Using a variable outside of the while loop (scope)

Viewed 645 times

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 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;
?>

Answers

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 +=

Friday, October 21, 2022

ihategals143-3210bfd39f62

Build an array up as you iterate with the while loop.

$result = mysql_query("SELECT * FROM `Departments`");
$results = array();
while($row = mysql_fetch_assoc($result))
{
   $results[] = $row;
}

Alternatively, if you used PDO, you could do this automatically.

Monday, October 24, 2022

ritwick_dey

The variable i that checks the condition is the one you declared in main() not the one inside the loop.

Both are different variables you are confusing them as one, the compiler doesn't get confused as easily as you were.

Inside the loop i refers to the one you declared inside the { } but outside the { } the i refers to the one declared in main()

Friday, September 23, 2022

michael_edenfield

The pipe operator creates a subshell, see BashPitfalls and BashFAQ. Solution: Don't use cat, it's useless anyway.

#!/bin/bash
postPriority=0
while read namesInFile
do   
    postPrioity=500
    echo "weeeeeeeeee ---> $postPrioity <--- 1"
done < /files.txt
echo "weeeeeeeeee ---> $postPrioity <--- 2"

Friday, December 16, 2022

nhat_nguyen

Try:

<?

$database = sqlite_open("thenew.db", 0999, $error);
if(!$database) die($error);

$query = "SELECT * FROM users";
$results = sqlite_query($database, $query);
if(!$results) die("Canot execute query");

$data = array();

while($row = sqlite_fetch_array($results)) {
  $data[] = $row['uid'] . " " . $row['username'] . " " . $row['xPos'] . " " . $row['yPos'];
}
echo json_encode(array("response"=>$data));

sqlite_close($database);

?>

Friday, November 25, 2022

gwatson

Only authorized users can answer the search term. Please sign in first, or register a free account.

Not the answer you're looking for? Browse other questions tagged :

php

loops

scope

while-loop

Using a variable outside of the while loop (scope)

Answers

php,loops,scope,while-loop