Hi Jim, echo cannot display the content of an array or object. try print_r() or var_dump() - Frank > Ok, > > > > I'm a PHP newbie and I'm having a problem. All I want to do is parse > some information out of a database and send someone and email stating > that their posting has been approved. > > > > I've verified that my SQL is indeed returning the correct value for > $key, but when I go to fetch the array for the row so that I can extract > the pertinent tidbits of information from that row, I get nada. My > troubleshooting echo statements give me the following out put. I've > included the actual snippet of code below. > > > > 54 > SELECT * FROM jobs WHERE pk_job = '54' > Resource id #4 > Array > > 0 > > > > > > //use the date to find the record we're looking for approved but not yet > posted > > $findkey = "SELECT pk_job from jobs WHERE (display = 'Yes') AND > (approval_stamp = '$matchnow') AND (posted IS NULL)"; > > list($getkey) = mysql_fetch_row(mysql_query ($findkey)); > > $key = ($getkey); > > echo $key; //note this is returning the correct and expected value > > echo "<br>"; > > // now that we have the key lets get the row > > $query = "SELECT * FROM jobs WHERE pk_job = '$key'"; > > echo $query; //note this is apparently behaving correctly and gives me a > correct result when executed from the command line > > echo "<br>"; > > $result = mysql_query($query); > > echo $result; > > echo "<br>"; > > //make array > > $row = mysql_fetch_array ($result); > > echo $row; > > > > I'm pretty sure that this is a "DUH!" moment, but I'm not sure quite > what it is that I've done. I copied the code from another thing that I > wrote a while back that's working fine and only changed SQL statements > to reflect their new purpose in life. > > > > TIA, > > > > Jimi > > > > > > If computers get too powerful, we can organize them into a committee -- > that will do them in. -- Bradley's Bromide > > > > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
