retrieve an image in folder from info out of database

[jsd219]

Hello, i am having problems displaying an image once i have uploaded
it. i upload it to a folder and store its name in a database. below is
the code to display the image. I have figured out the the code is not
grabbing the name from the database but i can not figure out why. can
anyone help?

//Retrieves data from MySQL
$data = mysql_query("SELECT IMAGE_NAME FROM item") or
die(mysql_error());

//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
//Outputs the image and other data
Echo "<img src=http://www.abcxyz.com/pics/".$info['photo'] ."<br>";
}

God bless
jason

[ZeldorBlat]

On Jun 21, 5:18 pm, jsd219 <[e-mail]:

Hello, i am having problems displaying an image once i have uploaded
it. i upload it to a folder and store its name in a database. below is
the code to display the image. I have figured out the the code is not
grabbing the name from the database but i can not figure out why. can
anyone help?

//Retrieves data from MySQL
$data = mysql_query("SELECT IMAGE_NAME FROM item") or
die(mysql_error());

//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
//Outputs the image and other data
Echo "<img src=http://www.abcxyz.com/pics/".$info['photo'] ."<br>";

}

God bless
jason

Because you're using $info['photo'] when the column is called
IMAGE_NAME. So you probably want to use $info['IMAGE_NAME'] instead.

[jsd219]

On Jun 21, 4:36 pm, ZeldorBlat <[e-mail]:

On Jun 21, 5:18 pm, jsd219 <[e-mail]:

Hello, i am having problems displaying an image once i have uploaded
it. i upload it to a folder and store its name in a database. below is
the code to display the image. I have figured out the the code is not
grabbing the name from the database but i can not figure out why. can
anyone help?

//Retrieves data from MySQL
$data = mysql_query("SELECT IMAGE_NAME FROM item") or
die(mysql_error());

//Puts it into an array
while($info = mysql_fetch_array( $data ))
{
//Outputs the image and other data
Echo "<img src=http://www.abcxyz.com/pics/".$info['photo'] ."<br>";

}

God bless
jason

Because you're using $info['photo'] when the column is called
IMAGE_NAME. So you probably want to use $info['IMAGE_NAME'] instead.

That worked, Thank you