javascript - Isset 不适用于 ajax 调用

Question

我正在制作一个简单的页面，用户可以在其中上传图像而无需刷新整个页面。但是 if(isset($_post[oneimgtxt])) 不起作用..这是我上传图像的服务器端代码:

<?php
$maxmum_size = 3145728; //3mb 
$image_type_allowed = array(IMAGETYPE_GIF, IMAGETYPE_JPEG, IMAGETYPE_PNG, IMAGETYPE_BMP);
if ($_SERVER["REQUEST_METHOD"] == "POST") {
    if(isset($_POST["oneimgtxt"])){//<!------------------ this line is not working
        if((!empty($_FILES[$_FILES['upimage']['tmp_name']])) && ($_FILES["upimage"]['error'] == 0)){
            $file=$_FILES['upimage']['tmp_name'];
            $image_count = count($_FILES['upimage']['tmp_name']);
            if($image_count == 1){
                $image_name = $_FILES["upimage"]["name"];
                $image_type = $_FILES["upimage"]["type"];
                $image_size = $_FILES["upimage"]["size"];
                $image_error = $_FILES["upimage"]["error"];
                if(file_exists($file)){//if file is uploaded on server in tmp folder (xampp) depends !!
                    $filetype =exif_imagetype($file); // 1st method to check if it is image, this read first binary data of image..
                    if (in_array($filetype, $image_type_allowed)) {
                        // second method to check valid image
                        if(verifyImage($filename)){// verifyImage is function created in fucrenzione file.. using getimagesize
                            if($ImageSizes < $maxmum_size){//3mb 
                                $usr_dir = "folder/". $image_name;
                                move_uploaded_file($file, $usr_dir);
                            }else{
                                $error_container["1006"]=true;
                            }
                        }else{
                            $error_container["1005"]=true;
                        }
                    }else{
                        $error_container["1004"]=true;
                    }
                }else{
                    $error_container["1003"]=true;
                }
            }else{
                $error_container["1002"]=true;
            }
        }else{
            $error_container["1007"]=true;
        }
    }else{//this else of image issset isset($_POST["oneimgtxt"])
        $error_container["1001"]=true;//"Error during uploading image";
    }
    echo json_encode($error_container);
}
?>

在 chrome 检查元素中我得到了这个.. image这是我使用 ajax 的 js 代码...

$(".sndbtn").click( function(e){
        var form = $("#f12le")[0];
        var formdata = new FormData(form)
        $.ajax({
            type:'POST',
            //method:'post',
            url: "pstrum/onphotx.php",
            cache:false,
            data: {oneimgtxt : formdata},
            processData: false,
            contentType: false,
            success:function (e){console.log(e);}
        });
    });

这是html代码:

<form  method="post" id="f12le" enctype="multipart/form-data">
   <input type="file" name="upimage"/>
   <label for="imgr">Choose an Image..</label>
   <textarea placeholder="Write something about photo"></textarea>
   <input   type="button" name="addimagedata" value="Post" class="sndbtn"/>
</form>

感谢您的帮助。

Answer 1

您应该将 FormData 作为整个数据对象而不是另一个数据对象的一部分发送。所以，应该是这样的 -

$(".sndbtn").click( function(e){
    var form = $("#f12le")[0];
    var formdata = new FormData(form)
    $.ajax({
        type:'POST',
        //method:'post',
        url: "pstrum/onphotx.php",
        cache:false,
        data: formdata,
        processData: false,
        contentType: false,
        success:function (e){console.log(e);}
    });
});

Now, you should be able to access the form as it is. For example if you have any input with name inputxt inside the form, you should be able to access it with $_POST['inputxt']. And if you have any input type="file" with the name upimage, you need to access through $_FILES['upimage']. So, if you want to do isset() for that. You can do like this :

if(isset($_FILES['upimage'])){