Php sql数据验证中的循环问题

您好，这是我的用于考勤登录的PHP代码，但是当我删除while循环时，它会输入多个条目

请帮助我得到哪个循环更适合这个编码

当我移除while循环时，它工作得很好。但是，可以在考勤中输入多个条目

<?php

$conn = mysqli_connect("localhost", "Vijay", "vijay123", "test");
if (mysqli_connect_errno())
    {
        echo "Unable to connect the Server" . mysqli_connect_error();
    }

if ($_SERVER["REQUEST_METHOD"] == "POST")
    {
        // getting details from form
        $EmployeeNoA = mysqli_real_escape_string($conn, $_POST['EmployeeNoA']);
        $EmployeeNameA = mysqli_real_escape_string($conn, $_POST['EmployeeNameA']);
        $Shift = mysqli_real_escape_string($conn, $_POST['Shift']);
        $SignInDay = mysqli_real_escape_string($conn, $_POST['SignInDay']);
        $SignInDate = mysqli_real_escape_string($conn, $_POST['SignInDate']);
        $SignInTime = mysqli_real_escape_string($conn, $_POST['SignInTime']);

        if ($Shift == "0")
            {
                echo "<script>alert('Please Select the Shift!');</script>";
            }
        else
            {

                // $rowcount = mysqli_query($conn, "SELECT * From attend");
                // $rowCount = mysqli_num_rows($rowcount);

                $ver = mysqli_query($conn, "SELECT * FROM attend WHERE EmployeeNoA='$EmployeeNoA' && SignInDate='$SignInDate'");
                while ($view = mysqli_fetch_array($ver, MYSQL_ASSOC)) // **it is repeatedly running and store multiple data and error message. 
                    {
                            if ($SignInDate != $view['SignInDate'])
                            {
                                $sql = "INSERT INTO attend (EmployeeNoA, EmployeeNameA, Shift, Day, SignInDate, SignInTime) VALUES ('$EmployeeNoA', '$EmployeeNameA', '$Shift', '$SignInDay', '$SignInDate', '$SignInTime')";

                                if (!mysqli_query($conn, $sql))
                                    {
                                        echo mysqli_error($conn);
                                    }
                                else
                                    {
                                        echo "<script>alert ('You have Signed In!');</script>";
                                    }
                            }
                        else
                            {
                                echo "<script>alert ('You have ALREADY Signed In!');</script>";
                            }
                    }
            }
    }

?>

---

看起来您只是想测试第一个查询是否返回任何行。使用：

$ver = mysqli_query($conn, "SELECT COUNT(*) AS count FROM attend WHERE EmployeeNoA='$EmployeeNoA' && SignInDate='$SignInDate'");
$row = mysqli_fetch_assoc($ver);
if ($row['count'] == 0) {
    sql = "INSERT INTO attend (EmployeeNoA, EmployeeNameA, Shift, Day, SignInDate, SignInTime) VALUES ('$EmployeeNoA', '$EmployeeNameA', '$Shift', '$SignInDay', '$SignInDate', '$SignInTime')";
    if (!mysqli_query($conn, $sql)) {
        echo mysqli_error($conn);
    } else {
        echo "<script>alert ('You have Signed In!');</script>";
    }
} else {
    echo "<script>alert ('You have ALREADY Signed In!');</script>";
}

$ver=mysqli_query（$conn，“选择COUNT（*）作为参加会议时的计数，其中EmployeeNoA='$EmployeeNoA'&&signendate='$signendate'）；
$row=mysqli\u fetch\u assoc（$ver）；
如果（$row['count']==0）{
sql=“在出席（EmployeeNoA、EmployeeNameA、Shift、Day、SignInDate、SignInTime）值（“$EmployeeNoA”、“$EmployeeNameA”、“$Shift”、“$SignInDay”、“$SignInTime”）中插入”；
if（！mysqli_query（$conn，$sql））{
echo mysqli_错误（$conn）；
}否则{
echo“警报（'您已登录！'）；”；
}
}否则{
echo“警报（'您已经登录！'）；”；
}
尝试替换（&O）and@RakeshSharma为什么？在MYSQL中，它们不是一样吗？根据它们的等效性，我已经尝试过了。它不起作用……测试
if（$signendate！=$view['signenda']）
将永远不会成功，因为查询中的
WHERE
子句只返回与变量匹配的日期。
$ver = mysqli_query($conn, "SELECT COUNT(*) AS count FROM attend WHERE EmployeeNoA='$EmployeeNoA' && SignInDate='$SignInDate'");
$row = mysqli_fetch_assoc($ver);
if ($row['count'] == 0) {
    sql = "INSERT INTO attend (EmployeeNoA, EmployeeNameA, Shift, Day, SignInDate, SignInTime) VALUES ('$EmployeeNoA', '$EmployeeNameA', '$Shift', '$SignInDay', '$SignInDate', '$SignInTime')";
    if (!mysqli_query($conn, $sql)) {
        echo mysqli_error($conn);
    } else {
        echo "<script>alert ('You have Signed In!');</script>";
    }
} else {
    echo "<script>alert ('You have ALREADY Signed In!');</script>";
}