Can´t add values to an array after array_shift

Question

I add values to an associative array with this code:

$tokenarray += [$datetime => $newtoken];

This works fine. But after I sort the array and shift it with that code:

$sortedarray = krsort($tokenarray, 1);
$shiftedarray = array_shift($sortedarray);
$shiftedarray += [$datetime => $newtoken];
$tokenarrayjson = json_encode($shiftedarray);

This error appears:

Fatal error: Uncaught Error: Unsupported operand types in SITE Stack trace: #0 {main} thrown in SITE on line

$shiftedarray += [$datetime => $newtoken]; <- This line throes the error

Can someone tell my why please? Does array_shift make an object out of my array and if so, how can I prevent it?

Regards, Andreas

Nick · Accepted Answer · 2019-02-05 22:27:35Z

4

krsort does not return the sorted array, it sorts its argument in-place and returns true/false dependent on whether it succeeds or not. The same applies to array_shift, which returns the value shifted off the array, not the array post shift: You need to do this instead:

$sortedarray = $tokenarray;
krsort($sortedarray, SORT_NUMERIC);
$shiftedarray = $sortedarray;
array_shift($shiftedarray);
$shiftedarray += [$datetime => $newtoken];
$tokenarrayjson = json_encode($shiftedarray);

If you don't actually need the intermediate arrays, you can simplify that to:

krsort($tokenarray, SORT_NUMERIC);
array_shift($tokenarray);
$tokenarray += [$datetime => $newtoken];
$tokenarrayjson = json_encode($tokenarray);

Note

Since your keys are numeric strings, array_shift will interpret them as numbers and renumber your array starting from 0. To avoid this, use unset on the first key of the array (found using key) instead:

krsort($tokenarray, SORT_NUMERIC);
unset($tokenarray[key($tokenarray)]);
$tokenarray += [$datetime => $newtoken];
$tokenarrayjson = json_encode($tokenarray);

Also note that you should use SORT_NUMERIC instead of 1, just in case the value changes in a future version of PHP.

Demo on 3v4l.org

edited Feb 5, 2019 at 22:27

answered Feb 5, 2019 at 22:14

Nick

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2 Comments

Andreasschnetzer Over a year ago

Nick, im sorry for bothering you, but can you tell me if there is a way to echo a variable of php in the browsers console? I work with javascript ajaxs so when I test it I don´t see whats going on in the php-file.

Nick Over a year ago

@Andreasschnetzer When I want to do that I usually include an extra field in the JSON I return from my ajax call (e.g. 'status' => 'here is a message') and then in my JS I use something like if ('status' in result) alert(result.status);

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