PHP connect mysql 显示未定义的变量

Question

我有一个问题，同时从我的php连接phpmyadmin数据库。 下面的代码用于表单，

<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，上面显示了错误：

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(( 失败了怎么办？ 修： 仅当设置了参数时，才具有$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'];
}

电子邮件和消息也是如此。但是我不建议命名这样的输入

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不会有这些字段，因为您的命名方式不同(并且两次(，因此您还需要修复它。