Question

我知道stackoverflow上有很多这样的问题，但没有一个卖掉了我的问题。

我正在制作活动日历，目前正在构建添加活动页面。提交页面时没有显示错误，但表单信息没有上传到数据库，我不知道为什么我检查了语法并且看不到任何错误。

*请求loginform.php和core.php请求显示$ _SESSION ['user_id']的位置。来自码： loginform.php

<?php
    $con = mysql_connect("localhost","pytsuemg_brodie","brodie");
    $db = mysql_select_db("pytsuemg_brodie");

    if (isset($_POST['username'])&&isset($_POST['password'])) {
        $username = $_POST['username'];
        $password = $_POST['password'];
        $password_hash =md5($password);

        if (!empty($username)&&!empty($password)) {
            $query="SELECT `AccountID` FROM `Parent` WHERE `Username`='$username' AND `Password`='$password_hash'";

            if ($query_run = mysql_query($query)) {
                $query_num_rows = mysql_num_rows($query_run);

                if ($query_num_rows==0) {
                    echo 'Invalid username/password combination';
                } else
                if ($query_num_rows==1) {
                    $user_id = mysql_result($query_run,0,'AccountID');
                    $_SESSION['user_id']=$user_id;

                    header('Location: profilepage.php');
                }

            }

        } else {
            echo 'You must supply a username and password.';
        }

    }

    ?>
<form action="login.php" method="POST">
Username: <input type="text" name="username">
Password: <input type="password" name="password">
<input type="submit" value="Log In">
</form>

core.php

<?php
    ob_start();
    session_start();
    $current_file = $_SERVER['SCRIPT_NAME'];
    function loggedin() {

        if (isset($_SESSION['user_id'])&&!empty($_SESSION['user_id'])) {
            return true;
        } else {
            return false;
        }

    }


    ?>

addevent.php

 require 'core.php';
    $con=mysqli_connect("localhost","pytsuemg_brodie","brodie","pytsuemg_brodie");
    if (isset($_POST['DateStart'])&&isset($_POST['DateEnd'])&&isset($_POST['TimeStart'])&&isset($_POST['TimeEnd'])&&isset($_POST['EventType'])&&isset($_POST['Description'])&&isset($_POST['Children'])&&isset($_POST['Location'])&&isset($_POST['MealsGiven'])){

    $DateStart = $_POST['DateStart'];
    $DateEnd = $_POST['DateEnd'];
    $TimeStart = $_POST['TimeStart'];
    $TimeEnd = $_POST['TimeEnd'];
    $EventType = $_POST['EventType'];
    $Description = $_POST['Description'];
    $Children = $_POST['Children'];
    $Location = $_POST['Location'];
    $MealsGiven = $_POST['MealsGiven'];

    if (!empty($DateStart)&&!empty($DateEnd)&&!empty($TimeStart)&&!empty($TimeEnd)&&!empty($EventType)&&!empty($Description)&&!empty($Children)&&!empty($Location)&&!empty($MealsGiven)){



    $sql = "INSERT INTO Events (DateStart, DateEnd, TimeStart, TimeEnd, Location, EventType, MealsGiven, description,Children, AccountID) VALUES(
     '$DateStart','$DateEnd','$TimeStart','$TimeEnd','$Location','$EventType','$MealsGiven','$Description','$Children','".$_SESSION['user_id']."')";

     mysqli_query($con, $sql);

     /* commit transaction */
    if (!mysqli_commit($con)) {
        print("Transaction commit failed\n");
        exit();
    }
    } else {
                    echo'All fields are required';
                }
    }
    /* close connection */

    mysqli_close($con);
    ?>
    <br>

    <form action="addevent.php" method="post">
     <p> Date:<label for="from"> From</label>
    <input type="text" id="from" name="DateStart">
    <label for="to">to</label>
    <input type="text" id="to" name="DateEnd"></p>
      <p>Start Time:<input type="text" name="TimeStart" /></p>
      <p>End Time:<input type="text" name="TimeEnd" /></p>
      <p>Event Type :
        <select name="EventType">
    <option value="Change over">Changeover</option>
    <option value="Parents Evening">Parents Evening</option>
    <option value="After School Activity">After School Activity</option>
    <option value="Weekend Activity">Weekend Activity</option>
    <option value="Holiday">Holiday</option>
        </select> </p>

    <p>Description (max 100 Characters) <br><textarea rows="2" cols="50" maxlength="100" name="Description"></textarea></p>
    <p>Children Involved: <br><textarea rows="1" cols="50" maxlength="32" name="children"></textarea></p>
      <p>Location: <input type="text" name="Location" /></p> 
      <p>Meals Given: <select name="MealsGiven">
    <option value="Yes">Yes</option>
    <option value="No">No</option>
    </select></p>



        <p><input type="submit" value="Submit"/></p>
    </form>

Answer 1

此name="children"和$_POST['Children']不一样。

您需要将其更改为name="Children"

POST变量区分大小写。

请密切关注Brad's comment （要认真对待）非常值得付出努力。

This article读得很好。

我注意到您使用md5()存储密码（根据您的loginform.php文件），这被视为“太快”。这不再被认为是安全的。考虑使用一些最新的密码存储技术，例如PHP的password_hash()函数（如果在PHP 5.5上）或crypt()

您的loginform.php文件使用mysql_*个功能，而addevent.php正在使用mysqli_*个功能。我建议您对所有文件使用相同的SQL函数，mysqli_*使用prepared statements或PDO。

将数据添加到数据库PHP

1 个答案: