введите значение как значение массива

Question

Я хочу получить записи и отобразить их как массив. select * from album_comment where SUB_ID='$id' отображает строку, отображаемую на изображении.

$sql = "select * from album_comment where SUB_ID='$id'"; 
$query = mysql_query($sql) or sqlerrorhandler("(" . mysql_errno() . ") " . mysql_error(), $sql, __LINE__); 
while ($comment_fet = mysql_fetch_assoc($query)) { 
    $content = $comment_fet['CONTENT_VALUE']; 
    $datas = json_decode($content); 
} 
echo $get_like = json_encode($content); 

Я хочу, чтобы получить значение как

$datas=[{"name":"hello","commentuser":"desc 259 desc","id":8,"date":"2015-12-09T05:40:12.773Z","displayDate":"Wed Dec 09 2015 11:10:12 GMT+0530 (India Standard Time)8","Like":0,"Unlike":0,"rating":0,"reportAbuse":0}, 

{"name":"hnnb","commentuser":"hjh ghj ghj ghj","id":68,"date":"2015-12-09T06:00:44.718Z","displayDate":"Wed Dec 09 2015 11:30:44 GMT+0530 (India Standard Time)68","Like":0,"Unlike":0,"rating":0,"reportAbuse":0}, 

{"name":"56844","commentuser":"689 498 4849 48","id":45,"date":"2015-12-09T06:03:03.178Z","displayDate":"Wed Dec 09 2015 11:33:03 GMT+0530 (India Standard Time)45","Like":0,"Unlike":0,"rating":0,"reportAbuse":0}, 

{"name":"dcfg","commentuser":"bbnbvbvbnbb","id":60,"date":"2015-12-09T06:49:10.875Z","displayDate":"Wed Dec 09 2015 12:19:10 GMT+0530 (India Standard Time)60","Like":0,"Unlike":0,"rating":0,"reportAbuse":0}] 

Answer 1

Измените код, как это,

while ($comment_fet = mysql_fetch_assoc($query)) { 
    $content = $comment_fet['CONTENT_VALUE']; 
    $datas[] = json_decode($content);// capture in array 
} 
echo $get_like = json_encode($datas);// encode array here 

Answer 2

вам нужно сохранить все это в массив как

$datas = array(); 
while ($comment_fet = mysql_fetch_assoc($query)) { 
    $content = $comment_fet['CONTENT_VALUE']; 
    $datas[] = json_decode($content); 
} 
echo $get_like = json_encode($datas); 

Answer 3

Просто поместите $ datas [] вместо $ datas in while loop.

$sql = "select * from album_comment where SUB_ID='$id'"; 
$query = mysql_query($sql) or sqlerrorhandler("(" . mysql_errno() . ")  " . mysql_error(), $sql, __LINE__); 
while ($comment_fet = mysql_fetch_assoc($query)) { 
    $content = $comment_fet['CONTENT_VALUE']; 
    $datas[] = json_decode($content); 
} 
echo $get_like = json_encode($content);