Question

目前我在用户上传照片的表单上使用AJAX，然后进入php检查文件，并根据成功与否返回消息。我将结果回显到段落标记中，所以此刻我的PHP会将用户标题到我想要的页面，但是它会在标签内容中加载页面，我在回显结果。

我无法真正使用AJAX的成功来做到这一点，因为即使向用户显示错误消息它也是成功的

这是我的AJAX

$("#avatar_form").submit(function (event) {

                    //disable the default form submission

                    //grab all form data  
                    var formData = new FormData($(this)[0]);

                    $.ajax({
                        url: 'photo_system.php',
                        type: 'POST',
                        data: formData,
                        async: false,
                        cache: false,
                        contentType: false,
                        processData: false,
                        success: function (result) {
                            console.log(result);
                            $('#status').html(result);
                        }
                    });

                    event.preventDefault();
                });

PHP

$result = "";
if (isset($_FILES["avatar"]["name"]) && $_FILES["avatar"]["tmp_name"] != "") {
    $fileName = $_FILES["avatar"]["name"];
    $fileTmpLoc = $_FILES["avatar"]["tmp_name"];
    $fileType = $_FILES["avatar"]["type"];
    $fileSize = $_FILES["avatar"]["size"];
    $fileErrorMsg = $_FILES["avatar"]["error"];
    $kaboom = explode(".", $fileName);
    $fileExt = end($kaboom);
    list($width, $height) = getimagesize($fileTmpLoc);
    if ($width < 10 || $height < 10) {
        $result = "That image has no dimensions";
        echo $result;
        exit();
    }
    $db_file_name = rand(100000000000, 999999999999) . "." . $fileExt;
    if ($fileSize > 1048576) {
        $result = "Your image file was larger than 1mb";
        echo $result;
        exit();
    } else if (!preg_match("/\.(gif|jpg|png)$/i", $fileName)) {
        $result = "Please only JPG, GIF or PNG images";
        echo $result;
        exit();
    } else if ($fileErrorMsg == 1) {
        $result = "An unknown error occurred";
        echo $result;
        exit();
    }
    $sql = "SELECT profilePicture FROM User WHERE username='$log_username' LIMIT 1";
    $query = mysqli_query($db_conx, $sql);
    $row = mysqli_fetch_row($query);
    $avatar = $row[0];

    //delete old pic if set
    if ($avatar != "") {
        $picurl = "users/$log_username/$avatar";
        if (file_exists($picurl)) {
            unlink($picurl);
        }
    }
    //move file from temp folder to users folder
    $moveResult = move_uploaded_file($fileTmpLoc, "users/$log_username/$db_file_name");
    if ($moveResult != true) {
        $result = "File upload failed";
        echo $result;
        exit();
    }
    include_once("image_resize.php");
    //replace original file with resized version
    $target_file = "users/$log_username/$db_file_name";
    $resized_file = "users/$log_username/$db_file_name";
    $wmax = 400;
    $hmax = 600;
    img_resize($target_file, $resized_file, $wmax, $hmax, $fileExt);
    $sql = "UPDATE User SET profilePicture='$db_file_name' WHERE username='$log_username' LIMIT 1";
    $query = mysqli_query($db_conx, $sql);
    mysqli_close($db_conx);
    header("location: userprofileyh.php?u=$log_username");
    $result = "Uploaded successfully";
    echo $result;
    exit();
}
?>

HTML表格

$avatar_form .= '<form id="avatar_form" method="post" enctype="multipart/form-data">';
    $avatar_form .= '<h1>Change avatar</h1>';
    $avatar_form .= '<input type="file" name="avatar" required>';
    $avatar_form .= '<p><input type="submit" value="Upload"></p>';
    $avatar_form .= '<p id="status"></p>';
    $avatar_form .= '</form>';

Answer 1

如果上传失败，您可以通过在php中抛出异常来使用错误块。

例如，而不是

SELECT * FROM `mrbs_entry` 
WHERE DATE_SUB(FROM_UNIXTIME(`end_time`), INTERVAL 14 HOUR) = FROM_UNIXTIME(`start_time`);

你做了

if ($moveResult != true) {
    $result = "File upload failed";
    echo $result;
    exit();
}

并在前端抓住它：

if ($moveResult != true) {
    throw new Exception();
}

Answer 2

也许尝试从php脚本中传回更多关于它是否真的成功的信息。然后在ajax select中编写自己的逻辑，以解析它是否成功或是否发生错误。然后你也可以用它来做重定向。

示例：

<?php 
$result['err'] = 'error';
$result['msg'] = 'File upload failed';

echo json_encode($result);
?>

然后用jQuery解析JSON。

使用PHP或JQuery直接转到页面

2 个答案: