hi all thanks again for all th direction you guys have given me..
i have a partial backend done for uploading graphics to a directory
and information into mysql. Now what i'm having problems with is
getting the file path with the file name to upload to the sql
database.
Right now i have 2 unique qualifiers carrying over (jobnum, custID).
I'm querying the database to find those matching qualifier and i need
to fill in the file name into the directorytile field in the db. here
is what i have now (excuse the sloppy code i'm a newbie!):
<?php
mysql_connect ('localhost', 'acme', 'acme') ;
mysql_select_db ('olcg');
$query = mysql_query("SE LECT directorytile FROM itl WHERE
jobnum='$jobnum ' AND custID='$custID '") or die (mysql_error()) ;
if (isset($_POST['submit'])) {
$sql = "INSERT INTO itl SET
directorytile=' $file_name'";
}
?>
<?php
if($file_name !="")
{
copy ("$file", "/home/guestdir/www/olcg/$file_name")
or die("Could not copy file");
}
else { die("No file specified"); }
?>
again thanks so much!!!!! 1 1569 ph*********@com cast.net (Mikey P) wrote: hi all thanks again for all th direction you guys have given me..
i have a partial backend done for uploading graphics to a directory and information into mysql. Now what i'm having problems with is getting the file path with the file name to upload to the sql database.
Right now i have 2 unique qualifiers carrying over (jobnum, custID). I'm querying the database to find those matching qualifier and i need to fill in the file name into the directorytile field in the db. here is what i have now (excuse the sloppy code i'm a newbie!): ... mysql_select_d b ('olcg'); $query = mysql_query("SE LECT directorytile FROM itl WHERE jobnum='$jobnu m' AND custID='$custID '") or die (mysql_error()) ;
if (isset($_POST['submit'])) { $sql = "INSERT INTO itl SET directorytile=' $file_name'"; }
INSERT is only used to add an entirely new record. To update the fields in
an existing record, you use UPDATE. Further, an SQL connection has no
concept of a "current record", so you have to tell it exactly which
record(s) to update. Hence:
$sql = "UPDATE itl SET directorytile=' $file_name' " .
"WHERE jobnum='$jobnum ' AND custID='$custID ';";
--
- Tim Roberts, ti**@probo.com
Providenza & Boekelheide, Inc. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics |
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