将辅助数据输入到现有数据库中

Question

I need some help. 我需要协助。 I have written a script to put first and last name into a database. 我写了一个脚本，将名字和姓氏放入数据库中。 This works correctly. 这可以正常工作。 Then I have written a script to display these names along with 4 text fields per name where student points can by typed in and then stored in the DB. 然后，我编写了一个脚本来显示这些名称以及每个名称4个文本字段，可以在其中键入学生分数，然后将其存储在数据库中。 The names from the DB are displayed correctly and the text fields display correctly however, when I try to put the numbers in the fields it does not put the numbers in the DB and generates "undefined index" errors. 来自数据库的名称正确显示，并且文本字段正确显示，但是，当我尝试将数字放在字段中时，它没有将数字放入数据库中并生成“未定义索引”错误。 I have worked on this for a while but am just not getting it. 我已经为此工作了一段时间，但没有得到。 Thanks for your help. 谢谢你的帮助。 My code is below. 我的代码如下。 Thank you. 谢谢。

<html>
<body>
<form action="pts_summary.php" method="post">
<table border="1">
<tr>
<th>Student Name</th>
<th>First Hour</th>
<th>Second Hour</th>
<th>Third Hour</th>
<th>Fourth Hour</th>
</tr>
<br>

<?php

$hour1 = $_POST['hour1'];
$hour2 = $_POST['hour2'];
$hour3 = $_POST['hour3'];
$hour4 = $_POST['hour4'];


$con=mysqli_connect("localhost","root","","srrdb");                         
if (mysqli_connect_errno())                                                 
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * From students");                      


while($row = mysqli_fetch_array($result))                                   
{
    echo "<tr>"."<td>".$row['fname']."&nbsp".$row['lname']."</td>".     
    "<td>".'<input type="text" name="hour1">'."</td>".                  
    "<td>".'<input type="text" name="hour2">'."</td>".                  
    "<td>".'<input type="text" name="hour3">'."</td>".                  
    "<td>".'<input type="text" name="hour4">'."</td>"."</tr>";          
}

if (isset ($_POST['submit']))
{
$sql="INSERT INTO students (hour1, hour2, hour3, hour4) 
VALUES ('".$hour1."','".$hour2."','".$hour3."','".$hour4."')";
}   

mysqli_close($con);                                                         

?>                                                                          

</table>
<br><input type="submit" value="SUBMIT" name="submit">
</form>
</body>
</html>

Answer 1

You're trying to grab post data before you even check if the submit button was pressed. 您甚至在检查是否已按下“提交”按钮之前都尝试获取帖子数据。 If the submit button wasn't pressed, you won't have values in any of the $_POST['hour#'] fields, and that will throw an undefined index error. 如果未按下提交按钮，则$ _POST ['hour＃']字段中将没有值，这将引发未定义的索引错误。 Throw those lines AFTER the submit check like so. 像这样在提交支票之后扔那些行。

if (isset ($_POST['submit']))
{
    $hour1 = $_POST['hour1'];
    $hour2 = $_POST['hour2'];
    $hour3 = $_POST['hour3'];
    $hour4 = $_POST['hour4'];

    $sql="INSERT INTO students (hour1, hour2, hour3, hour4) 
    VALUES ('".$hour1."','".$hour2."','".$hour3."','".$hour4."')";
}

Answer 2

Your undefined index notices are caused by using $_POST[...] without checking if they are set. 未定义的索引通知是由于使用$_POST[...]而引起的，而不检查它们是否已设置。 Your data is not inserting into your database, as you are only setting the INSERT query - 您的数据未插入数据库，因为您仅设置INSERT查询-

$sql="INSERT INTO students...

but you never execute a query. 但您永远不会执行查询。

mysqli_query($con,$sql);

try - 尝试-

if (isset ($_POST['submit'])){
    // put these inside isset() to prevent undefined index notices
    $hour1 = $_POST['hour1'];
    $hour2 = $_POST['hour2'];
    $hour3 = $_POST['hour3'];
    $hour4 = $_POST['hour4'];

    $sql="INSERT INTO students (hour1, hour2, hour3, hour4) 
    VALUES ('".$hour1."','".$hour2."','".$hour3."','".$hour4."')";
    //missing the query line
    // Insert or die with error message
    $update = mysqli_query($con,$sql) or die(mysqli_error($con));
}

Also, you are using unsanitized $_POST data so you are open to SQL Injection. 另外，您使用的是未经消毒的$_POST数据，因此您可以使用SQL注入。 Either sanitize using mysqli_real_escape_string() or better yet use prepared statements - http://php.net/manual/en/mysqli.quickstart.prepared-statements.php 无论是使用消毒mysqli_real_escape_string()或更好的使用准备好的语句- http://php.net/manual/en/mysqli.quickstart.prepared-statements.php

将辅助数据输入到现有数据库中

问题描述

2 个解决方案

解决方案1
0 2013-06-14 18:14:20

解决方案2
0 2013-06-14 18:31:29

将辅助数据输入到现有数据库中

问题描述

2 个解决方案

解决方案1 0 2013-06-14 18:14:20

解决方案2 0 2013-06-14 18:31:29

解决方案1
0 2013-06-14 18:14:20

解决方案2
0 2013-06-14 18:31:29