doukongpao0903 2011-12-02 08:44
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json_decode(json_encode(索引数组))给出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
    }
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1条回答 默认 最新

  • doufei3152 2011-12-02 08:49
    关注

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

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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