Question

我有一个可以由用户编辑的表单。我希望使用Ajax在不重新加载页面的情况下更新表单。新值将替换旧值。对于输入字段，我从php SQL查询中读取值。因此，用户将能够第一眼看到细节。问题是这个代码运行时没有错误，也没有结果。我试过console.log。

我从while循环中获取用户值。

if(isset($_GET['edit_user'])){
    $the_user_id =  $_GET['edit_user'];
    $query = "SELECT * FROM users WHERE user_id = $the_user_id ";
    $select_users_query = mysqli_query($connection,$query);  

    while($row = mysqli_fetch_assoc($select_users_query)) {
        $user_id        = $row['user_id'];
        $user_firstname = $row['first_name'];
        $user_lastname  = $row['last_name'];
        $user_contact = $row['mobile'];


      $_SESSION["id"] = $user_id;
    }  
}

Ajax代码：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
<script>
$(document).ready(function(){
    $('#update').click(function(){
        var user_id =  $("#user_id").val();
        var firstname = $("#firstname").val();
        var lastname = $("#lastname").val();
        var contact = $("#mob").val();
        var dataString = 'firstname='+firstname + 'lastname='+lastname + 'contact='+contact+'user_id='+user_id;

        if(firstname=='' || lastname=='' || contact=='') {
            alert("Please fill all fields");
        } else {
            $.ajax({
                type: "POST",
                url: "update.php",
                data: dataString,
                cache: false,
                success: function(html){
                    alert(html);
                }
            });
        }
        return false;
    });
}); 
</script>

我希望能够在不重页加载的情况下编辑表单。

<form method="post" name="form">
    <div style="position:relative; left:120px;">
        <p>Title: <select><option value="Mr">Mr</option> <option value="Mrs">Mrs</option><option value="Miss">Miss</option><option value="Ms">Ms</option><option value="Dr">Dr</option></select></p>
        <p>First name *: <input type="text" id="firstname" name="firstname" value="<?php echo $user_firstname; ?>" style="width:50%;"></p>
        <p>Last name *: <input type="text" id="lastname" name="lastname" value="<?php echo $user_lastname; ?>" style="width:50%;"></p>
        <p>Contact telephone number *: <input type="text" id="mob" name="contact" value="<?php echo $user_contact; ?>" style="width:50%;"></p>
                    <p><input type="hidden" id="user_id" name="user_id" value="<?php echo $_SESSION['user_id']; ?>" style="width:50%;"></p>
        <button class="btn btn-primary btn-lg" id="update" name="update" role="navigation" type="submit" style="border-radius:0px;">Save & continue</button>
    </div>
</form>

update.php

         <?php require_once("includes/db.php"); ?>

             if(isset($_POST['update'])){

              $firstname = mysqli_real_escape_string($_POST['firstname']);
             $lastname =  mysqli_real_escape_string($_POST['lastname']);
             $contact =  mysqli_real_escape_string($_POST['contact']);
             $user_id =  mysqli_real_escape_string($_POST['user_id']);

              $query = "UPDATE users
          SET first_name   ='". $_POST['firstname'] . "',
              last_name ='". $_POST['lastname'] . "',
              mob    ='". $_POST['contact'] . "'

         WHERE
             user_id = '". $_POST['user_id'] . "'";



        $edit_user_query = mysqli_query($connection,$query);



}

Answer 1

您没有设置用户ID。确保在您的ajax请求中发送userid或根据给定的数据获取它

最新情况：＆GT;你需要将你的jquery改为

 var dataString = 'firstname='+firstname + '&lastname='+lastname + '&contact='+contact;

我会将update.php更改为以下代码

<?php
require_once("includes/db.php");

//get user_id from session data or from call to db to combine session id with user_id
//this is depending on where you store this data for a currently logged in user
$userid = $_SESSION['user_id'];

if (isset($_POST['firstname'], $_POST['lastname'], $_POST['contact'])) {

$firstname = mysqli_real_escape_string($_POST['firstname']);
$lastname = mysqli_real_escape_string($_POST['lastname']);
$contact = mysqli_real_escape_string($_POST['contact']);

$query = "UPDATE users
SET first_name   ='" . $_POST['firstname'] . "',
last_name ='" . $_POST['lastname'] . "',
mob    ='" . $_POST['contact'] . "'

WHERE
user_id = '" . $userid . "'";


$edit_user_query = mysqli_query($connection, $query);
} 
else {
echo "invalid response";
}

?>

使用AJAX和PHP更新表单字段

1 个答案: