I am having one problem with the PHP json_encode function. It encodes numbers as strings, e.g.
array('id' => 3)
becomes
"{ ["id": "3", ...)
When js encounters these values, it interprets them as strings and numeric operations fail on them. Does anyone know some way to prevent json_encode
from encoding numbers as strings? Thank you!
This question is related to
php
javascript
json
The following test confirms that changing the type to string causes json_encode() to return a numeric as a JSON string (i.e., surrounded by double quotes). Use settype(arr["var"], "integer") or settype($arr["var"], "float") to fix it.
<?php
class testclass {
public $foo = 1;
public $bar = 2;
public $baz = "Hello, world";
}
$testarr = array( 'foo' => 1, 'bar' => 2, 'baz' => 'Hello, world');
$json_obj_txt = json_encode(new testclass());
$json_arr_txt = json_encode($testarr);
echo "<p>Object encoding:</p><pre>" . $json_obj_txt . "</pre>";
echo "<p>Array encoding:</p><pre>" . $json_arr_txt . "</pre>";
// Both above return ints as ints. Type the int to a string, though, and...
settype($testarr["foo"], "string");
$json_arr_cast_txt = json_encode($testarr);
echo "<p>Array encoding w/ cast:</p><pre>" . $json_arr_cast_txt . "</pre>";
?>
$rows = array();
while($r = mysql_fetch_assoc($result)) {
$r["id"] = intval($r["id"]);
$rows[] = $r;
}
print json_encode($rows);
Like oli_arborum said, I think you can use a preg_replace
for doing the job. Just change the command like this :
$json = preg_replace('#:"(\d+)"#', ':$1', $json);
I also had the same problem processing data from the database. Basically the problem is that the type in the array to convert in json, is recognized by PHP as a string and not as integer. In my case I made a query that returns data from a DB column counting row. The PDO driver does not recognize the column as int, but as strings. I solved by performing a cast as int in the affected column.
You can use (int) if any issue occurs!! It will work fine.
Just run into the same problem and was the database returning the values as strings.
I use this as a workaround:
$a = array(
'id' => $row['id'] * 1,
'another' => ...,
'ananother' => ...,
);
$json = json_encode($a);
That is multiplying the value by 1 to cast it into a number
Hope that helps someone
try
$arr = array('var1' => 100, 'var2' => 200);
$json = json_encode( $arr, JSON_NUMERIC_CHECK);
But it just work on PHP 5.3.3. Look at this PHP json_encode change log http://php.net/manual/en/function.json-encode.php#refsect1-function.json-encode-changelog
I'm encountering the same problem (PHP-5.2.11/Windows). I'm using this workaround
$json = preg_replace( "/\"(\d+)\"/", '$1', $json );
which replaces all (non-negative, integer) numbers enclosed in quotes with the number itself ('"42"' becomes '42').
See also this comment in PHP manual.
Casting the values to an int or float seems to fix it. For example:
$coordinates => array(
(float) $ap->latitude,
(float) $ap->longitude
);
For sake of completeness (as I can't add comments yet), let me also add this detail as another answer:
(Edit: To be read after realizing that the source data (i.e. in the OP's case, database result set) could be the problem (by returning numeric columns as strings), and json_encode() in fact was not the source of the problem)
Manual pages of both "mysql_fetch_array":
Returns an array of strings that corresponds to the fetched row,
... and "mysql_ fetch_ row":
Returns an numerical array of strings that corresponds to the fetched row
clearly states that; the entries in the returned array will be strings.
(I was using the DB class in phpBB2 (yes I know, it's obsolete!), and "sql_fetchrow()" method of that class uses "mysql_fetch_array()")
Not realizing that, I also ended up finding this question, and understanding the problem! :)
As Pascal Martin stated above in his follow-up comments, I believe a solution that takes care of the "incorrect type" problem at the source (i.e. by using the "mysql_field_type()" function and doing the casting right after fetch, (or other fetch methods like "object"?)) would be the better in general.
Well, PHP json_encode() returns a string.
You can use parseFloat() or parseInt() in the js code though:
parseFloat('122.5'); // returns 122.5
parseInt('22'); // returns 22
parseInt('22.5'); // returns 22
So Pascal MARTIN isn't getting enough credit here. Checking for numeric values on every JSON return is not feasable for an existing project with hundreds of server side functions.
I replaced php-mysql with php-mysqlnd, and the problem went away. Numbers are numbers, strings are strings, booleans are boolean.
I, likewise was reading from a DB (PostgreSQL) and everything was a string. We loop over each row and do things with it to build up our final results array, so I used
$result_arr[] = array($db_row['name'], (int)$db_row['count']);
within the loop to force it to be an integer value. When I do json_encode($result_arr)
now, it correctly formats it as a number. This allows you to control what is and is not a number coming from your database.
EDIT:
The json_encode()
function also has the ability to do this on the fly using the JSON_NUMERIC_CHECK
flag as a second argument to it. You need to be careful using it though as shown in this users example in the documentation (copied below): http://uk3.php.net/manual/en/function.json-encode.php#106641
<?php
// International phone number
json_encode(array('phone_number' => '+33123456789'), JSON_NUMERIC_CHECK);
?>
And then you get this JSON:
{"phone_number":33123456789}
json_encode serializes some data structure in JSON format to be send across the network. Therefore all content will be of the type string. Just like when you receive some parameter from $_POST or $_GET.
If you have to make numeric operations on the values sent, just convert them to int first (with the intval() function in PHP or parseInt() in Javascript) and then execute the operations.
it is php version the problem, had the same issue upgraded my php version to 5.6 solved the problem
Note that since PHP 5.3.3, there's a flag for auto-converting numbers (the options parameter was added in PHP 5.3.0):
$arr = array( 'row_id' => '1', 'name' => 'George' );
echo json_encode( $arr, JSON_NUMERIC_CHECK ); // {"row_id":1,"name":"George"}
Source: Stackoverflow.com