讓說我提交數據的形式與下面的代碼在Javascript中捕獲表單響應?
var xhr = new XMLHttpRequest(), formData = new FormData();
formData.append("img", img);
formData.append("user", localStorage["username"]);
formData.append("pass", localStorage["password"]);
xhr.onreadystatechange = function (event) {
if (xhr.readyState === 4 && xhr.status === 200) {
var value = xhr.responseText; // value should equal "1234"
alert("value = " + value);
}else{
alert("none");
}
};
xhr.open("POST", "http://joubin.me/uploads3/upload_file.php", true);
xhr.send(formData);
upload.php的步驟完成後,它重定向到另一個名爲頁面giveid.php,它顯示的僅僅是文本字符串用一個id
說1234
我如何使用JavaScript捕獲這個確切的ID。
請記住,不同的upload.php重定向在giveid.php上會有不同的id號碼嗎?
我查看了xmlhtml響應,但無法弄清楚。
這裏是什麼樣的形式去
$password = $_REQUEST['pass'];
$username = $_REQUEST['user'];
$image = $_REQUEST['img'];
echo $password;
echo "<br/>";
echo $username;
echo "<br/>";
echo $image;
$con = mysql_connect("localhost","ya","right");
if (!$con)
{
die('Could not connect: ' . mysql_error());
echo "could not connect";
}
$asql = "SELECT * FROM `ashkan`.`users` where user='$username' and pass='$password';";
$result = mysql_query($asql);
echo $result;
// Mysql_num_row is counting table row
$count=mysql_num_rows($result);
echo $count;
echo 11;
if($count == 1){
$sql = "INSERT INTO `ashkan`.`goog` (`user`, `pass`, `img`) VALUES ('$username', '$passwo$
}
mysql_query($sql);
mysql_close($con);
header('Location: giveid.php') ;
,這裏是giveid.php
1234
的內容任何幫助將是巨大的。
感謝
有沒有你不使用JavaScript庫的理由如jQuery處理Ajax請求? –
^柺杖沒有修復 –
沒有必要重新發明輪子。使用'$ .post(url,fn)' – Ibu