Question

我正在尝试使用php进行分析。我不知道怎么检查。它唯一能检查的是formno，firstname和lastname。

<?php
                        $FrmN = $_POST["formno"];
                        $FN = $_POST["firstname"];
                        $LN = $_POST["lastname"];

                        $con = mysql_connect("localhost","root","");
                        if (!$con)
                          {
                          die('Could not connect: ' . mysql_error());
                          }

                        mysql_select_db("vianney300", $con);

                        $sql = "INSERT INTO profile(FormNo, FirstName, LastName)
                        VALUES ('$FrmN', '$FN', '$LN')";

                        if (!mysql_query($sql,$con))
                          {
                          die('Error: ' . mysql_error());
                          }
                        echo "<br>1 record added!";
                        mysql_close($con);
                         ?>

Answer 1

尝试以下代码

<?php
    $FrmN = $_POST["formno"];
    $FrnN=$mysqli->real_escape_string($FrmN);
    $FN = $_POST["firstname"];
    $FN=$mysqli->real_escape_string($FN);
    $LN = $_POST["lastname"];
    $LN=$mysqli->real_escape_string($LN);
    $con = mysqli_connect("localhost","root","");
    if (!$con)
    {
    die('Could not connect: ' . mysqli_error());
    }
    mysqli_select_db("vianney300", $con);
    $sql = "INSERT INTO profile(FormNo, FirstName, LastName)
    VALUES ('{$FrmN}', '{$FN}', '{$LN}')"";
    if (!mysqli_query($sql,$con))
    {
     die('Error: ' . mysqli_error());
    }
    echo "<br>1 record added!";
    mysqli_close($con);
    ?>

检查我的数据库中是否存在此配置文件

1 个答案: