javascript - AJAX回调失败

Question

我正在使用ajax update将数据写入数据库，如果成功，它会返回一些文本，但是它不起作用

html

<div class="container">
        <div class="row">
            <div class="col-md-6 col-md-offset-3">
                <form>
                    <input type="text" name="name" value=""  placeholder="name..." >
                    <br>
                    <input type="text" name="email"  value="" placeholder="email..." >
                    <br>
                    <button id="form-btn" class="btn btn-default" onclick="update()">click me</button>
                </form>
            </div>
        </div>
        <div class="row">
            <div class="col-md-12"><span id="result">here</span></div>
        </div>
    </div>

js

function update() {

  if (window.XMLHttpRequest) {
    // code for IE7+, Firefox, Chrome, Opera, Safari
    xmlhttp = new XMLHttpRequest();
  } else {  // code for IE6, IE5
    xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
  }

  xmlhttp.onreadystatechange = function() {
    if (xmlhttp.readyState = 4 && xmlhttp.status == 200) {
      // document.getElementById("result").innerHTML = xmlhttp.responseText;
      alert(xmlhttp.responseText);
    }
  }

  xmlhttp.open("POST","ajax_form.php",true);
  xmlhttp.setRequestHeader("Content-type","application/x-www-form-urlencoded");
  var str = $("form").serialize();
  xmlhttp.send(str);
}

php

  $name = $_POST["name"];
$email = $_POST["email"];

if ($sql_insert = $conn->prepare("INSERT INTO $tablename (name, email) VALUES (?, ?)")) {
    # Bind the variables to the parameter as strings
    if (!$sql_insert->bind_param("ss", $name, $email)) {
        echo "binding parameter fail";
    }

# Execute the statement
if (!$sql_insert->execute()) {
    echo "insert fail: " . $sql_insert->error .  "<br>";

} else {
    echo "success";
}

# Close the prepared statement
$sql_insert->close();               
}

我期望的结果是:当用户提交表单时，它会通知用户在 <span id='result'></span> 中更新成功，但它只是更新数据而不通知。

Answer 1

由于声誉问题，我无法对您的帖子发表评论。但是你的代码有错误。您错过了一个等号

if (xmlhttp.readyState = 4 && xmlhttp.status == 200) 

应该是

if (xmlhttp.readyState == 4 && xmlhttp.status == 200)