PHP - New numbered var for every object in array [duplicate]

Question

I have the following scenario:

Array with multiple waypoints
Want a new variable for every waypoint
Number of waypoints is not fixed

The array stores the address data for the waypoint like street, number, zip, town etc.

I can loop and print the output of the array in foreach loop like:

foreach ($waypoints as $waypoint) {
  echo $waypoint->street
  echo $waypoint->nb
  echo $waypoint->zip
  echo $waypoint->town
}

What I'm trying to do is, to get a new variable for each waypoint. E.g:

$wp1 = Data from waypoint 1
$wp2 = Data from waypoint 2

What I have tried:

$waypointCount = count($waypoints);    

for ($i = 1; $i < $waypointCount; $i++) {
  $wp[$i] = $waypoints->street.' '.$waypoints->nb.' '.$waypoints->zip.' '.$waypoints->town.' '.$waypoints->state;
}

My idea was to count the number of waypoints, set a new variable for each waypoint number and store the corresponding waypointdata in the new variable. I'm kinda stuck on how to create the $wp[i] variables and assign the data to it. Does it need to be a combination with a for and a foreach loop?

Looking for some help to get me on the right direction. Thanks!

…why tho…?! You already have an array of waypoints. You can access it as $waypoints[0] etc. What's the advantage of extracting that to individual variables?! (Hint: there is none.) — deceze
– deceze ♦, Commented Jul 29, 2020 at 14:25

zeterain · Accepted Answer · 2020-07-29 14:26:33Z

0

It looks like you need to access the correct index of your $waypoints array in your loop. Also, make sure you start your loop with $i = 0 or else you'll skip the first element.

$waypointCount = count($waypoints);    

for ($i = 0; $i < $waypointCount; $i++) {
    $wp[$i] = $waypoints[$i]->street.' '.$waypoints[$i]->nb.' '.$waypoints[$i]->zip.' '.$waypoints[$i]->town.' '.$waypoints[$i]->state;
}

answered Jul 29, 2020 at 14:26

zeterain

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3 Comments

InsaneTeddy Over a year ago

Totally missed the index there, now it works as expected. Thanks!

deceze Over a year ago

@InsaneTeddy Note that you're not creating variables $wp1 etc. with this; you're creating an array $wp.

InsaneTeddy Over a year ago

Yes, noticed that. But helped alot to solve my problem. Thank you for the replies.

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