Question

我是PHP的新手，我正在尝试制作一个在我的数据库中的表中插入名称，电子邮件和ID的小表单，但它不起作用 index.php：

    <form action="getData.php" method="post">
      Name
      <input type="text" name="name" maxlength="40" value="<?=$name;?>" />
      Email
      <input type="text" name="email" maxlength="50" value="<?=$email;?>" />
      <input type="submit" name="formSubmit" value="Submit" />
   </form>

访问getdata.php：

<?php
if($_POST['formSubmit'] == "Submit"){
$name = $_POST['name'];
$email = $_POST['email'];
$servername = "localhost";
$username = "tj_testdb";
$password = "tjtorin04";
$dbname = "tj_testdb";

$conn = new mysqli($servername,$username,$password,$dbname);

if($conn->connect_error){
  echo "Connections failed: " . $conn->connect_error;
}

$sql = "INSERT INTO data(id,name,email) VALUES(0,$name,$email)";

  if($conn->query($sql)===TRUE){
    echo "inserted data";
  }
  else{
    echo "failed";
  }

$conn->close();
}
?>

Answer 1

如果您的ID自动增加，则无需将其放在SQL行中，HTML代码中的变量（名称和电子邮件）也未定义

PHP表单在执行时不会将数据插入数据库

1 个答案: