jQuery / ajax upload image and save to folder

2020-03-23 18:26发布

站内问答 / PHP
1912 3 6

UPDATE CODE BELOW

I found some code that is able to upload an image and display its thumbnail. However, I would like to save the images to a particular folder as well.

What jQuery code or ajax code can I use to save the original image to a folder of my choice?

Here is the live demo: http://jsfiddle.net/dn9Sr/2/

Here is the full code:

<html>
<head>
<script src="http://code.jquery.com/jquery-1.8.3.js" type="text/javascript"></script>

<style>
.input-file-row-1:after {
    content: ".";
    display: block;
    clear: both;
    visibility: hidden;
    line-height: 0;
    height: 0;
}

.input-file-row-1{
    display: inline-block;
    margin-top: 25px;
    position: relative;
}

#preview_image {
  display: none;
  width: 90px;
  height: 90px;
  margin: 2px 0px 0px 5px;
  border-radius: 10px;
}

.upload-file-container { 
    position: relative; 
    width: 100px; 
    height: 137px; 
    overflow: hidden;   
    background: url(http://i.imgur.com/AeUEdJb.png) top center no-repeat;
    float: left;
    margin-left: 23px;
} 

.upload-file-container-text{
    font-family: Arial, sans-serif;
    font-size: 12px;
    color: #719d2b;
    line-height: 17px;
    text-align: center;
    display: block;
    position: absolute; 
    left: 0; 
    bottom: 0; 
    width: 100px; 
    height: 35px;
}

.upload-file-container-text > span{
    border-bottom: 1px solid #719d2b;
    cursor: pointer;
}

.one_opacity_0 {
  opacity: 0;
  height: 0;
  width: 1px;
  float: left;
}
</style>
<script>
$( document ).ready(function() {

   function readURL(input, target) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();
            var image_target = $(target);
            reader.onload = function (e) {
                image_target.attr('src', e.target.result).show();
            };
            reader.readAsDataURL(input.files[0]);
         }
     }

    $("#patient_pic").live("change",function(){
        readURL(this, "#preview_image")
    });

});
</script>

</head>

<body>
<form name="" method="post" action="#" class="feedback-form-1">
    <fieldset>
        <div class="input-file-row-1">
            <div class="upload-file-container">
                <img id="preview_image" src="#" alt="" />
                <div class="upload-file-container-text">
                    <div class = 'one_opacity_0'>
                        <input type="file" id="patient_pic" label = "add" />
                    </div>
                    <span> Add Photo </span>
                </div>
            </div>
        </div>
    </fieldset>
</form>
</body>

</html>

UPDATE: I think I am on the right track. I am close but I don't know what data to send from the jQuery. I added a php scrit and its getting a call back as success but I am not sending the right var. I think if I just send the right val I can get it. CODE:

<script>
$( document ).ready(function() {

   function readURL(input, target) {
        if (input.files && input.files[0]) {
            var reader = new FileReader();
            var image_target = $(target);
            reader.onload = function (e) {
                image_target.attr('src', e.target.result).show();


                 $.ajax({
            type:'POST',
            url: 'theUpload.php',
            data: input.files[0],
            success:function(data){
                console.log("success");
                console.log(data);
                alert(data);
            },
            error: function(data){
                console.log("error");
                console.log(data);
            }
        });



            };
            reader.readAsDataURL(input.files[0]);








         }
     }

    $("#patient_pic").live("change",function(){
        readURL(this, "#preview_image")
    });

});
</script>

3条回答
何必那么认真
2楼-- · 2020-03-23 18:38

You need a server side language to store the uploaded image to specified folder. PHP is one of them, so I highly recommend to take a look of it.

AJAX is just for executing server side calls without refreshing the page.

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做自己的国王
3楼-- · 2020-03-23 18:44

Try this one.

Script:

function uploadFile(){
  var input = document.getElementById("file");
  file = input.files[0];
  if(file != undefined){
    formData= new FormData();
    if(!!file.type.match(/image.*/)){
      formData.append("image", file);
      $.ajax({
        url: "upload.php",
        type: "POST",
        data: formData,
        processData: false,
        contentType: false,
        success: function(data){
            alert('success');
        }
      });
    }else{
      alert('Not a valid image!');
    }
  }else{
    alert('Input something!');
  }
}

HTML:

<html>
<head>
    <script src="http://code.jquery.com/jquery-latest.min.js" type="text/javascript"></script>    
</head>
<body>
    <input type="file" id="file" />
    <button onclick="uploadFile();">Upload</button>
</body>
</html>

upload.php:

<?php
$dir = "image/";
move_uploaded_file($_FILES["image"]["tmp_name"], $dir. $_FILES["image"]["name"]);
?>
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家丑人穷心不美
4楼-- · 2020-03-23 18:56

John's answer is good but file = input.files[0] doesn't work for me

file = input[0].files[0]

works.

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