Question

我尝试通过Ajax将HTML表单中的数据发送到php代码并且它没有得到响应

html表单从用户条目获取user_id然后通过处理ajax代码的java脚本函数发送它并将user_id发送到php代码以通过$user_id = $_GET['user_id'];获取user_id并通过user_id搜索然后显示php中的内容其他html代码中的代码显示div内容显示来自php的document.getElementById("content").innerHTML=xmlhttp.responseText;响应

＆＃13;

function showUser(str) {
  if (str == "") {
    document.getElementById("content").innerHTML = "";
    return;
  }
  if (window.XMLHttpRequest) {
    xmlhttp = new XMLHttpRequest();
  } else {
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
  }
  xmlhttp.onreadystatechange = function() {
    if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
      document.getElementById("content").innerHTML = xmlhttp.responseText;
    }
  }
  xmlhttp.open("POST", "http://localhost/gpms/admin_modify_user.php?user_id=" + str, true);
  xmlhttp.send();
}

＆＃13;

<head>
  <script src="http://localhost/gpms/admin_user.js"></script>
</head>

<body>
  <form method=post>
    <label> Enter User ID: </label>
    <input id="user_id" type=text name=user_id>
    <br><br>
    <input id="modify" type=submit value=Modify onclick="<script>showUser(user_id);</script>">
  </form>

</body>

＆＃13;

<?php

    $user_id = $_GET['user_id'];
    
    //create connection
    $conn = mysqli_connect("localhost","root","","gpms");

    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }

    $sql = "SELECT * FROM user WHERE user_id = '$user_id' ";
    $result = mysqli_query($conn, $sql);

    if (mysqli_query($conn, $sql)) {
        $row = mysqli_fetch_assoc($result);
        if ($row == 0) {
            echo "No Results";
        }
        else {
            $id = $row['user_id'];
            $name = $row['user_name'];
            $password = $row['user_password'];
	    $email = $row['user_email'];
            $department = $row['user_department'];
            
            echo "<div id = demo>";
                echo "<table>";
                    echo "<tr>";
                        echo "<td> ID </td><td> Name </td><td> Password </td><td> E-mail </td><td> Department </td>";
                    echo "</tr>";
                    echo "<tr>";
                        echo '<td> '. $id .' </td><td> '. $name .' </td><td> '. $password .' </td><td> '. $email .' </td><td> '. $department .' </td>';
                    echo "</tr>";				
                echo "</table>";
            
                echo "<button onclick = 'editUser(\"$id\",\"$name\",\"$password\",\"$email\",\"$department\")' > Edit </button>";
                echo "<button onclick = 'deleteUser(".$id.")' > Delete </button>";
            echo "</div>";
        }
    } 
    else {
        echo "Error: " . $sql . "<br>" . mysqli_error($conn);
    }
	
    $conn->close();
    ?>

＆＃13;

Answer 1

您的onclick属性错误。它应该只是Javascript代码，它不应该有<script>...</script>。

<input id="modify" type=submit value=Modify onclick="showUser(user_id);">

如果你查看过你的Javascript控制台，你会看到它在你点击时抱怨语法错误。

在PHP中，这一行：

if (mysqli_query($conn, $sql)) {

应该是：

if ($result) {

否则，您执行两次相同的查询，这是不必要的。

通过Ajax获取PHP代码的响应

1 个答案: