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我使用腳本輸出上傳的圖片預覽。它工作正常。我只想在一個div中顯示圖像,並在另一個div中顯示錯誤或成功按摩。有沒有機會做到這一點?這是代碼。ajax圖片上傳和預覽表格
的Java腳本
<script type="text/javascript">
$(document).ready(function(){
$('#photoimg').live('change', function(){
$("#preview").html('');
$("#preview").html('<img src="loader.gif" alt="Uploading...."/>');
$("#imageform").ajaxForm(
{
target: '#preview'
}).submit();
});
});
</script>
HTML代碼
<?php
include('db.php');
session_start();
$session_id='1'; // User login session value
?>
<div id="output"></div>
<form id="imageform" method="post" enctype="multipart/form-data" action='ajaximage.php'>
Upload image <input type="file" name="photoimg" id="photoimg" />
</form>
<div id='preview'>
</div>
PHP代碼
include('db.php');
session_start();
$session_id='1'; // User session id
$path = "uploads/";
$valid_formats = array("jpg", "png", "gif", "bmp","jpeg");
if(isset($_POST) and $_SERVER['REQUEST_METHOD'] == "POST")
{
$name = $_FILES['photoimg']['name'];
$size = $_FILES['photoimg']['size'];
if(strlen($name))
{
list($txt, $ext) = explode(".", $name);
if(in_array($ext,$valid_formats))
{
if($size<(1024*1024)) // Image size max 1 MB
{
$actual_image_name = time().$session_id.".".$ext;
$tmp = $_FILES['photoimg']['tmp_name'];
if(move_uploaded_file($tmp, $path.$actual_image_name))
{
mysql_query("UPDATE users SET profile_image='$actual_image_name' WHERE uid='$session_id'");
echo "<img src='uploads/".$actual_image_name."' class='preview'>";
echo "<span class=ok-msg">Image has been uploaded..!</span>";
}
else
echo "<span class=error-msg">failed<span>";
}
else
echo "<span class=error-msg">Image file size max 1 MB</span>";
}
else
echo "<span class=error-msg">Invalid file format..</span>";
}
else
echo "<span class=error-msg">Please select image..!</span>";
exit;
}
我喜歡來顯示所有的按摩(錯誤味精,OK-MSG)和div輸出和圖像在相同的地方div預覽。誰能告訴我如何做到這一點。提前致謝。
謝謝你的工作很棒 – maxlk