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Using a variable outside of the while loop (scope)

开发者 https://www.devze.com 2023-04-03 02:02 出处:网络
Small problem regarding scope in PHP, I can\'t seem to call the variable $report outside of the while loop. I have tried various things, including return. This doesn\'t work, the only two functions th

Small problem regarding scope in PHP, I can't seem to call the variable $report outside of the while loop. I have tried various things, including return. This doesn't work, the only two functions that work here are if I echo the variable $report inside the loop, or if I print it. Which I do not want to do, although it solves the problem, but 开发者_开发百科I don't want random gibberish on the user's screen.

I have been looking around for the last 15 or so minutes, and I haven't seen any problems quite like this one on here.

Any help would be appreciated.

<?
require "functions2.php";
require "members.php";
$query = "SELECT MAX(DOCid) as prevDOCid from reports";
$result = mysql_query($query);

while ($row = mysql_fetch_array($result)) {
    $prevDOCid = $row[prevDOCid];

$thisDOCid = $prevDOCid+1;
$report = "a"."b".$thisDOCid;


}
echo $report;
?>


You could try to define the variable before the loop, e.g.

$report = "";
while ($row = mysql_fetch_array($result)) {
    $report .= "a"."b".$row["prevDOCid"]+1;
}
echo $report;

I hope this helps you!

Edit Use .= not +=

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