如何使用
PHP将MysqL数据库表转换为JSON数据.有没有办法做到这一点？
<?PHP $host = "emriphone.db.6420177.hostedresource.com"; $user = "emriphone"; $pass = "Light12-"; $database = "emriphone"; $linkID = MysqL_connect($host,$user,$pass) or die("Could not connect to host."); MysqL_select_db($database,$linkID) or die("Could not find database."); $sth = MysqL_query("SELECT * FROM ProviderAppointmentListings"); $rows = array(); while($r = MysqL_fetch_assoc($sth)) { $rows[] = $r; } print json_encode($rows); ?>
试试这样：
$query = MysqL_query("SELECT * FROM table"); $rows = array(); while($row = MysqL_fetch_assoc($query)) { $rows[] = $row; } print json_encode($rows);
if (!function_exists('json_encode'))
{
function json_encode($a=false)
{
if (is_null($a)) return 'null';
if ($a === false) return 'false';
if ($a === true) return 'true';
if (is_scalar($a))
{
if (is_float($a))
{
// Always use "." for floats.
return floatval(str_replace(",",".",strval($a)));
}
if (is_string($a))
{
static $jsonReplaces = array(array("\\","/","\n","\t","\r","\b","\f",'"'),array('\\\\','\\/','\\n','\\t','\\r','\\b','\\f','\"'));
return '"' . str_replace($jsonReplaces[0],$jsonReplaces[1],$a) . '"';
}
else
return $a;
}
$isList = true;
for ($i = 0,reset($a); $i < count($a); $i++,next($a))
{
if (key($a) !== $i)
{
$isList = false;
break;
}
}
$result = array();
if ($isList)
{
foreach ($a as $v) $result[] = json_encode($v);
return '[' . join(',',$result) . ']';
}
else
{
foreach ($a as $k => $v) $result[] = json_encode($k).':'.json_encode($v);
return '{' . join(',$result) . '}';
}
}
}
