javascript - 使用ajax、jquery和PHP上传带有其他信息的图像

Question

我在使用 ajax + php 上传图像信息时遇到问题。尝试添加 dataResult.file = form_data; 时收到非法调用消息。正如您在下面的代码中看到的。

我将非常感谢任何帮助!我对此很困惑，并且在 stackoverflow 上进行了大量搜索，寻找可以帮助我的答案。

function uploadImage (upFile,userId,description,category) {
    console.log("We have reached the point where we connect to the service to upload the image...");
    var dataResult = {};
    var form_data = new FormData();
    form_data.append('file',upFile[0]);
    console.log("Is the information on the form data variable?");
    console.log(form_data);
    dataResult.file = form_data;
    dataResult.idUser = userId;
    dataResult.description = description; 
    dataResult.nameCat = category;

    console.log("Description " + dataResult.description);
    console.log("Id User " + dataResult.idUser);
    console.log("name Cat " + dataResult.nameCat);

    console.log("This is the information that is being sent thorugh ajax");
    console.log(dataResult);

    $.ajax({
            url: 'http://localhost/App/www/classes/actionsPhotos/uploadPhoto.php',
            type: 'post',
            data: dataResult,
            success: function (data) {
                console.log("All is good and working.");
                console.log(data);
            },
            error: function(errorInfo){
                console.log("There was an error: ");
                console.log(errorInfo);
            }
        });
}

Answer 1

好吧，我通过稍微切换代码解决了这个问题，这是为遇到同样问题的任何人提供的新代码。

函数uploadImage(upFile,userId,描述,类别){ console.log(“我们已经连接到服务来上传图像了......”); var form_data = new FormData(); form_data.append('文件',upFile[0]); form_data.append('idUser',userId); form_data.append('描述',描述); form_data.append('nameCat',类别);

$.ajax({
        url: 'http://localhost/App/www/classes/actionsPhotos/uploadPhoto.php',
        type: 'post',
        data: form_data,
        cache: false,
        processData: false,
        contentType: false,
        success: function (data) {
            console.log("All is good and working.");
            console.log(data);
        },
        error: function(errorInfo){
            console.log("There was an error: ");
            console.log(errorInfo);
        }
    });

}