Question

我使用以下表格：

<form id="dataForm" method="post">
  <h2 id="formheader"> Update Product Description</h2>
    <div>
      <label>Product Name:</label>
      <input class="inputForm" id="orginalName" type="text" name="Name">
    </div>
    <div>
      <label>New Description:</label>
      <input class="inputForm" id="newDescription" type="text" name="newDescription">
    </div>
    <div id="theSubmit">
      <button id="editDescription">Submit</button>
    </div>
  </form>

并使用以下简单的php，当与action=editProductDes.php一起使用时...

 $Name = $_POST['Name'];
 $Description = $_POST['newDescription'];

 if($Name !="" && $Description !=""){
 $sql = "UPDATE PRODUCTS SET P_Description = '$Description' WHERE P_NAME = '$Name'";
 $conn->exec($sql);

然后当我使用下面的java脚本时，数据没有通过，我无法理解为什么因为我有一个类似的函数和表单，JavaScript工作正常，任何人都可以看到为什么数据没有通过？

function editDescription(){
    xmlhttp = new XMLHttpRequest();
    var name = document.getElementById("orginalName");
    var Description = document.getElementById("newDescription");

    var data_seen = false;
        // this is a flag to record whether any data has been seen. Used in the guard ofthe alert statement.
    if (name.value !="" && Description.value !="" ){
        data_seen = true;
        xmlhttp.open("POST","editDescription.PHP",true);
        xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
        xmlhttp.send("Name=" + name.value + "&Description=" + Description.value);
    }
    if (!data_seen) {
        alert("please enter some data");
    }
   }

submitButton = document.getElementById("editDescription");
submitButton.addEventListener("click", editDescription);

Answer 1

您要发布到editDescription.PHP而不是editProductDes.php

更改以下内容：

xmlhttp.open("POST","editProductDes.php",true);

您还发布了帖子中的数据，而不是您希望它在PHP代码中的名称（Description而不是newDescription） - 更改：

xmlhttp.send("Name=" + name.value + "&newDescription=" + Description.value);

为什么数据没有通过javascript传递给php

1 个答案: