Question

实际上我想通过将我的表单（包含我的上传文件）发送到php上传处理文件，使用$ .post异步上传照片。.serializeArray()不会做必要的事情。有人可以提出任何建议吗？

HTML：

<form action="upload1.php" method="post" id="usrform" enctype="multipart/form-data" >
        <div class="input-group text-center ">
        <input type="file" class="form-control btn btn-default" name="fileToUpload" id="fileToUpload"data-toggle="tooltip" data-placement="left" title="Choose your video to upload by clicking on the choose file  button"></div>
        <br><div class="form-group"><strong><span class="glyphicon glyphicon-pencil" style="font-size:13px" ></span> Caption:</strong> <input form="usrform" name="post"id="textarea1" type="text" class="form-control" name="image" placeholder="Write something" data-toggle="tooltip" data-placement="top" title="">
        </div>
        <div class="text-right">
        <button type="submit" id="sub"class="btn btn-primary">POST</button>
        </div></form>

的javascript：

$('#sub').click(function(){$.post('porthome_.php',$("#usrform").serializeArray(),function(info){

            clearInput();$("#myModal1 .close").click();

           });});

           $('#usrform').submit(function(){return false;});

PHP：

<?php
 require("../includes/config.php"); 

$target_dir = "uploads/";
$target_file = $target_dir . basename($_FILES["fileToUpload"]["name"]);
$uploadOk = 1;
$imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);
// Check if image file is a actual image or fake image
if(isset($_POST["submitfile"])) {
    $check = getimagesize($_FILES["fileToUpload"]["tmp_name"]);
    if($check !== false) {
        //apologize( "File is an image - " . $check["mime"] . ".","portimage_.php","Error");
        $uploadOk = 1;
    } else {
        apologize( "File is not an image.","portimage_.php","Error");
        $uploadOk = 0;
    }
}
// Check if file already exists
if (file_exists($target_file)) {
    apologize( "Sorry, file already exists.Try changing the name of your image file.","portgallery_.php","Error");
    $uploadOk = 0;
}
// Check file size
if ($_FILES["fileToUpload"]["size"] > 5000000) {
    apologize( "Sorry, your file is too large.","portimage_.php","Error");
    $uploadOk = 0;
}
// Allow certain file formats
if($imageFileType != "jpg" && $imageFileType != "png" && $imageFileType != "jpeg"
&& $imageFileType != "gif" ) {
    apologize( "Sorry, only JPG, JPEG, PNG & GIF files are allowed.","portimage_.php","Error");
    $uploadOk = 0;
}
// Check if $uploadOk is set to 0 by an error
if ($uploadOk == 0) {
    apologize( "Sorry, your file was not uploaded.","portimage_.php","Error");
// if everything is ok, try to upload file
} else {
    if (move_uploaded_file($_FILES["fileToUpload"]["tmp_name"], $target_file)) {


    apologize( "The file ". basename( $_FILES["fileToUpload"]["name"]). " has been uploaded.","porthome_.php","Success");

    } else {
        apologize( "Sorry, there was an error uploading your file.","portimage_.php","Error");
    }
}?>

如果有其他方法可以这样做，我们将不胜感激。

Answer 1

我不使用jQuery，所以对于你目前使用的$.post()函数没有提供真正的指导，但是在普通的，普通的javascript你可以这样做。

<form action="upload1.php" method="post" id="usrform" enctype="multipart/form-data" >
    <div class="input-group text-center ">
        <input type="file" class="form-control btn btn-default" name="fileToUpload" id="fileToUpload"data-toggle="tooltip" data-placement="left" title="Choose your video to upload by clicking on the choose file  button">
    </div>
    <br>
    <div class="form-group">
        <strong><span class="glyphicon glyphicon-pencil" style="font-size:13px" ></span> Caption:</strong>
        <input form="usrform" name="post"id="textarea1" type="text" class="form-control" name="image" placeholder="Write something" data-toggle="tooltip" data-placement="top" title="">
    </div>
    <div class="text-right">
        <button type="submit" id="sub" class="btn btn-primary">POST</button>
    </div>
</form>

<script>
    function uploadfiles(e){
        var data=new FormData( document.getElementById('usrform') );
        var xhr=new XMLHttpRequest();
        xhr.onload=function(e){
            /* You probably would want to do more than popup an alert here! */
            alert( this.response );
        }
        xhr.onerror=function(e){
            alert(e);
        }
        xhr.open( 'POST', 'upload1.php', true );
        xhr.send( data );
    }
    document.addEventListener('DOMContentLoaded',function(){
        document.getElementById('sub').onclick=uploadfiles;
    },false );
</script>

如何在jquery中使用$ .post将上传的文件发送到任何php文件进行处理？

1 个答案: