Question

通过Ajax上传多个图像。我正在使用fd，.append（）来发送图像，但是当我在另一个页面上进行恢复时，它会抛出错误：move_uploaded_file（upload_documents / 2017-10-04-04：41：39-aa - Copy - Copy.html）：无法打开流：

中的参数无效

所以我怎么能解决这个错误请帮助..

HTML代码： -

 <input type="file" class="form-control" name="txtBusinessImageName[]" multiple id="txtDocuName" >

     <label>Select Parent</label>
          <select class="form-control" id="txtParentDocFolder">
            <option value="0">--none--</option>
    </select>

  <button type="button" class="btn btn-success pull-right btnUploadDocClick" >Upload</button>

Ajax代码： -

<script>

$(".btnUploadDocClick").on("click",function(e){

    var total_img=$("#txtDocuName").get(0).files.length;
    if(total_img==0){
        alert("Pleas select an image");
    }
    else{
        alert(total_img);       
         for(var i=0; i<=total_img-1; i++){
             var f_name=$("#txtDocuName").get(0).files.item(i).name;
             var f_size=$("#txtDocuName").get(0).files.item(i).size;
            console.log(f_name);
            console.log(f_size);
         }       
            var fd = new FormData();
            var file_data = $("#txtDocuName")[0].files;

            var selfolder=$("#txtParentDocFolder option:selected").val();

            fd.append("file0", file_data.length+1);

            for(var i = 0;i<file_data.length;i++){
                fd.append("file"+(i+1), file_data[i]);                          
            }
            fd.append("parent_name", selfolder);
            $.ajax({
                url: 'upload_files_page.php',
                data: fd,
                contentType: false,
                processData: false,
                type: 'POST', async : true,
                success: function(data){
                    //alert("Sucessfully Saved");
                    //fetch_product_image();
                }
            }); 
      }; 
});
</script>

PHP代码： -

<?php
    session_start();
    if(isset($_SESSION['login'])){
        include("root/db_connection.php");
        $userID=str_replace("'","",$_SESSION['login']);
        $parent_name=str_replace("'","",$_REQUEST['parent_name']);

        $total=$_REQUEST['file0'];
         for($i=1; $i<$total; $i++)
         {
             $ctrl_id="file".$i;
             $file_name=$_FILES[$ctrl_id]['name'];

             $tmp_file_name= $file_name;

             $file_ext=pathinfo($tmp_file_name,PATHINFO_EXTENSION);

             $a1_file=str_replace(".".$file_ext,"",$tmp_file_name);


             $currentDate=date("Y-m-d-h:i:s");
             $orignal_file_name=$currentDate."-".$a1_file;
             $orignal_file_name=$orignal_file_name.".".$file_ext;


             $insQ=$db->query("insert into user_document(file_name,file_tmp_name,folder_id,created_by,created_on) values('$orignal_file_name','$file_name',$parent_name,$userID,now())") or die("");

             if($insQ==TRUE){
                 move_uploaded_file($_FILES[$ctrl_id]['tmp_name'],"upload_documents/".$orignal_file_name);

                 echo "Sucessfully Inserted !...";
             }
             else{
                 echo "0";
             }
         }  
    }
    else{
        echo "0";
    }
?>

Answer 1

使用此代码将图像插入本地目录..它适用于我。用于将图像路径插入到DB和映像到本地目录的简单且具有挑战性的代码。希望这对你有用。

$targetfolder = !empty(@$_FILES['file']['name']) ? $targetfolder . basename( @$_FILES['file']['name']) : false;         
$file_name="your-domain/images/".@$_FILES['file']['name'];
if(move_uploaded_file(@$_FILES['file']['tmp_name'], $targetfolder))
 {
Code for Insert the value here in DB as path= $file_name
        }else
    {
        echo "<script type='text/javascript'>alert('Problem Uploading File!')</script>";
    }

move_uploaded_file（）：无法打开流：

1 个答案: