图片上传到mysql数据库无法正常工作无法正常、图片上传、数据库、工作

2023-09-10 21:56:40 作者：爱或者不爱都抵不过相爱

我加载使用jQuery FileDialog的图像文件。然后发送一个Ajax请求的PHP文件将图像上传到数据库中。这里被我发送Ajax请求

I am loading an image file using jquery filedialog. And then sending an ajax request to a php file to upload the image to database. Here is my ajax request being sent

$("#imgfile").on("change",function(e){
                    var file_data = $('#imgfile').prop('files')[0]; 
                e.preventDefault();
                var form_data = new FormData();                  
                form_data.append('file', file_data);
                //alert(form_data); 
                $.ajax({
            url: 'changeProfileImg.php', // point to server-side PHP script 
            contentType: "image/jpeg",  // what to expect back from the PHP script, if anything
            cache: false,
            dataType: "image",
            processData: false,
            data: form_data,                         
            type: 'post',
            success: function(php_script_response){
                alert(php_script_response); // display response from the PHP script, if any
            }
            }); 
                });

下面是我的PHP code

Here is my php code

include_once 'dbconnect.php';
    if ( 0 < $_FILES['file']['error'] ) {
        echo 'Error: ' . $_FILES['file']['error'] . '<br>';
    }
    else {
        //move_uploaded_file($_FILES['file']['tmp_name'], 'uploads/' . $_FILES['file']['name']);
        $file=$_FILES['file']['name'];
        $id=$_SESSION['id'];
        $sql="UPDATE  mainadmin set photo='$file' where id='$id'";
        mysql_query($sql);
        echo 'success';
        $_SESSION['photo']=$file;
    }

但我看到一些garbase的事情被插入到数据库中不显示图像。我怎么能做到这一点的工作。

But i see some garbase thing is being inserted into database not showing the image. How can i do it working.

推荐答案

以下网址，code见我认为这是对您有所帮助。

See below URL and code I think this is helpful to you.

Simple jQuery的进度条百分比

这是我的测试code文件上传有进度条百分比

This is my tested code file upload with progress bar percentage

但你可以管理或修改按您的要求。

But u can manage or modify as per Your requirement.

<!doctype html>
<head>
<title>File Upload Progress Demo #1</title>
<style>
body { padding: 30px }
form { display: block; margin: 20px auto; background: #eee; border-radius: 10px; padding: 15px }

.progress { position:relative; width:400px; border: 1px solid #ddd; padding: 1px; border-radius: 3px; }
.bar { background-color: #B4F5B4; width:0%; height:20px; border-radius: 3px; }
.percent { position:absolute; display:inline-block; top:3px; left:48%; }
</style>
</head>
<body>
    <h1>File Upload Progress Demo #1</h1>
    <code>&lt;input type="file" name="myfile"></code>
        <form action="upload.php" method="post" enctype="multipart/form-data">
        <input type="file" name="uploadedfile"><br>
        <input type="submit" value="Upload File to Server">
    </form>

    <div class="progress">
        <div class="bar"></div >
        <div class="percent">0%</div >
    </div>

    <div id="status"></div>

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.7/jquery.js"></script>
<script src="http://malsup.github.com/jquery.form.js"></script>
<script>
(function() {

var bar = $('.bar');
var percent = $('.percent');
var status = $('#status');

$('form').ajaxForm({
    beforeSend: function() {
        status.empty();
        var percentVal = '0%';
        bar.width(percentVal)
        percent.html(percentVal);
    },
    uploadProgress: function(event, position, total, percentComplete) {
        var percentVal = percentComplete + '%';
        bar.width(percentVal)
        percent.html(percentVal);
    },
    complete: function(xhr) {
     bar.width("100%");
    percent.html("100%");
        status.html(xhr.responseText);
    }
}); 

})();       
</script>

</body>
</html>

我的PHP code

my php code

<?php
$target_path = "uploads/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']); 

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename( $_FILES['uploadedfile']['name']). 
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
}
?>

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