2017-06-15 89 views

回答

0

這裏是使用jquery ajax進行文件上傳的代碼。 HTML和jQuery代碼:

<form name="fileupload" id="fileupload" onsubmit="return submitFile()"> <input type="file" id="files" name="files[]" multiple><br/> <input type="submit" value="Submit" /> </form> <script type="text/javascript"> function submitFile() { var formData = new FormData($("#fileupload")[0]); $.ajax({ url: "ajax-upload.php", type: "post", data: formData, dataType: "json", processData: false, contentType: false, success: function (response) { $(response.filenames).each(function(index, value){ console.log("File Name " + value); }); } }); return false; } </script>

這裏是PHP代碼上傳多個文件。

$filenames = array(); for ($cnt = 0; $cnt < count($_FILES['files']["name"]); $cnt++) { if (move_uploaded_file($_FILES['files']["tmp_name"][$cnt], "images/" . $_FILES['files']["name"][$cnt])) { $filenames[] = $_FILES['files']["name"][$cnt]; } } echo json_encode(array("filenames" => $filenames));

相關問題