CODEWITHSUNDEEP
phpmysqljson
Hassan
Hassan

Reputation: 313

json_decode(json_encode(an indexed array)) gives NULL

i encode the user entred testarea content with json_encode here is how it storred in mysql (there is a return to line after "aaa")

{"content":"aaa
bbb"}

now when i get the the content from database and trying to decoded it using json_decode, i get NULL for this, instead of what expected.

what wrrong? a bug in PHP?

EDIT 1: more details

$data =array('content'=>$textareaText);
$addres_insert = "INSERT INTO `rtable` (`data`) VALUES ('".json_encode($data)."');";
$rows = mysql_query($addres_insert);

then to get the content 

$query = "SELECT * FROM  `rtable` WHERE id =  '".$id."'";
        $rows = mysql_query($query);
    if ( $rows ){
        $row = mysql_fetch_array($rows);
        $res['data'] = json_decode($row['data']);//i tryed substr($row['data'],3) but didn't work
    }

Upvotes: 1

Views: 1160

Answers (1)

Treffynnon
Treffynnon

Reputation: 21553

JavaScript and JSON do not allow line returns to be contained within a string. You need to escape them.

json_encode() should escape them automatically for you.

Here is the output of my playing with your JSON code supplied on the PHP interactive shell:

php > $json = '{"content":"aaa
php ' bbb"}';
php > var_dump(json_decode($json, true));
NULL

As you can see when I escape your line return it works just fine:

php > $json = '{"content":"aaa\n bbb"}';
php > var_dump(json_decode($json, true));
array(1) {
  ["content"]=>
  string(8) "aaa
 bbb"
}

This is also further discussed in a previous question relating to a similar problem: Problem when retrieving text in JSON format containing line breaks with jQuery

Upvotes: 4

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