我想從數據庫接收數據基於用戶輸入的名稱,一切工作正常我確實看到屏幕上的值,沒有錯誤這裏是PHP代碼:爲什麼我的json_encode沒有給出正確的值PHP
<?php
$dbhost = "localhost";
$username = "root";
$password = "";
mysql_connect($dbhost,$username,$password);
@mysql_select_db("trynew") or die(mysql_error());
$user ="mon";
$query = "SELECT * FROM trynewtable where name = '$user' ";
$all_result = array();
$result = mysql_query($query);
if($result==FALSE)
{
die(mysql_error());
}
while($row = mysql_fetch_array($result))
{
$all_result[] = $row;
}
header('Content-Type: application/json');
$jsondata = json_encode($all_result);echo $jsondata;
mysql_close();
?>
但我對HTML DIV看到輸出是:
存儲在數據庫中[{"0":"1","id":"1","1":"mon","Name":"mon","2":"26","Age":"26","3":"F","Gender":"F"}]
實際數據是: 在數據庫中的實際數據
請讓我知道我做錯了什麼?
你爲什麼要壓制錯誤?你爲什麼還在使用'mysql_'?你應該從MySQL [正式](http://php.net/manual/en/migration55.deprecated.php)棄用。使用[MySQLi](http://php.net/manual/en/book.mysqli.php)或[PDO](http://wiki.hashphp.org/PDO_Tutorial_for_MySQL_Developers)**與** [已準備好的語句]( http://php.net/manual/en/pdo.prepared-statements.php)。 – Script47