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我想用ajax上传文件。通过ajax上传文件
form enctype="myltipart/form-data" id="pastTest-form" method="POST" action="upload.php">
<div class="feedback-form-inputs col-5">
<div class="input-field">
<select id="type" required>
<option value="quiz">Quiz</option>
<option value="midterm">Midterm</option>
<option value="final">Final</option>
</select>
</div>
<div class="input-field">
<input type="text" id="professor"/>
</div>
<div class="input-field">
<input type="text" id="name"/>
</div>
<div class="input-field">
<input type="file" id="uploaded_file" name="file" accept="" required />
</div>
</div><!-- END feedback-form-inputs -->
<div class="clear"></div>
<input type="submit" value="submit" onclick="submit() />
<div id="upload-status"> </div>
</form>
我的打开ajax的函数在外部文件中。
function addPastTest1(cid){
// form variables
var type = _("type").value;
var professor = _("professor").value;
var name = _("name").value;
var fileSelect = _('uploaded_file');
var status = _('upload-status');
event.preventDefault();
// Update status text.
status.innerHTML = 'Uploading...';
// Get the selected files from the input.
var file = fileSelect.files[0];
var FileName = file.name;
var FileSize = file.size;
var allowed = ["msword", "pdf","pages"];
var found = false;
// check if the extension of the file match the allowed ones
allowed.forEach(function(extension) {
if (file.type.match('application/'+extension)) {
found = true;
}
})
if (FileSize >10204){
status.innerHtml = 'File must be less than 1mb in size';
}
if (found==true){
// Create a new FormData object.
var formData = new FormData();
// Add the file to the request.
formData.append('file', file, FileName);
// Set up the request.
var ajax = ajaxObj("POST", "ajaxResponse.php");
ajax.onreadystatechange = function(){
if (ajaxReturn(ajax)==true){
if (ajax.responseText=='failed'){
status.innerHtml = "failed to upload file";
}
else {
status.innerHtml = 'uploaded';
alert(ajax.responseText);
}
}
}
ajax.send(formData); //ajax.send("f="+formData+"&t="+type+"&p="+professor+"&n="+name+"&cid="+cid+"&fn="+FileName);
}
}
所以我发送formData到php。但此时我无法从表单数据中提取文件并将其上传到服务器。 这里是我的PHP
// Ajax calls this code to add a past test
if (isset($_FILES['file']){
$file = $_FILES['file'];
$path = 'files/'.$type.'/'.$fileName;
$moveResult = move_uploaded_file($file, $path);
if ($moveResult != true) {
echo "ERROR: File not uploaded. Try again.";
//unlink($fileTmpLoc); // Remove the uploaded file from the PHP temp folder
exit();
}
$path = 'files/'.$type.'/'.$fileName;
$sql = "INSERT into past_papers VALUES ('$name', '$type', '$cid', '$professor','$path')";
$query = mysqli_query($db_conx,$sql);
if (mysqli_affected_rows($db_conx)>0){
echo "success";
exit();
}
else {
echo "failed sql";
exit();
}
}
?>
我也想抓住与文件的输入和共同处理它们。上传文件,并将输入插入数据库。
您应该使用搜索功能,吨后都与此有关,例如 HTTP ://stackoverflow.com/questions/166221/how-can-i-upload-files-asynchr rq = 1 也 http://stackoverflow.com/search?q=php+ajax+upload –
您的上传处理是完全无效的。你需要检查'$ _FILES ['file'] ['error']'** FIRST **,以查看上传是否被巧妙执行并成功。你只是假设什么都不会出错,这是完全错误的。 –