我有麻烦试图用Plupload保存上传文件的文件名。这里是Plupload部分:plupload文件名到数据库问题
echo<<<_END
<div id="container">
<div id="filelist">No runtime found.</div>
<br />
<a id="pickfiles" href="javascript:;">[Select file]</a>
<a id="uploadfiles" href="javascript:;">[Upload it]</a>
</div>
<script type="text/javascript">
// Custom example logic
function $(id) {
return document.getElementById(id);
}
var uploader = new plupload.Uploader({
runtimes : 'gears,html5,flash,silverlight,browserplus',browse_button : 'pickfiles',container: 'container',max_file_size : '1mb',
url : 'upload.php',
resize : {width : 320, height : 240, quality : 90},
flash_swf_url : 'plupload/js/plupload.flash.swf',
silverlight_xap_url : 'plupload/js/plupload.silverlight.xap',
filters : [{title : "Image files", extensions : "jpg,jpeg,gif,png"}]
});
uploader.bind('Init', function(up, params) {
$('filelist').innerHTML = "<div>Current runtime: " + params.runtime + "</div>";
});
uploader.bind('FilesAdded', function(up, files) {
for (var i in files) {
$('filelist').innerHTML += '<div id="' + files[i].id + '">' + files[i].name + ' (' + plupload.formatSize(files[i].size) + ') <b></b></div>';
}});
uploader.bind('UploadProgress', function(up, file) {
$(file.id).getElementsByTagName('b')[0].innerHTML = '<span>' + file.percent + "% </span>";
});
$('uploadfiles').onclick = function() {
uploader.start();
return false;
};
uploader.init();
</script>
_END;
但我不知道如何插入文件名到我的数据库:
$post_img = file??
$topicid = mysql_insert_id();
$sql = "INSERT INTO
posts(post_content,
post_date,
post_topic,
post_by,
post_img
)
VALUES
(
'" . mysql_real_escape_string($_POST['post_content']) . "',
NOW(),
" . $topicid . ",
" . $_SESSION['userid'] . ",
" . $post_img . ")";
$result = mysql_query($sql);
if(!$result) {
//something went wrong, display the error
echo 'An error occured while inserting your post. Please try again later.<br /><br />' . mysql_error();
$sql = "ROLLBACK;";
$result = mysql_query($sql);
}
我得到的错误“,而将你的文章时出错。请稍后再试'
什么mysql_error()或mysql_errno更好()返回? –
列计数与第1行的值计数不匹配。我将值更改为$ _FILES ['$ fileName'],但仍然是同样的问题。谢谢 – cbarg
但无论如何,如果我从Insert Into中取出post_img,请创建一个变量$ post_img = $ _FILES ['$ fileName'];并在稍后回显$ post_img,它不会给出任何输出: -/ – cbarg