PHP Array into JS Array

Question

Admittedly, my understanding of arrays is junior at best (although before this I thought I had them figured out).

Essentially, I am trying to convert a PHP array to a JS array. The purpose for this is that ultimately I'd like to create a graph/chart using jquery (I've tried using php charts, but they haven't worked for me to this point).

Long story short, every time I try and convert my PHP array to a JS array, the output is "null". This may be due to my understanding of arrays...

//this query gets me the count of each type of variabletype

$sql = "SELECT variabletype, COUNT(variabletype) AS value_occurrence FROM variable GROUP BY variabletype ORDER BY value_occurrence DESC";
$vars_query = mysqli_query($con,$sql) or die(mysqli_error($con));

 while($vars = mysqli_fetch_array($vars_query, MYSQLI_ASSOC)) 
    {

//this is a sub-query that gets me how many of those records are considered "positive"
$times_positive_qry = mysqli_query($con, "SELECT variable.variabletype, COUNT(value.valueid) AS positive_occurrence FROM variable INNER JOIN value On variable.variableid=value.variableid WHERE variable.variabletype = '" .$vars['variabletype']. "' AND value.valuelift>0.00 AND value.valuesignificant=1 GROUP BY variable.variabletype ORDER BY positive_occurrence DESC");
$times_positive = mysqli_fetch_array($times_positive_qry, MYSQLI_ASSOC);
$pos = $times_positive['positive_occurrence'];

}

//this is the code that is supposed to take my php array and turn it into JS array.
 echo "<script type='text/javascript'>";
 $php_array = $pos;
 $js_array = json_encode($php_array);
 echo "var javascript_array = ". $js_array . ";\n";
 echo "</script>";

Again, my understanding of arrays are junior, but I think it may have to do with my use of $pos. Is this not an array? or am I missing the mark somewhere else?

Any help would be much appreciated!

try a var_dump on $pos and see what the type is. That'll answer your question why it's returning null from json_encode — Tularis
– Tularis, Commented Mar 17, 2014 at 15:15
On a sidenote, you might want to read up on arrays in the official PHP manual, so your understanding of them increases :) — Tularis
– Tularis, Commented Mar 17, 2014 at 15:16
your while loop assigns a scalar value to $pos, so it's definitely not an array. and i'd bet the code is not executed at all (mysqli_fetch_array($vars_query, ...) returns no rows), so $pos is not defined, and you've shot yourself in the foot with insufficient error_reporting. — just somebody
– just somebody, Commented Mar 17, 2014 at 15:17

larsAnders · Accepted Answer · 2014-03-17 15:49:33Z

3

You could explicitly declare $pos as an array like this:

$pos = array();

Inside the while loop, you can ensure that it's an array and that you don't lose data while iterating over the loop by using brackets like so:

$pos[] = $times_positive['positive_occurrence'];

That syntax automatically creates an indexed array. If you don't use that syntax, then the value of $pos will be only the value assigned on the final loop of the while statement.

edited Mar 17, 2014 at 15:49

answered Mar 17, 2014 at 15:15

larsAnders

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3 Comments

user2828701 Over a year ago

I get the following error: "Fatal error: [] operator not supported for strings in /blah/blah/blah/filename.php. I added $pos[] = $times_positive['positive_occurrence']; in my while loop. Any thoughts here?

larsAnders Over a year ago

That probably means the $pos is declared with a value somewhere earlier in the code, so it's already a string variable. Make sure to add the explicit declaration $pos = array(); just before the while loop.

larsAnders Over a year ago

Great! Did you mean to accept the other answer though?

GolezTrol · Accepted Answer · 2014-03-17 15:16:41Z

1

You have this line:

$pos = $times_positive['positive_occurrence'];

It assignes a value to the variable $pos. The value you are assigning is probably not an array, but besides, you overwrite $pos every time in the loop.

So I think you meant to do this:

$pos[] = $times_positive['positive_occurrence'];

That will treat $pos as an array and appends every new value to the end of the array.

You may want to start your script by writing $pos = array();. That way, you make sure $pos already exists and is a valid (empty) array in case the query doesn't return any result.

answered Mar 17, 2014 at 15:16

GolezTrol

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2 Comments

user2828701 Over a year ago

I get the following error: "Fatal error: [] operator not supported for strings in /blah/blah/blah/filename.php. I added $pos[] = $times_positive['positive_occurrence']; in my while loop. Any thoughts here?

GolezTrol Over a year ago

That means $pos was already a string before entering the loop. Maybe you've accidentally recycled the same variable name? If you add $pos = array(); before the loop, you should be fine, although for readability it's better to use a more descriptive variable name, and not to use the same name for two different purposes.

vnimbala · Accepted Answer · 2014-03-17 16:10:33Z

0

this might be the solution you want.

before while loop just declare $pos = array();
and in the while loop use $pos as $pos[]

and your javascript looks like this:

<script type='text/javascript'>
<?php
//this is just and example you can replace it with $pos[]
$php_array = array('abc','def','ghi');
?>
var js_array =<?php echo json_encode($php_array);?>;//do not add quotes to this line
alert(js_array[0]);
</script>

answered Mar 17, 2014 at 16:10

vnimbala

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