Question

index.php

<html>
<head>
    <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
    <script type="text/javascript">
        $(function(){
            $('#submit').click(function(){
                var s = $('#submit').val();
                var nm = $('#name').val();
                var em = $('#mail').val();

                console.log("Starting ajax");
                $.ajax({
                    url: "./records.php",
                    type: "post",
                    data: {
                        submit:s,
                        name:nm,
                        mail:em
                    },
                    success:function(data) {
                        alert('done');
                    }
                });
            });
        });
    </script>
</head>
<body>
    <?php 
        if(isset($_GET['id'])) {
            $id = $_GET['id'];
            $query = mysqli_query($conn, "SELECT FROM form WHERE my_id = '".$id."' ");
            $q = mysqli_fetch_array($query);
        }
     ?>
    <form>
        <table>
            <tr>
                <td>Name</td>
                <td><input type="text" name="name" id="name"></td>
            </tr>
            <tr>
                <td>Email</td>
                <td><input type="text" name="mail" id="mail"></td>
            </tr>

            <tr>
                <td></td>
                 <?php 
                 if(isset($_GET['id']) > 1) 
                {
                ?>
                <input type = "submit" class = "btn btn-primary" style="width:49%" value = "Save" name = "submit">
                <?php
                 } else {
                ?>
                <input type = "hidden" name = "uid" id = "uid">
                <input type = "submit" class = "btn btn-primary" style="width:49%" value = "Update" name = "submit">
                <?php 
                } 
                ?>
            </tr>
        </table>
    </form>
</body>
</html>

如何使用相同的按钮进行插入和更新，如何使用PHP，jQuery和Ajax制作单个表单。因此解决了这个问题，使它成为了用于插入和更新的单个按钮，并且让我知道了如何用一个按钮将其制成一种形式并执行插入和更新操作

Answer 1

if(isset($_POST['submit']) && !isset($_POST['uid']) ) {
    $name = $_POST['name'];
    $email = $_POST['mail'];
    $gen = $_POST['gender'];
    $age = $_POST['age'];
    //$hob =implode(",",$_POST['hobbies']);
    $hob = implode(",",$_POST['hobbies']);
    $pass = $_POST['pass'];
    $cpass = $_POST['cpass'];

    $query = mysqli_query($conn, "INSERT INTO form(name, email, gender, age, hobbies, pass, cpass)VALUES('".$name."', '".$email."', '".$gen."', '".$age."', '".$hob."', '".$pass."', '".$cpass."') ");
    if($query)
     {
        echo "Insert";
    } 
    else
     {
        echo "Fail";
    }
}

if(isset($_POST['submit']) && isset($_POST['uid'])) {
    $id = $_POST['uid'];
    $name = $_POST['name'];
    $email = $_POST['mail'];
    $gen = $_POST['gender'];
    $age = $_POST['age'];
    //$hob =implode(",",$_POST['hobbies']);
    //$hob = implode(",",$_POST['hobbies']);
    $hob = implode(",",$_POST['hobbies']);
    $pass = $_POST['pass'];
    $cpass = $_POST['cpass'];
    $query = mysqli_query($conn, "UPDATE form SET name = '".$name."', email = '".$email."', gender = '".$gen."', age = '".$age."', hobbies = '".$hob."', pass = '".$pass."', cpass = '".$cpass."' WHERE my_id = '".$id."' ");
    if($query)
     {
        echo "Update";
    } 
    else
     {
        echo "Fail";
    }
}

如何使用相同的按钮进行插入和更新

1 个答案: