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Views: 353 | Replies: 5
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Join Date: Feb 2008
Posts: 296
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Posting Whiz in Training
is my code correct? i am adding elements to my db and the elemtents which have been add will be displayed on a different page. i am using an upload page which creates a thumb nail of an uploaded image and then stores the path for this image in a variable called $thumb_name. the code i have used to enter the image into the db is
and the code to display the image on the next page looks like this
php Syntax (Toggle Plain Text)
and the code to display the image on the next page looks like this
php Syntax (Toggle Plain Text)
Last edited by peter_budo : Mar 18th, 2008 at 6:11 am. Reason: Keep It Organized - please use [code] tags
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Join Date: Nov 2007
Location: Bangalore, India
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Everything is correct.. But this is wrong.
This should be
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$result = mysql_query("SELECT * FROM images WHERE 'Broad1'='$thumb_name'");
{
echo $row['Broad1'];
echo "<br />";
}
?>
php Syntax (Toggle Plain Text)
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
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Join Date: Dec 2007
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Junior Poster
And
Should be
No quotes around Broad1.
php Syntax (Toggle Plain Text)
php Syntax (Toggle Plain Text)
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Join Date: Nov 2007
Location: Bangalore, India
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Originally Posted by mwasif 
And
Should bephp Syntax (Toggle Plain Text)
No quotes around Broad1.php Syntax (Toggle Plain Text)
Ah! correct.. I couldn't spot that one!
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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Join Date: Feb 2008
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Posting Whiz in Training
i have correct the few lines of code which were wrong but it is now saying that the table i have created does not exist. the code i have used to create the table is
[code=php]include "db.php";
$con = mysql_pconnect("$dbhost","$dbusername","$dbpasswd")
or die ("QUERY ERROR: ".mysql_error());
$db = mysql_select_db("$database_name", $connection)
or die("QUERY ERROR: ".mysql_error());
// Create table in $database_name database
mysql_select_db("$database_name", $connection);
$sql = "CREATE TABLE images (imageID int NOT NULL AUTO_INCREMENT, Broad1 varchar(50), Broad2 varchar(50), Broad3 varhar(50))";
mysql_query("INSERT INTO images (Broad1) VALUES ('$thumb_name')");[code]
were have i gone wrong?
[code=php]include "db.php";
$con = mysql_pconnect("$dbhost","$dbusername","$dbpasswd")
or die ("QUERY ERROR: ".mysql_error());
$db = mysql_select_db("$database_name", $connection)
or die("QUERY ERROR: ".mysql_error());
// Create table in $database_name database
mysql_select_db("$database_name", $connection);
$sql = "CREATE TABLE images (imageID int NOT NULL AUTO_INCREMENT, Broad1 varchar(50), Broad2 varchar(50), Broad3 varhar(50))";
mysql_query("INSERT INTO images (Broad1) VALUES ('$thumb_name')");[code]
were have i gone wrong?
Last edited by peter_budo : Mar 18th, 2008 at 6:13 am. Reason: Keep It Organized - please use [code] tags
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Join Date: Nov 2007
Location: Bangalore, India
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You aren't executing the create table query.
After this, you need to have mysql_query($sql);
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$sql = "CREATE TABLE images (imageID int NOT NULL AUTO_INCREMENT, Broad1 varchar(50), Broad2 varchar(50), Broad3 varhar(50))";
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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