priest_004

Ok folks, I have sorted the FATAL ERROR now it is telling me.. Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in X:\xampp\htdocs\search.php on line 8 I think I might just have to go back to the drawing board for this one.... any clues on this?? code now reads.. <?php mysql_connect ("localhost", "user", "password") or die (mysql_error()); mysql_select_db ("it_homehelp_test") or die (mysql_error()); $term = $_POST['term']; $sql = mysql_query("select * from it_homehelp_test where ClientName1 like '%term%' "); while ($row = mysql_fetch_array($sql)){ echo 'Client Name' .$row['ClientName1']; echo 'Address' .$row['Address1']; echo 'Phone' .$row['Tel1']; } ?>

Hi Dan, and Rifts, I see it, I missed the " from the front and end... dough!! thanks for pointing that out. --------------------------------------- ok new problem... now it says... Fatal error: Function name must be a string in X:\xampp\htdocs\search.php on line 7 I thought it was a string. this is starting to hurt my head now... Same line.. ??? what have i missed this time?

Hi folks, I am a complete n00b at php and mysql. I am teaching myself from books and the WWW, but alas I am stuck... the error I get is: Parse error: syntax error, unexpected T_STRING in X:\xampp\htdocs\search.php on line 7 here is the code: <?php mysql_connect ("localhost", "user", "password") or die (mysql_error()); mysql_select_db ("it_homehelp_test") or die (mysql_error()); $term = $_POST['term']; $sql = $mysql_query(select * from it_homehelp_test where ClientName1 like '%term%'); <<<------this is line 7 while ($row = mysql_fetch_array($sql)){ echo 'Client Name:' .$row['ClientName1']; echo 'Address:' .$row['Address1']; echo 'Phone:' .$row['Tel1']; } ?> Any help you can offer would be great. I can also post the ".html" file that creates the search bar if it is needed. Thanks