Transform php array into a list of values

Question

I have a php array of key => value, and I would like to store all the values into a list (like a python list) of values. I need it cause I have to pass this list of values to a JQuery script.

Basically I have:

$var = Array ( [0] => 5 [1] => 7 [2] => 9 [3] => 10 )

and I would like to have something like

$var = [5, 7, 9, 10];

When I call from Jquery my variable:

var newvariable = "<?php echo $var ?>";

The final goal is to have a Jquery list like:

var newvariable = [5,7,9,10];

Is this possible in php?

Check out json_encode(): php.net/manual/en/function.json-encode.php — JustBaron
– JustBaron, Commented Feb 20, 2017 at 12:16
There's no difference between those two arrays. An array whose keys are sequential integers starting from 0 is the same as an ordinary indexed array. — Barmar
– Barmar, Commented Feb 20, 2017 at 12:19
@Barman no, basically my jQuery var should be: var values = [5, 7, 9, 10]; — LucaP
– LucaP, Commented Feb 20, 2017 at 12:23

Garf365 · Accepted Answer · 2017-02-20 13:03:13Z

1

$var = array(5,4,7,5,6,8);
echo json_encode($var);

edited Feb 20, 2017 at 13:03

Garf365

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answered Feb 20, 2017 at 12:31

Dilip Suthar

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Comments

Aman Kumar · Accepted Answer · 2017-02-20 12:24:45Z

0

Check this PHP code :

<?php
  $var = array(5, 7, 9, 10);
  $data = json_encode($var) ;
  echo $data ; //[5,7,9,10]
?>

answered Feb 20, 2017 at 12:24

Aman Kumar

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Comments

Hugo Martins · Accepted Answer · 2017-02-20 12:27:15Z

-1

I'm assuming you have a PHP array such as this:

$var = array ( 5, 7, 9, 10 );

This would print, with var_dump:

array(4) {
  [0]=>
  int(5)
  [1]=>
  int(7)
  [2]=>
  int(9)
  [3]=>
  int(10)
}

PHP has a list() language constructor [1]. I don't believe this is what you want though.

Since I'm not sure exactly how this is constructed, I'd suggest looking at the JSON functions [2] PHP has available which might be of help when interacting with jQuery.

Not a definitive answer but I hope it helps.

[1] http://php.net/manual/en/function.list.php
[2] http://php.net/manual/en/ref.json.php

answered Feb 20, 2017 at 12:27

Hugo Martins

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Comments

Joeri · Accepted Answer · 2024-04-12 19:42:50Z

-3

Not sure what is going on here.

$var = Array ( [0] => 5 [1] => 7 [2] => 9 [3] => 10 )

Is not valid PHP or valid JS. It would need to look something like:

$var = [0 => 5, 1 => 7, 2 => 9, 3 => 10];

It creates an Array with given keys.

array_keys($var) Will extract the keys.
array_values($var) Will extract the values.

To get a string that looks like

var newvariable = [5,7,9,10];

you can use

var newvariable = "<?php echo json_encode($var) ?>";

But that's just because json_encode creates a string from any variable be it a string, object or boolean.

More strict would be something like:

var newvariable = "[<?php echo join(',', array_values($var)); ?>]";

answered Apr 12, 2024 at 19:42

Joeri

2,33826 silver badges26 bronze badges

1 Comment

Your Common Sense Over a year ago

Array ( [0] => 5 [1] => 7 [2] => 9 [3] => 10 ) is just a print_r output. and you may learn how does json_encode work on the respective manual page

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