> -----Original Message----- > From: Ian Fingold [mailto:[EMAIL PROTECTED] > Sent: 03 June 2003 16:58 > > Ok I'm trying to write this function to up date some fields > in my database > so i'm grabbing some info from a query and throwing it into > an array (with > mysql_fetch_array). > > one of the values i'm puting into that array is the table id. I'm then > comparing the table id with the value of $i in my loop. Now > my problem is > when I loop through and the table id is not equal to $i I > need to add 1 to > $i but I need to move the array pointer back one. > > In the code below I used prev() but i'm getting warnings > about the "Passed > variable is not an array". > http://55yardline.cflmain.com/55main/2003/fantasy/tests/update.php > > I'm thinking that prev() must not view an array fetch from a > database the > same or something. But what else can I use besides prev() to > rewind the > pointer by 1? > > Thanks, > > > <?php > > include "function.php"; > connect1(); > > //query roster database and put results into an array. > $num1 = mysql_query("SELECT * FROM roster"); > $num2 = mysql_fetch_array($num1); > //set i to zero > $i = 0; > > //start loop to determine if $i matches the current roster id > do { > $i = $i + 1; > if ($num2["ros_id"] == $i) { > //if a match is found, query the fan_roster database > where the play_id = > $i > //and put the results into a array > $upd = mysql_query("SELECT * FROM fan_roster WHERE play_id='$i'"); > $updr = mysql_fetch_array($upd); > //loop through the passing field and add them up > do { > $passing = $passing + $updr["pass_yrd1"]; > } while($updr = mysql_fetch_array($upd)); > > //Print feedback > echo "Player:$i total Passing: $passing $updr[play_id]<br>"; > > //update the roster table with the total $passing where ros_id=$i > $upd1 = mysql_query("UPDATE roster SET pass_yrd='$passing' WHERE > ros_id='$i'"); > //reset passing for next loop > $passing = 0; > } else { > //if there isn't a match, add 1 to $i > $i = $i + 1; > > //print feedback > echo "$i no player<br>"; > > //put the array pointer back one for next loop > $num2 = prev($num2["ros_id"];
(1) prev() takes an array as its argument, not an individual element, so prev($num2) would be syntactically correct (but I doubt it's what you really mean). (2) anyway, this whole statement is pointless as you immediately do: > } > } while($num2 = mysql_fetch_array($num1)); Which promptly overwrites the value of $num2 you just put there! What are you *really* trying to do? > ?> > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, visit: http://www.php.net/unsub.php > -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
