Question

我将以下文件中的一些代码更改为JS，并进行了AJAX调用以使动画更容易。但是，由于我将部分PHP移动到另一个文件并进行了AJAX调用，因此我的第二个查询出现问题。

我要做的是显示用户名。然后，一旦单击按钮，它就会对这些用户名进行洗牌并显示它们。一切正常。

我的第二个查询的作用是获取该信息并将其发送到不同的数据库表。这是失败的，我收到以下错误：

注意：未定义的索引：位于第235行的/home4/pfarley1/public_html/sundayfundayleague.com/userCreator.php中的ID

警告：第235行/home4/gddgg/public_html/example.com/userCreator.php中为foreach（）提供的参数无效

LINE 235：

foreach ($_POST['id'] as $i => $shuffle_id) {

警告：mysqli_fetch_assoc（）要求参数1为mysqli_result，第245行/home4/gfdgd/public_html/example.com/userCreator.php中给出布尔值

LINE 245

$order_row = mysqli_fetch_assoc($stmt);


$con = mysqli_connect("localhost", "", "", "");
$query = mysqli_query($con, "SELECT * FROM users WHERE `group` = 3");

echo 'Users to be given draft order: <br>';
$array = array();
while ($row = mysqli_fetch_assoc($query)) {
    $array[] = $row;
    echo $row['firstname'] . ' ' . $row['lastname'] . '<br>';
}
?>
<form method="POST" name="form">
    <input type="submit" value="Create Draft Order" name="shuffle">
</form>

    Shuffled results: <br>
    <div class="shuffle_results" id="results"></div>
     <img id='paperBag' src="http://www.thecuriouscaterpillar.co.uk/media/catalog/product/cache/1/image/9df78eab33525d08d6e5fb8d27136e95/b/a/bag_to_white.jpg" width="200px" />
    <form method="post">

         <input type="submit" value="Finalize Draft Order" name="insert">
        </form>

 <?php

if (isset($_POST['insert'])) {
    $con = mysqli_connect("localhost", "", "", "");
    if (mysqli_connect_errno()) {
        printf("Connect failed: %s\n", mysqli_connect_error());
        exit();
    }
    $stmt2 = $con->prepare("INSERT INTO user_players (user_id, firstname, lastname, username, email) VALUES (?, ?, ?, ?, ?)");
    if ( false===$stmt2 ) {
         // Check Errors for prepare
        die('Add to user players prepare() failed: ' . htmlspecialchars($con->error));
    }
    $stmt2->bind_param('issss', $shuffle_id, $shuffle_firstname, $shuffle_lastname, $shuffle_username, $shuffle_email);

    foreach ($_POST['id'] as $i => $shuffle_id) {
        $shuffle_firstname = $_POST['firstname'][$i];
        $shuffle_lastname = $_POST['lastname'][$i];
        $shuffle_username = $_POST['username'][$i];
        $shuffle_email = $_POST['email'][$i];
        $stmt2->execute() or
            die('Add to user players execute() failed: ' . htmlspecialchars($stmt2->error));
    }

JS从我的php文件中获取Shuffle结果

   $('#results').append('<div class="result">' +
                        '<div class="shuffle_results">' + data[i].firstname + ' ' + data[i].lastname + '</div>' + 
                        '<input type="hidden" name="firstname[]" value="' + data[i].firstname + '">' +
                        '<input type="hidden" name="lastname[]" value="' + data[i].lastname + '">' +
                        '<input type="hidden" name="id[]" value="' + data[i].id + '">' +
                        '<input type="hidden" name="username[]" value="' + data[i].username + '">' +
                        '<input type="hidden" name="email[]" value="' + data[i].email + '">' +
                    '</div>');
           var $this = $('.shuffle_results:last'); 
    $this.show().animate({ 
    'left': 0 + 'px', 
    'bottom': + '0px' 
    //$(document).height() - (lineheight * data.length)
    }, { 
    duration: time 
    });
     i++;
       } else { 
    clearInterval(interval); 
    } 
    }, 3000); 
    };
$(function(){  
    $('form[name="form"]').on('submit', function(e){
        e.preventDefault();
       $.post('shuffle_results.php', function(data){ 
        var o = $.parseJSON(data); 
        displayResults(o); 
        });
        });

PHP文件

$con = mysqli_connect("localhost", "", "", "");
$query = mysqli_query($con, "SELECT * FROM users WHERE `group` = 3");
$array = array();

while ($row = mysqli_fetch_assoc($query)) { 
$array[] = array( 
'firstname' => $row['firstname'], 
'lastname' => $row['lastname'], 
'id' => $row['id'], 
'username' => $row['username'], 
'email' => $row['email'] 
); 

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

    }
}
shuffle($array);
    echo json_encode($array);

有没有人知道为什么这会失败，因为我将php代码移动到另一个文件并且现在正在进行AJAX调用？

我怎样才能获得这些变量..它们和以前一样？我只是动了一下。

INSERT查询未通过发送

0 个答案: