duancuan7057 2015-12-25 11:12
浏览 49
已采纳

Isset不适用于ajax调用

I am making a simple page where user can upload a image without refreshing the whole page. But if(isset($_post[oneimgtxt])) is not working.. here is my serverSide Code that upload image :

<?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);
}
?>

in chrome inspect element i got this.. image and this is my js code with ajax...

$(".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);}
        });
    });

Here is html code:

<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>

Thanks for any help.

  • 写回答

3条回答 默认 最新

  • duanna2026 2015-12-25 11:28
    关注

    You should send your FormData as a whole data object not a part of another data object. So, it should be like this -

    $(".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'])){
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论
查看更多回答(2条)

报告相同问题?

悬赏问题

  • ¥15 做个有关计算的小程序
  • ¥15 MPI读取tif文件无法正常给各进程分配路径
  • ¥15 如何用MATLAB实现以下三个公式(有相互嵌套)
  • ¥30 关于#算法#的问题:运用EViews第九版本进行一系列计量经济学的时间数列数据回归分析预测问题 求各位帮我解答一下
  • ¥15 setInterval 页面闪烁,怎么解决
  • ¥15 如何让企业微信机器人实现消息汇总整合
  • ¥50 关于#ui#的问题:做yolov8的ui界面出现的问题
  • ¥15 如何用Python爬取各高校教师公开的教育和工作经历
  • ¥15 TLE9879QXA40 电机驱动
  • ¥20 对于工程问题的非线性数学模型进行线性化