将记录插入mysql echo表？

Question

最近，我为联系人启动了echo mysql表，现在我需要一个按钮来添加新的联系人。 我尝试了很多事情，这就是我的结果（请记住，所有内容都位于同一文件中）：

PHP：

//create record
if (isset($_POST['submitc'])) {
$empresa = $_POST['empresa'];
$contato = $_POST['contato'];
$telefone = $_POST['telefone'];
$email = $_POST['email'];
$sql = $conn->query("INSERT INTO Contacts (empresa, contato, email, phone) 
VALUES ('$_POST[empresa]', '$_POST[contato]', '$_POST[telefone]',      
'$_POST[email]')");

if(!$sql) {
echo ("Could not create" .mysqli_error());
  }
}

形成：

<form method=post>
  <div class='input-field'>
  <i class='material-icons prefix'>work</i>
      <input id='first_name' name='empresa' type='text' class='validate'>
      <label for='first_name'>Empresa</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>account_circle</i>
      <input id='first_name' name='contato' type='text' class='validate'>
      <label for='first_name'>Contato</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>phone</i>
      <input id='first_name' name='telefone' type='text' class='validate'>
      <label for='first_name'>Telefone</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>email</i>
      <input id='first_name' name='email' type='text' class='validate'>
      <label for='first_name'>E-mail</label>
    </div>
  </form>

</div>
<div class='modal-footer'>
<button class='green darken-4 waves-effect waves-light btn' type='submit'       
name='submitc' value='Add'>Criar</button>
   </form>

如您在PHP部分中所看到的，我已经尝试了很多类似使用vars，$ _ POST的方法，但是当我单击Submit按钮时，什么也没有发生，甚至没有错误告诉我一些事情。 请注意，我正在尝试为所有4列增加价值。 怎么了

（请忽略这些div，我使用的是Materialize，那只是CSS）

Answer 1

您的代码中几乎没有错误。

1）首先，您先关闭表单，然后再提交按钮。 删除</form> 。 希望你能得到一些结果。

2）像这样<form action="post">开始更新您的表单。

3）像这样更新您的表单过帐检查

 if (!empty($_POST)) {
 echo "test<br/>";
 $empresa = $_POST['empresa'];
 $contato = $_POST['contato'];
 $telefone = $_POST['telefone'];
 $email = $_POST['email'];
 $sql = $conn->query("INSERT INTO Contacts (empresa, contato, email, phone) 
 VALUES ('$_POST[empresa]', '$_POST[contato]', '$_POST[telefone]',      
 '$_POST[email]')");

 if(!$sql) {
 echo ("Could not create" .mysqli_error());
 }
 }

Answer 2

我认为这就是您要尝试做的。

if (isset($_POST['submitc'])) {
$empresa = $_POST['empresa'];
$contato = $_POST['contato'];
$telefone = $_POST['telefone'];
$email = $_POST['email'];
$sql = "INSERT INTO Contacts (empresa, contato, email, phone) VALUES('$empresa', '$contato', '$telefone', '$email')";
$insert = $conn->query($sql);
      if ( $insert) {
          header('Location: name.php');
      }else {
          echo "Error: " . $sql . "<br>" . mysqli_error($conn);
      }
}

FORM REPLACED THIS 
<form method="post" action="">
  <div class='input-field'>
  <i class='material-icons prefix'>work</i>
      <input id='first_name' name='empresa' type='text' class='validate'>
      <label for='first_name'>Empresa</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>account_circle</i>
      <input id='first_name' name='contato' type='text' class='validate'>
      <label for='first_name'>Contato</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>phone</i>
      <input id='first_name' name='telefone' type='text' class='validate'>
      <label for='first_name'>Telefone</label>
    </div>

    <div class='input-field'>
    <i class='material-icons prefix'>email</i>
      <input id='first_name' name='email' type='text' class='validate'>
      <label for='first_name'>E-mail</label>
    </div>

<div class='modal-footer'>
<button class='green darken-4 waves-effect waves-light btn' type='submit'       
name='submitc' value='Add'>Criar</button>
</div>
</form>

Answer 3

//create record
if (isset($_POST['submitc'])) {
$empresa = $_POST['empresa'];
$contato = $_POST['contato'];
$telefone = $_POST['telefone'];
$email = $_POST['email'];

$sql = $conn->query("INSERT INTO `contacts`(`empresa`, `contato`, `email`, `phone`) 
VALUES('$empresa', '$contato', '$email', '$telefone');");

if(!$sql) {
echo ("Could not create" .mysqli_error());
  }
}

 <form method="post"> <div class='input-field'> <i class='material-icons prefix'>work</i> <input id='first_name' name='empresa' type='text' class='validate'> <label for='first_name'>Empresa</label> </div> <div class='input-field'> <i class='material-icons prefix'>account_circle</i> <input id='first_name' name='contato' type='text' class='validate'> <label for='first_name'>Contato</label> </div> <div class='input-field'> <i class='material-icons prefix'>phone</i> <input id='first_name' name='telefone' type='text' class='validate'> <label for='first_name'>Telefone</label> </div> <div class='input-field'> <i class='material-icons prefix'>email</i> <input id='first_name' name='email' type='text' class='validate'> <label for='first_name'>E-mail</label> </div> <div class='modal-footer'> <button class='green darken-4 waves-effect waves-light btn' type='submit' name='submitc' value='Add'>Criar</button> </div> </form> 

Answer 4

改成：

<form method="post" action="post.php">

其中post.php是您的POST操作文件，并添加到该文件的顶部

if($_SERVER['REQUEST_METHOD'] == "POST") 
{
.... YOU CODE ...
}