Question

我的数据库中有一个错误..代码中写入了从第41行到50的未定义索引。在我的网站中，我尝试将表单中的数据插入到PhpMyAdmin数据库中，一切正常除了这个......

错误是：注意：未定义的索引：第41-50行的C：\ xampp \ htdocs \ ers \ Database.php中的服务

Database.php：

<html>
<body>
<?php
$con = mysql_connect("localhost","root","123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

// some code

?>


<table border="1">
    <tr >
        <td>Finding</td>
        <td>ServiceType</td>
        <td>Title</td>
        <td>RootCause</td>
        <td>RiskRating</td>
        <td>Impact</td>
        <td>Efforts</td>
        <td>Likelihood</td>
        <td>Finding</td>
        <td>Implication</td>
                <td>Recommendation</td>
                        <td>Report</td>
    </tr>


<?php





mysql_select_db ( "ers_1", $con);
$sql="INSERT INTO findings (ServiceType_ID, Title, RootCause_ID, RiskRating_ID, Impact_ID, Efforts_ID, Likelihood_ID, Finding,Implication,  Recommendation, Report_ID) VALUES (

    '$_POST[Services]',
    '$_POST[title]',
    '$_POST[RootCause]',
    '$_POST[RiskRating]',
    '$_POST[impact]',
    '$_POST[Efforts]',
    '$_POST[likelihood]',
    '$_POST[Finding]',
    '$_POST[Implication]',
    '$_POST[Recommendation]',
    '1'
    )";



$result = mysql_query("SELECT * FROM findings");

while($row = mysql_fetch_assoc($result)) 
  {
  echo "<tr>";
  echo "<td>" . $row['Finding_ID'] . "</td>";                                           
  echo "<td>" . $row['ServiceType_ID'] . "</td>";
  echo "<td>" . $row['Title'] . "</td>";
  echo "<td>" . $row['RootCause_ID'] . "</td>";
  echo "<td>" . $row['RiskRating_ID'] . "</td>";
  echo "<td>" . $row['Impact_ID'] . "</td>";
   echo "<td>" . $row['Efforts_ID'] . "</td>";
    echo "<td>" . $row['Likelihood_ID'] . "</td>";
    echo "<td>" . $row['Finding'] . "</td>";
    echo "<td>" . $row['Implication'] . "</td>";

    echo "<td>" . $row['Recommendation'] . "</td>";
    echo "<td>" . $row['Report_ID'] . "</td>";
  //echo "<td><a href='edit.php'>[EDIT]</a> <a href='delete_risk.php?risk_no=" . $row['risk_no'] . "'>[DELETE]</a></td>";
  echo "</tr>";
  }



mysql_close($con);
?> 

<input type="button" value="Back" onclick="window.location.href='option_Frame.php'" />

</body>
</html

＆GT;

Answer 1

听起来好像没有发布$ _POST ['Services']而不是数据库问题。

Answer 2

检查您的HTML;可能没有name="Services"的输入元素。

如何解决未定义的索引

2 个答案: