我正在尝试使用ajax和php上传文件而不刷新页面并提交。
如果我只执行preg_match
,我的代码就可以运行并提醒有效消息,但是当我添加需要使用$_FILES[$filrec]["tmp_name"]
的其余验证时,它不会提醒我有效消息。
这里有什么问题?如果没有使用以下方法提交表单,是不是可以上传文件?
有许多不同的建议和示例以及更复杂的javascript或jquery方法,但我正在尝试简单地使用ajax并将其余部分留给PHP。这可能与我的轰鸣声ajax功能?
Javascript :
var fileselected = $("#browse").val().replace(/C:\\fakepath\\/i, '');
setTimeout(function() {
$.ajax({
type: 'POST',
url: "ajax/extval.php",
data: {fileprs: fileselected},
dataType: 'json',
cache: false,
success: function(resuval) {
// file validation result
if (resuval === "valid"){
alert ("valid")
PHP :
<form id="upload" method="post" class="<?php echo $orvalue."<separator>".$user_id ?>" action="" enctype="multipart/form-data">
<div id="formcontent">
<label class="required" for="unitprice" title="Unit price"><input type="text" id="unitprice" name="unitprice" />Unit price</label>
<label class="required" for="qty" title="How many pcs"><input type="text" id="qty" name="qty" />Quanity</label>
<label class="required" for="express" title="Express within China"><input type="text" id="express" name="express" />Express</label>
<label class="required" for="linkURL" title="Insert a full URL http:\\"><input type="text" id="linkURL" name="linkURL" />Link</label>
<label for="yourdesc" title="Describe your need clearly"><textarea id="yourdesc" name="yourdesc"></textarea>Description<li><font size="-2">You can type 400 letters only, you typed :</li><li id="lettercounter"></li>letters</font></label>
<label for="formsubmit" class="nocontent"><input type="button" id="submitButton" href="#" class="progress-button" value="Add to order" /><strong>Note:</strong> Items marked <img src="../../images/required.jpg" alt="Required marker" width="20" height="20" /> are required fields</label>
</div>
</form>
PHP:
$filrec =mysql_real_escape_string($_POST['fileprs']);
if(preg_match("/\.(gif|png|jpg|JPG|jpeg|bmp|BMP)$/", $filrec))
{
$fileType = exif_imagetype($_FILES[$filrec]["tmp_name"]);
$allowed = array(IMAGETYPE_GIF, IMAGETYPE_JPEG, IMAGETYPE_PNG);
$allin = "valid";
echo json_encode($allin);
}
赞赏
答案 0 :(得分:0)
您可以使用以下PHP代码来获取从Ajax收到的文件:
$ data = split(“,”,file_get_contents('php:// input'));
$ img_data = base64_decode($ data [1]);
file_put_contents('uploads /'。$ _SERVER ['HTTP_X_FILENAME'], $ img_data);
答案 1 :(得分:0)
您可以使用Follow Ajax POST Request这将为您提供帮助
<script>
$(document.body).on('click','.postDefects',function(){
var form_data = new FormData();
var defect = $(this).closest('tr').find( "input[name='defect_id']" ).val();
var txt_defect=$("#text_defect").val();
var upload_defect = document.getElementById("upload_defect").files[0];
form_data.append("upload_defect",upload_defect);
form_data.append("defect_id",defect_id);
form_data.append("txt_defect",txt_defect);
console.log(form_data);
$.ajax({
url:"postsample_defects.php",
method:"POST",
data: form_data,
contentType: false,
cache: false,
processData: false,
beforeSend:function(){
$('#uploaded_image').html("<label class='text-success'>Image Uploading..</label>");
},
success:function(data)
{
$('#uploaded_image').html(data);
}
});
});
</script>