Question

我有问题，我从php连接phpmyadmin数据库。以下代码适用于表单，

<div id="wb_element_instance53" class="wb_element">
  <form class="wb_form wb_mob_form" method="post"><input type="hidden" name="wb_form_id" value="18498be5"><textarea name="message" rows="3" cols="20" class="hpc"></textarea>
    <table>
      <tr>
        <th class="wb-stl-normal">Name&nbsp;&nbsp;</th>
        <td><input type="hidden" name="wb_input_0" value="Name"><input class="form-control form-field" type="text" value="" name="wb_input_0" required="required"></td>
      </tr>
      <tr>
        <th class="wb-stl-normal">Email&nbsp;&nbsp;</th>
        <td><input type="hidden" name="wb_input_1" value="E-mail"><input class="form-control form-field" type="text" value="" name="wb_input_1" required="required"></td>
      </tr>
      <tr class="area-row">
        <th class="wb-stl-normal">Message&nbsp;&nbsp;</th>
        <td><input type="hidden" name="wb_input_2" value="Message"><textarea class="form-control form-field form-area-field" rows="3" cols="20" name="wb_input_2" required="required"></textarea></td>
      </tr>
      <tr class="form-footer">
        <td colspan="2"><button type="submit" class="btn btn-default">Submit</button></td>
      </tr>
    </table>
  </form>
  <script type="text/javascript">

然后，我尝试使用下面的PHP代码连接phpmyadmin数据库，

<?php
/*
$connect=mysqli_connect('localhost','root','','Contact_db') ;

if(mysqli_connect_errno($connect))
{
        echo 'Failed to connect';
}

// create a variable

if (isset($_POST['name'])) {
    $name = $_POST['name'];
}

if (isset($_POST['email'])) {
    $email = $_POST['email'];
}
if (isset($_POST['message'])) {
    $message = $_POST['message'];
}



$sql ="INSERT INTO contac_ds ('Name','email','message') VALUES ('$name','$email,'$message')";

//Execute the query

mysqli_query($connect,$sql);

?>

但是，上面显示错误：

Notice: Undefined variable: name in this line "$sql ="INSERT INTO contac_ds ('Name','email','message') VALUES ('$name','$email,'$message')";"
Notice: Undefined variable: email in in this line "$sql ="INSERT INTO contac_ds ('Name','email','message') VALUES ('$name','$email,'$message')";"
Notice: Undefined variable: message in in this line "$sql ="INSERT INTO contac_ds ('Name','email','message') VALUES ('$name','$email,'$message')";"

Answer 1

如果isset（）失败怎么办？修理：只有在设置了参数时才有$ sql ..

if (isset($_POST['name']) && isset($_POST['email']) &&   isset($_POST['message']) ){

    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['message'];
    $sql ="INSERT INTO contac_ds ('Name','email','message') VALUES ('$name','$email,'$message')";

    //Execute the query

    mysqli_query($connect,$sql);
}

Answer 2

问题是$ _POST搜索输入的名称。你的名字是wb_input_0，试试这个：

if (isset($_POST['wb_input_0'])) {
      $name = $_POST['wb_input_0'];
  }

电子邮件和消息也一样。但是，我不建议将输入命名为

Answer 3

试试这个：

$email ='';
$name ='' ;
$message ='';
print_r($_POST);//to review is all vars in form.
if (isset($_POST['name'])) {
    $name = $_POST['name'];
}

if (isset($_POST['email'])) {
    $email = $_POST['email'];
}
if (isset($_POST['message'])) {
    $message = $_POST['message'];
}

Answer 4

if (isset($_POST['name'])) {
    $name = $_POST['name'];
}else{
    $name = '';
}

if (isset($_POST['email'])) {
    $email = $_POST['email'];
}else{
    $email = '';
}

if (isset($_POST['message'])) {
    $message = $_POST['message'];
}else{
    $message = '';
}

Answer 5

帮自己一个忙，准备好你的陈述：

$sql ="INSERT INTO contac_ds ('Name','email','message') VALUES (?,?,?)";
$stmt = mysqli_prepare($connect, $sql);
$name="";
if (isset($_POST['name'])) {
    $name = $_POST['name'];
}
$email="";
if (isset($_POST['email'])) {
    $email = $_POST['email'];
}
$message="";
if (isset($_POST['message'])) {
    $message = $_POST['message'];
 } 

mysqli_stmt_bind_param($stmt,"sss",$name,$email,$message);
mysqli_stmt_execute($stmt);

请注意，您当前的$_POST不会包含这些字段，因为您对它们的命名方式不同（两次），因此您还需要修复它。

PHP连接mysql显示未定义的变量

5 个答案: