Question

我一直在努力向我的sql数据库添加数据我尝试了很多不同的方法，但仍然遇到了同样的错误。

注意：未定义的索引：第63行的N：\ ftp \ compc \ d11os \ Project \ addModule.php中的ModuleId

注意：未定义的索引：第64行的N：\ ftp \ compc \ d11os \ Project \ addModule.php中的标题

注意：未定义的索引：第65行的N：\ ftp \ compc \ d11os \ Project \ addModule.php中的CreditLevel

注意：未定义的索引：第66行的N：\ ftp \ compc \ d11os \ Project \ addModule.php中的CreditPoints

注意：未定义的索引：第67行的N：\ ftp \ compc \ d11os \ Project \ addModule.php中的状态

我在做什么呢？我知道它可能很小但是我看不到它并真的需要帮助吗？

这是我正在使用的代码

 <?php
$con=mysqli_connect("localhost","ROOT","ROOT","ROOTdb");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
$ModuleId = mysqli_real_escape_string($con, $_POST['ModuleId']);
$Title = mysqli_real_escape_string($con, $_POST['Title']);
$CreditLevel = mysqli_real_escape_string($con, $_POST['CreditLevel']);
$CreditPoints = mysqli_real_escape_string($con, $_POST['CreditPoints']);
$Status = mysqli_real_escape_string($con, $_POST['Status']);
$Award = mysqli_real_escape_string($con, $_POST['Award']);


$sql="INSERT INTO module(ModuleId, Title, CreditLevel, CreditPoints, Status, Award)
VALUES ('$ModuleId', '$Title', '$CreditLevel' ,'$CreditPoints', '$Status', '$Award')";

if (!mysqli_query($con,$sql)) {
  die('Error: ' . mysqli_error($con));
}
echo "1 record added";

mysqli_close($con);
?>

使用

的Html代码

<html>
<head>
<link rel="stylesheet" type="text/css" href="style.css">

        <img src="logo.jpg" alt="University of Ulster Logo" width="332" height="132">

    <h1 style="font-family:Bell MT;color:blue;font-size:28px;">Add a New Module.</h1>

</head>
 <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">

 <table border="0">

  <tr><td>Module:</td><td>

 <input type="text" name="name" maxlength="60" required>

 </td></tr>

  <tr><td>Title:</td><td>

 <input type="text" name="telephoneNo" maxlength="60" required>

 </td></tr>

 <tr><td>CreditLevel:</td><td>

 <input type="text" name="email" maxlength="60" required>

 </td></tr>

 <tr><td>CreditPoints:</td><td>

 <input type="text" name="course" maxlength="60" required>

 </td></tr>

 <tr><td>Status:</td><td>

 <input type="text" name="Staus" maxlength="100" required>

 </td></tr>

  <tr><td>Award:</td><td>

 <input type="text" name="Award" maxlength="100" required>

 </td></tr>

 <td colspan="2" style="text-align:center">
<input type="submit" name='submit' value='Submit'>

 </form>

Answer 1

问题出现了，因为您使用的变量未在$_POST数组中定义。

尝试类似

的内容

if (isset($_POST['ModuleId']))
{
    $ModuleId = mysqli_real_escape_string($con, $_POST['ModuleId']);
} else {
    $ModuleId = -1; // Default value or Error flag
    echo "Module Id is not specified.";
}

以及其他变量等等。

在插入之前检查所有值是否存在可能是个好主意。例如，

if ($ModuleId != -1) {
    $sql="INSERT INTO module(ModuleId, Title, CreditLevel, CreditPoints, Status, Award)
    VALUES ('$ModuleId', '$Title', '$CreditLevel' ,'$CreditPoints', '$Status', '$Award')";

    if (!mysqli_query($con,$sql)) {
      die('Error: ' . mysqli_error($con));
    }
    echo "1 record added";
}

mysqli_close($con);

//编辑

您的问题似乎源于不检查页面是否已回复。首次加载页面时，它会尝试插入数据 - 但由于您尚未提交表单，因此无需插入数据。

用简单的支票包裹你的PHP代码

<?php
if (isset($_POST['submit']))
{
    $con=mysqli_connect("localhost","ROOT","ROOT","ROOTdb");
    // Check connection
    if (mysqli_connect_errno()) {
      echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }

    // escape variables for security
    $ModuleId = mysqli_real_escape_string($con, $_POST['ModuleId']);
    $Title = mysqli_real_escape_string($con, $_POST['Title']);
    $CreditLevel = mysqli_real_escape_string($con, $_POST['CreditLevel']);
    $CreditPoints = mysqli_real_escape_string($con, $_POST['CreditPoints']);
    $Status = mysqli_real_escape_string($con, $_POST['Status']);
    $Award = mysqli_real_escape_string($con, $_POST['Award']);


    $sql="INSERT INTO module(ModuleId, Title, CreditLevel, CreditPoints, Status, Award)
        VALUES ('$ModuleId', '$Title', '$CreditLevel' ,'$CreditPoints', '$Status', '$Award')";

    if (!mysqli_query($con,$sql)) {
      die('Error: ' . mysqli_error($con));
    }
    echo "1 record added";

    mysqli_close($con);
}
?>

使用php插入sql数据库

1 个答案: