无法将JSON解析为Java(android)
我的PHP文件:无法将JSON解析为Java(android),java,php,android,json,Java,Php,Android,Json,我的PHP文件: <?php include("ConnectDatabase.php"); $Username = mysql_real_escape_string($_POST['Username']); $Password = mysql_real_escape_string($_POST['Password']); $q = mysql_query("SELECT Username, Password FROM Users where Usern
<?php
include("ConnectDatabase.php");
$Username = mysql_real_escape_string($_POST['Username']);
$Password = mysql_real_escape_string($_POST['Password']);
$q = mysql_query("SELECT Username, Password FROM Users
where Username = '".$Username."' and
Password = '".$Password."'", $con);
if(mysql_num_rows($q) > 0){
$row = mysql_fetch_assoc($q);
print json_encode($row);
}else{
print "0";
}
?>
我认为它们最终为空,因为第一行不起作用。它不是您生成的JSONArray
try {
JSONObject root = new JSONObject(result);
username = root.getString("Username");
password = root.getString("Password");
} catch (JSONException e) {
Log.e("log_tag", "Error parsing data "+e.toString());
}
类似的东西。我认为它们最终为空,因为第一行不起作用。它不是您生成的JSONArray
try {
JSONObject root = new JSONObject(result);
username = root.getString("Username");
password = root.getString("Password");
} catch (JSONException e) {
Log.e("log_tag", "Error parsing data "+e.toString());
}
类似的东西。谢谢,我知道了,只要重新打开日食,我能问更多吗?我想知道什么时候需要使用JSONArray?谢谢谢谢我知道了,重新打开日食我能问更多吗?我想知道什么时候需要使用JSONArray?谢谢