Question

您好我正在尝试从与PHP代码相同的页面上的表单更新数据库中的数据，而不重定向/重新加载页面。

我尝试过这个教程但是没有用：http://www.formget.com/form-submit-without-page-refreshing-jquery-php /

更新代码：

<?php
include "db.php";
session_start();
$value=$_POST['name'];
$query = mysqli_query($connection,"UPDATE users SET profiel='$value' WHERE username='$_SESSION['user']'");
?>

Profilecomplete.js：

$(document).ready(function() {
    $("#submit").click(function() {
        var name = $("#name").val();
        if (name == '') {
            alert("Insertion Failed Some Fields are Blank....!!");
        } else {
        // Returns successful data submission message when the entered information is stored in database.
            $.post("config/profilecomplete.php", {
                value: name
            }, function(data) {
                alert(data);
                $('#form')[0].reset(); // To reset form fields
            });
        }
    });
});

表格：

<form method="POST" id="form">
    <div class="input-field col s6">
        <input id="name" type="text" class="validate">
        <label for="name">Value</label>
    </div>
    <button type="submit" id="submit" class="btn-flat">Update</button>
</form>

Answer 1

即使您没有阻止表单提交按钮的默认行为，您的点击事件也会如何发生。将提交输入作为按钮或使用event.preventDefault()通过ajax提交表单。

<?php
include "db.php";
session_start();

if(isset($_POST)) {
    $value=$_POST['name'];
    $query = mysqli_query($connection,"UPDATE users SET profiel='$value' WHERE username='{$_SESSION['user']}'");
    echo ($query) ? "Updated" : "Not Updated";
    exit;
} else {
?>
<form method="POST" id="form">
    <div class="input-field col s6">
        <input id="name" type="text" class="validate" name="name">
        <label for="name">Value</label>
    </div>
    <button type="button" id="submit" class="btn-flat">Update</button>
</form>
<?php } ?>

<script type="text/javascript">
$(document).ready(function() {
    $("#submit").click(function() {
        var name = $("#name").val();
        if (name === '') {
            alert("Insertion Failed Some Fields are Blank....!!");
        } else {
            // Returns successful data submission message when the entered information is stored in database.
            $.post("config/profilecomplete.php", {name: name}, function(data) {
                alert(data);
                $('form')[0].reset(); // To reset form fields
            });
        }
    });
});
</script>

Answer 2

看起来问题，或者至少是其中一个问题，就在这一行;

$query = mysqli_query($connection,"UPDATE users SET profiel='$value' WHERE username='$_SESSION['user']'");

您打开和关闭单引号两次，此处

WHERE username='$_SESSION['user']'

请尝试使用此功能;

$query = mysqli_query($connection,"UPDATE users SET profiel='$value' WHERE username='" . $_SESSION["user"] . "'");

Answer 3

使用它，它已经有效了。

<强>的index.php

<form method="post">
    <input type="text" id="name">
    <input type="submit" id="submit">
</form>

<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.2.0/jquery.min.js"></script>
<script> 
$(document).ready(function() {
    $("#submit").click(function(e) {
        e.preventDefault();
        var nameInput = $("#name").val();
        var name = {
            'name' : nameInput
        }
        if (nameInput == '') {
            alert("Insertion Failed Some Fields are Blank....!!");
        } else {
            // Returns successful data submission message when the entered information is stored in database.
            $.post("config/profilecomplete.php", {value: name}, function(data) {
                alert(data);
                //$('#form')[0].reset(); // To reset form fields
            });
        }
    });
});
</script>

<强> profilecomplete.php

<?php

$_POST = array_shift($_POST); // array_shift is very important, it lets you use the posted data.

echo $_POST['name'];


?>

如果你想要一种更简单的方法。

尝试使用$.ajax()

PHP表单保留在同一页面上

3 个答案: