问：<input type="file" name="file" id="file" />
<div style="background:url() no-repeat">
<span id="uploaded_image" ><img src=" '.$row["carimg"].' " height="150" width="225" class="img-thumbnail" /></span></div>
这是ajax代码
<script>
$(document).ready(function(){
$(document).on('change', '#file', function(){
var name = document.getElementById("file").files[0]...
var form_data = new FormData();
var ext = name.split('.').pop().toLowerCase();
if(jQuery.inArray(ext, ['gif','png','jpg','jpeg']) == -1)
{
alert("Invalid Image File");
}
var oFReader = new FileReader();
oFReader.readAsDataURL(document.getEleme...
var f = document.getElementById("file").files[0]...
var fsize = f.size||f.fileSize;
if(fsize > 2000000)
{
alert("Image File Size is very big");
}
else
{
form_data.append("file", document.getElementById('file').files[0]...
$.ajax({
url:"up1.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>
这是PHP
<?php
//upload.php
$id = $_FILES["vid"];
if($_FILES["file"]["name"] != '')
{
$test = explode('.', $_FILES["file"]["name"]);
$ext = end($test);
$name = gen_random_string(6)."n" . '.' . $ext;
$location = 'assets/img/cars/' . $name;
move_uploaded_file($_FILES["file"]["tmp_name"], $location);
echo $id;
echo '<img src="'.$location.'" height="150" width="225" class="img-thumbnail" />';
$sql = "UPDATE vehicles SET carimg='".$location."' WHERE vid='".$id."'";
}
mysqli_query($connect, $sql);
?>
如何以这种php ajax形式发送多个数据，这个只能发送一个数据的数据请帮助
答：您可以在formData数据中附加/添加与文件相同的更多数据，例如：
-form_data.append（document.getelementById（someid）.vlaue（））或
form_data.append（$（'＃someid或任何其他选择器.value）;
或者您可以使用：$（'form'）。serialize（）; 如果数据格式相同
