Question

我写了这段代码

$stmt = $mysqli->prepare("INSERT INTO `verifyAccounts` (`CUsername`, `CPassword`) VALUES (?,?)");
$stmt->bind_param("ss", $CUsername, $CPassword);
$stmt->execute();
$stmt = $mysqli->prepare("SELECT `Started` FROM `verifyAccounts` WHERE `CUsername` = ?");
do {
    $stmt->bind_param("s", $CUsername);
    $stmt->execute();
    $stmt->bind_result($Startedvalue);
    echo $Startedvalue;
    sleep(2);
} while ($Startedvalue < 2);

if ($Startedvalue == 2) {
    echo "Correct";
} elseif ($Startedvalue == 3) {
    echo "Incorrect";
}
$stmt = $mysqli->prepare("DELETE FROM `verifyAccounts` WHERE `CUsername` = ?");
$stmt->bind_param("s", $CUsername);
$stmt->execute();

出于某种原因，这似乎并没有起作用，我从来没有回复说正确或不正确，并且帐户也没有被删除。看起来它似乎停留在Do While循环中，但$Startedvalue也永远不会回显。

这是我有的Ajax

 function verifyAccount(f) {
            f.preventDefault();
            var CUsername = $('#CUsername').val();
            var CPassword = $('#CPassword').val();
            $.ajax({
                type: 'POST',
                data: {
                    CUsername: CUsername,
                    CPassword: CPassword
                },
                url: 'verifyAccount.php',
                success: function(data) {
                    console.log(data);
                    if (data == "Correct") {
                        $("#verifyPasswordText").text('Valid Login');
                    } else if (data == "Incorrect") {
                        $("#verifyPasswordText").text('Invalid Login');
                        $("#verifyPasswordOKC").prop("disabled", false);
                    } else if (data == "CUsername") {
                        $("#verifyPasswordText").text('Fill out your OKC Username');
                        $("#verifyPasswordOKC").prop("disabled", false);
                    } else if (data == "CPassword") {
                        $("#verifyPasswordText").text('Fill out your OKC Password');
                        $("#verifyPasswordOKC").prop("disabled", false);
                    }           
                },
                error: function(xhr, err) {
                    console.log("readyState: " + xhr.readyState + "\nstatus: " + xhr.status);
                    console.log("responseText: " + xhr.responseText);
                }
            });
        }

Answer 1

很不知道以这种方式处理登录是否是个好主意，但

如果没有任何内容可以修改＆＃39;启动＆＃39;，您提供的代码中没有任何内容，那么您将被困在do..while

至于没有回声

确保在输出<?xml version="1.0" encoding="utf-8"?> <manifest xmlns:android="http://schemas.android.com/apk/res/android" package="myapp.com.tuberapp"> <uses-permission android:name="android.permission.INTERNET"/> <application android:allowBackup="true" android:icon="@mipmap/ic_launcher" android:label="@string/app_name" android:supportsRtl="true" android:theme="@style/AppTheme"> <activity android:name=".MainActivity" android:label="@string/app_name" android:theme="@style/AppTheme.NoActionBar"> <intent-filter> <action android:name="android.intent.action.MAIN" /> <category android:name="android.intent.category.LAUNCHER" /> </intent-filter> </activity> </application> </manifest>
在每个echo header( 'Content-type: text/html; charset=utf-8' );

我的Mysql查询不是比较值

1 个答案: