PHP表單保留在同一頁面上

Question

您好我正在嘗試從與PHP代碼相同的頁面上的表單更新數據庫中的數據，而無需重定向/重新加載頁面。

我嘗試了這個教程但是沒有用： http ： //www.formget.com/form-submit-without-page-refreshing-jquery-php /

更新代碼：

<?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

即使您沒有阻止表單提交按鈕的默認行為，您的點擊事件也會如何發生。 將提交輸入作為按鈕或使用event.preventDefault()通過ajax提交表單。

<?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

看起來問題，或至少是其中一個問題，就在這條線上;

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

您在這里打開和關閉單引號兩次

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

嘗試使用它;

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

Answer 3

使用它，它已經工作了。

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>

profilecomplete.php

<?php

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

echo $_POST['name'];


?>

如果你想要一個更簡單的方法。

嘗試使用$ .ajax（）