Displaying Help! Pllease

[Anon]

Hey all! I need some help. This code below is giving me parse errors on line 1 it says whenI try ti display it. I'm trying to retrieve the data from my SQL database to display an image from the article then the name of it below. That's all. Check it out below. Any help would be greatly appreicated. Thanks.

<?$connection = @mysql_connect("localhost", "----", "----");$db = @mysql_select_db("----");$query= mysql_query("SELECT article_posts.id,headline,main_page_image,UNIX_TIMESTAMP(date_stamp) as Time_Stamp,f_name,l_name FROM article_posts,article_staff_profiles WHERE article_posts.staff_id=article_staff_profiles.id AND article_posts.type='review' ORDER BY Time_Stamp DESC LIMIT 3");while($r=mysql_fetch_row($query)) {print "<a href=\"http://www.myurl.com/review.php3?review_id=$r[0]\"> <img src=\"$r[3]\" width=\"100\" height=\"61\" border=\"0\"><br><img border="0" src="review.jpg" width="100%" height="61"><img border="0" src="beneath.jpg" width="188" height="16"><br><b>$r[2] </b><br>\n";}mysql_close();?></td>

[Anon]

Hey All I can think is that you might need to make a new line for each part of the code eg:

<?
$connection = @mysql_connect ("localhost", "----", "----");

$db = @mysql_select_db("----");

also (i'm no expert on mysql) but you might want to remove the @ sign from the mysql_connect coz I never use it and it seems to work allright...

[Anon]

Hey. tryed it and it doesn't work. Not it says PARSE error on line 5?!?!? Line 5 being "$query= mysql_query("SELECT article_posts.id,headline,main_page_image FROM " What in the world is wrong with that?

<?$connection = mysql_connect("localhost", "----", "----");

$db = mysql_select_db("----");

$query= mysql_query("SELECT article_posts.id,headline,main_page_image FROM

article_posts,article_staff_profiles WHERE article_posts.staff_id=article_staff_profiles.id AND

article_posts.type='review' ORDER BY Time_Stamp DESC LIMIT

3");while($r=mysql_fetch_row($query)) {print "<a

href=\"http://www.myurl.com/reviews/review.php3?review_id=$r[0]\"> <img src=\"$r[3]\"

width=\"100\" height=\"61\" border=\"0\"><br><img border="0" src="review.jpg" width="100%"

height="61"><img border="0" src="beneath.jpg" width="188" height="16"><br><b>$r[2]

</b><br>\n";}mysql_close();?></td>

[Anon]

Hi Tim...

If I recall the @ sign before will supress the error messages which when first creating a script is important however once it is done I would put them back in.

Are you sure that your SQL works properly have you gone to a mysql command line and tried to run it?

K.