I have posted the following message a couple of times on the php-db
list and have not gotten
a response. I am hoping someone here is willing to help.
Thank you
John
Below is the output and code for my problem. When the database gets read,
it returns 0 rows. When it writes, it says it can't because of duplicate
entry. FLD_FileName and FDL_Path are the Primary key.
Two questions, what I am doing wrong and is there a better way to create a
string of text then the one used below.
TIA
John
Read_sql SELECT * FROM TBL_Picture WHERE FLD_FileName = "SailBoat.jpg"and
FLD_Path =" /var/www/html/slidecollection/WinterCarnival"
Result of read Resource id #2
Error Results 0
Number or Rows 0
Insert Sql INSERT INTO TBL_Picture VALUES
("SailBoat.jpg","/var/www/html/slidecollection/WinterCarnival",
"1234","gfgfdgfddg","gfdgfd")
Error Results Duplicate entry
'/var/www/html/slidecollection/WinterCarnival-SailBoat.jpg'
for key 1Result is of 2 is
//reading the database to see if record already exists
$result = mysql_select_db ("Picture_DB", $link) or die ("Could not get the
database");
$read_sql = "SELECT * FROM TBL_Picture WHERE FLD_FileName = \"";
$read_sql .= $fn_FileName;
$read_sql .= "\"and FLD_Path =\" ";
$read_sql .= $fn_Path;
$read_sql .= "\"";
echo " Read_sql ";
echo $read_sql;
$result = mysql_query($read_sql, $link) or die("could not read the table");
echo "Result of read ";
echo $result;
echo " Error Results ";
echo mysql_errno ($link);
$number_of_rows = mysql_num_rows($result);
echo " Number or Rows ";
echo $number_of_rows;
//if return no rows then we need to add the record
if ($number_of_rows == 0){
$insert_sql = "INSERT INTO TBL_Picture VALUES (\"";
$insert_sql .= $fn_FileName;
$insert_sql .= "\",\"";
$insert_sql .= $fn_Path;
$insert_sql .= "\",\"";
$insert_sql .= $fn_Year;
$insert_sql .= "\",\"";
$insert_sql .= $fn_DescriptiveText;
$insert_sql .= "\",\"";
$insert_sql .= $fn_PictureName;
$insert_sql .= "\")";
echo " Insert Sql ";
echo $insert_sql;
$result2 = mysql_query($insert_sql, $link);
// or die("could not write to the table ");
echo " Error Results ";
echo mysql_error ($link);
echo "Result is of 2 is";
echo $result2;
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