如何从FormData获取PHP文件中的数据

本文介绍了如何从FormData获取PHP文件中的数据的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我的格式如下:

<form id="f-comment" class="form" method="post" action="submit_img_comment.php">
    <textarea name="comment"></textarea>
    <input type="submit" value="Publish" data-params='{"imageid":<?php echo $imageid; ?>}'>
</form>

和以下javascript:

and the following javascript:

$(document).on("submit", ".form", function(e) {
    e.preventDefault();

    // what form are you submitting?
    var form = $("#" + e.target.id);

    var data = new FormData(this);
    var params = $("input[type=submit]", this).data("params"); // parameters to send along with data
    data.append("params", params);

    // data is ok
    console.log(params)

    $.ajax({
        type: form.attr("method"),
        url: "include/" + form.attr("action"),
        data: data,
        dataType: "json",
        contentType: false,
        processData: false,
        cache: false
    }).done(function(data) {
        alert(data['msg']);
    }).fail(function(data) {
        alert("Error: Ajax Failed.");
    }).always(function(data) {
        // always do the following, no matter if it fails or not
    })
});

在我的php文件(submit_img_comment.php)中，我无法获得评论，就像这样

in my php file (submit_img_comment.php) im able to get the comment, like this

$_POST['comment'];

但是，当我尝试获取imageid时，就像这样

But, when i try to get the imageid, like this

$_POST['imageid'];

我收到错误:未定义索引:imageid

注释是表单的一部分，但是imageid作为参数发送并附加在FormData中.

The comment is part of the form, but the imageid is send as a parameter and appended in FormData.

如何在我的php文件中获取imageid?

How do i get the imageid in my php file?

推荐答案

您看错了，添加到表格中的不是imageid而是params.另外，您要发送的是一个javascript对象，您需要首先将其转换为字符串.您将需要在JavaScript中执行以下操作:

You are look at this all wrong, what you have appended to your form is not imageid but params. Also, what you are sending is a javascript object, you need to convert this to a string first. You will need to do the following in JavaScript:

var data = new FormData(this);
var params = $("input[type=submit]", this).data("params"); // parameters to send along with data
var json_params = JSON.stringify(params); // This converts your javascript object into a string that you can send to the server.
data.append("params", json_params); // We are adding the string we have converted from the javascript object.

现在我们已经向服务器发送了JSON字符串，现在我们需要对其进行解码.要在php中解码JSON，我们使用:

Now that we have sent a JSON string to the server, we now need to decode it. To decode the JSON in php we use:

$params = json_decode($_POST['params']);

变量$params现在将是一个包含imageid作为属性的php对象.这意味着您的图片ID现在存储在$params->imageid变量中，例如echo $params->imageid将输出您的图像ID.

The variable $params will now be a php object which contains imageid as a property. This means that your image id is now stored in $params->imageid variable e.g. echo $params->imageid will output your image id.

正如@baboizk正确提到的那样，您应该在PHP中使用isset()来确保它确实存在，然后再使用它.例如

As @baboizk has rightly mentioned, you should use isset() in PHP to make sure it actually exists before using it. E.g.

$params = json_decode($_POST['params']);

// Check that imageid exists.
if(isset($params->imageid) == true) {
    // Do code that needs $params->imageid;
} else {
    // Fail gracefully.
}

这篇关于如何从FormData获取PHP文件中的数据的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！

imageID