Может ли кто-либо предоставить полный код сервера ajax и (php или asp) для загрузки изображения непосредственно с простого html на онлайн-сервер. На самом деле я пытаюсь загрузить изображение на сервере с помощью phonegap cli
<form id="uploadaadhar" action="http://server_url" method="post"> <input type="hidden" name="userid" > <br><label>Aadhar Number</label> <br><input type="number" name="aadhar" class="form-control" placeholder="XXXX XXXX XXXX"> <br><label>Choose File</label> <br><input name="files[]" class="form-control" type="file" /> <br><label>PAN Number</label> <br><input type="number" name="pan" class="form-control" placeholder="XXXX XXXX XXXX"> <br><span class="ak-text-indigo">Leave Pan Card Field Blank if you do not want to update your pan card</span> <br><br> <input type="submit" value="Upload" class="ak-btn ak-blue"> </form> <script> $("[type='hidden']").val(localStorage.getItem("userid")); $("#uploadaadhar").ajaxForm(function(data){ if(data !== ""){ $("[type='submit']").val(data); } }); </script> Вот пример демонстрации для загрузки изображения. Здесь data – это изображение и другие данные, которые вы хотите отправить с помощью post . url , где вы хотите опубликовать данные.
<html> <form> <input id="file-input" type="file" name="files" accept="image/*" /> <button type="button" onclick="uploadImage()">upload</button> </form> </html>
и ajax-функция
function uploadImage(){ url = "your web url where u want to upload image"; var formdata = new FormData(); formdata.append( 'files', $('#file-input')[0].files[0]); $.ajax({ type: "POST", url: url, data: formdata, processData: false, contentType: false, type: 'POST', success: function(msg){ // on success }, error: function(){ alert("failure"); }, async: false }); }
<form id="uploadaadhar" action= "http://app.365valueservices.com/365APP/settings.php" method="post"> <label>Aadhar Number</label> <input type="number" name="aadhar" class="form-control" placeholder="XXXX XXXX XXXX">`enter code here` <label>Choose File</label> <input name="files[]" class="form-control" type="file" /> <input type="submit" value="Upload" class="ak-btn ak-blue"> </form>
Javascript
<script> $("#uploadaadhar").ajaxForm(function(data){ if(data !== ""){ $("[type='submit']").val(data); } }); </script>
if(isset($_FILES['files'])) { if(is_array($_FILES)) { foreach ($_FILES['files']['name'] as $name => $value) { $file_name = explode(".", $_FILES['files']['name'][$name]); $allowed_ext = array("jpg","JPG", "jpeg", "JPEG", "png", "PNG", "gif","pdf","docx","doc"); if(in_array($file_name[1], $allowed_ext)) { $new_name = md5(rand()) . '.' . $file_name[1]; $sourcePath = $_FILES['files']['tmp_name'][$name]; $targetPath = "../httpdocs/docs/".$new_name; //$targetPath = "upload/".$new_name; move_uploaded_file($sourcePath, $targetPath); $filepath = "../docs/".$new_name; $sql = "update usertbl set aadhar = '".$_POST['aadhar']."' where userid = ".$_POST['userid']; sqlsrv_query($conn,$sql); if($_POST['pan'] != ""){ $sql2 = "update usertbl set panno = '".$_POST['pan']."' where userid = ".$_POST['userid']; sqlsrv_query($conn,$sql2); } $stmt = sqlsrv_query($conn,"select * from user_meta where userid = ".$_POST['userid']." and metakey = 'aadharpath'"); $row_count = sqlsrv_has_rows($stmt); if($row_count === false){ sqlsrv_query($conn,"insert into user_meta(userid,metakey,metavalue,Active) values (".$_POST['userid'].",'aadharpath','".$filepath."',1)"); } else { sqlsrv_query($conn,"update user_meta set metavalue = '".$filepath."' where metakey='aadharpath' and userid=".$_POST['userid']); } sqlsrv_query($conn,"insert into useractivity values(".$_POST['userid'].",'You had updated your aadhar details','" .date("dmY H:i:s"). "',1)"); echo "Updated Successfully"; } } } }
//I had tried $_FILES in var_dump and got this array(1) { ["files"]=> array(5) { ["name"]=> array(1) { [0]=> string(7) "bot.jpg" } ["type"]=> array(1) { [0]=> string(10) "image/jpeg" } ["tmp_name"]=> array(1) { [0]=> string(24) "C:\xampp\tmp\php1069.tmp" } ["error"]=> array(1) { [0]=> int(0) } ["size"]=> array(1) { [0]=> int(40287) } } }