[PHP-DB] trouble reading a database (fwd)

2001-04-23 Thread johndmiller 


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.jpgand
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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[PHP-DB] trouble reading a database

2001-04-22 Thread johndmiller 


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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