Question

所以我有2个不同的页面，1个带有HTML表单，1个带有PHP代码。因此，当我按表单所在的第1页上的按钮时，如何确保此代码运行？

<?php 

$conn = mysqli_connect("localhost", "root", "123", "soup") or die("Geen connectie gemaakt: ".mysqli_connect_error());
if(isset($_POST["submit"])) { 
  if (isset($_POST["Cluster_Name"]) && isset($_POST["Cluster_FQDN"]) && 
  isset($_POST["DBS"]) && isset($_POST["Local_AE"]) && 
  isset($_POST["AE_Title"]) && isset($_POST["Modality_Type"])) {
    $Cluster_Name = $_POST["Cluster_Name"];
    $Cluster_FQDN = $_POST["Cluster_FQDN"];
    $DBS = $_POST["DBS"];
    $Local_AE = $_POST["Local_AE"];
    $AE_Title = $_POST["AE_Title"];
    $Modality_Type = $_POST["Modality_Type"];

    $qryUpdate = "UPDATE tblcluster SET Cluster_Name = '$Cluster_Name', 
    Cluster_FQDN = '$Cluster_FQDN', DBS = '$DBS', Local_AE = '$Local_AE', 
    AE_Title = '$AE_Title', Modality_Type = '$Modality_Type' WHERE 
    Cluster_FQDN = '$Cluster_FQDN'";

    if(!mysqli_query($conn, $qryUpdate)) {
      echo "Gegevens zijn niet geupdated";
    } 
    else {
      echo "Gegevens zijn geupdate.";
    }
  }
}
?>

Answer 1

您可以在此处添加html表单代码或确保表单中的“提交”按钮具有名称属性，并且表单标签应具有操作和方法属性像这样：

<form action="your-php-filename.php" method="post">
<!-- OTHER FIELDS -->
<input type="submit" name="submit">
</form>

按下输入按钮时如何运行PHP代码？

1 个答案: