',
bath ='$POST[bath]',
sqft ='$POST[sqft]',
yrblt ='$POST[yrblt]',
acres ='$POST[acres]',
des ='$POST[des]',
sdes ='$POST[sdes]',
keyf ='$POST[keyf]'
WHERE id ='$_POST[id]'";

$result = @($sql,$connection) or die(mysql_error());

?>
<HTML>
<HEAD>
<TITLE>My Listing Management System:Listing Updated</TITLE>
</HEAD>
<BODY>
<h1>My Listing Management System</h1>
<h2>Modify a Listing -Listing Updated</h2>

The following information was successfully updated

Address:
 
<? echo "$_POST[addr]"; ?>

City:
 
<? echo "$_POST[city]"; ?>

State:
 
<? echo "$_POST[state]"; ?>

Zip Code:
 
<? echo "$_POST[zip]"; ?>

County:
 
<? echo "$_POST[cnty]"; ?>

Price:
 
<? echo "$_POST[price]"; ?>

Bedrooms:
 
<? echo "$_POST[bed]"; ?>

Bathrooms:
 
<? echo "$_POST[bath]"; ?>

Square Feet:
 
<? echo "$_POST[sqft]"; ?>

Year Built:
 
<? echo "$_POST[yrblt]"; ?>

Acreage:
 
<? echo "$_POST[acres]"; ?>

Detailed Description:
 
<? echo "$_POST[des]"; ?>

Short Description:
 
<? echo "$_POST[sdes]"; ?>

Key Features:
 
<? echo "$_POST[keyf]"; ?>

Picture #1:
<? echo "$pic1name"; ?>

</td>

</tr>
</table>

</BODY>
</HTML>

bpat1434

It might be better to check to see if the $_FILES['pic1'] array is empt() rather than ''. And you should only need to do it once, rather than twice and building a query in between.

// Uploading Files
if (!empty($_FILES[pic1]))
{
    $pic1name = $_FILES[pic1][name];
    @copy($_FILES[pic1][tmp_name], "/var/www/html/secure/pics/" .$_FILES[pic1][name]);

//build and issue query
    $pic1_sql = "UPDATE $table_name SET
pic1 ='$pic1name'
WHERE id ='$_POST[id]'";

$pic1_result = @mysql_query($pic1_sql,$connection) or die(mysql_error());
}

~Brett

justinlyman

Hey thank you for your response, I am new to PHP and need all the help that I can get.

I tryed changing my code like you suggested but it is still doing the same thing. It is really wierd becuase if I take that segment of code out completely then it doesn't change the record to an empty value. It is almost like it is disregarding the if statement. I was thinking maybe if I gave it an else statement with something else to do that it might fix the problem. Maybe like grab the current data and make it the $pic1name variable. What do you think? I really appreciate your help!
Thanks,
Justin

bpat1434

Originally posted by bpat1434

// Uploading Files
if (!empty($_FILES['pic1']))
{
    $pic1name = $_FILES['pic1']['name'];
    @copy($_FILES['pic1']['tmp_name'], "/var/www/html/secure/pics/" .$_FILES['pic1']['name']);

//build and issue query
    $pic1_sql = "UPDATE $table_name SET
pic1 ='$pic1name'
WHERE id ='".$_POST['id']."'";

$pic1_result = @mysql_query($pic1_sql,$connection) or die(mysql_error());
}

Try the quoted code above. I think it may be as small as that.

You could also replace if(!empty($_FILES . . with:

if( is_uploaded_file($_FILES['pic1']['tmp_name']) )
{
    // Do something here
}

~Brett

justinlyman

Hey thanks so much, you got it on the second try. I have been trying to get this figured out for a few days and this really helps me out.

~Justin

bpat1434

No problem.

~Brett