Question

我想将一个输入类型元素发送到没有表单提交的PHP页面，只能使用javascript。我尝试使用 $。ajax ，但没办法。我不知道它是语法还是逻辑错误。有人告诉我要做一个PHP页面而不是两个。

HTML页面（但是.php）

<!DOCTYPE html>
<html lang="en">
<head>
  <title>Bootstrap Example</title>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1">
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
  <script src="../../js/jquery-3.2.0.min.js"></script>
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
  <style>
      #prova{
          padding-right: 1%;
          padding-top: 1%; 
      }
    </style>
    <script>
function ajax_post(apri){
    var stile = "top=10, left=10, width=650, height=600, status=no, menubar=no, toolbar=no scrollbars=no";
    // Create our XMLHttpRequest object
    var hr = new XMLHttpRequest();
    // Create some variables we need to send to our PHP file
    var url = "sperem.php";
    var fn = document.getElementById("id_buyers").value;
    alert(fn);
    var vars = "postid="+fn;
    alert(vars);
    hr.open("POST", url, true);
    // Set content type header information for sending url encoded variables in the request
    hr.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
    // Access the onreadystatechange event for the XMLHttpRequest object
    hr.onreadystatechange = function() {
        if(hr.readyState == 4 && hr.status == 200) {
            var return_data = hr.responseText;
            document.getElementById("status").innerHTML = return_data;
        }
    }
    // Send the data to PHP now... and wait for response to update the status div
    hr.send(vars); // Actually execute the request
    /*window.open(apri, "", stile);*/
    document.getElementById("status").innerHTML = "processing...";
}
</script>

</head>
<body>
    <div class="container">
        <center><h1>Add Orders:</h1></center>
        <form class="form-inline"  name="ordine">
            <div class="form-group" id="prova">
                <label for="Buyers">ID_Buyers:</label>
                <input type="text" class="form-control" placeholder="Enter IDBUYERS" id="id_buyers">
            </div>

            <div class="form-group" id="prova">
                <input type="submit" id="bottone"  value="show" onClick="ajax_post('sperem.php')"/>
            </div>

        </form>
        <br><div id="status"></div>

    </div>
</body>
</html>

这是 Php页面：

<?php
    $dato = $_POST['postid'];
    echo($dato);
?>

Answer 1

从

更改此内容

input type="submit"

到

input type="button"

然后你可以获得你在文本框中给出的数据。

这里的问题是表单是在浏览器上运行时提交的，就像你想要运行ajax一样，你需要运行页面而不需要提交/重新加载。

希望它对你有所帮助。

将数据发送到PHP弹出页面

1 个答案: