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我一直在努力工作幾天,才能上傳到我的網站的管理部分,並且幾乎在那裏......! 我在我的數據庫中有一個表,名爲測試4個字段 - id(int),title(varchar),desc(varchar)和photo(varchar) - 照片字段表示服務器上圖像的來源。 我的代碼是:重複空行在文件上傳時輸入到mysql
<?php include 'dbc.php'; page_protect();
if(!checkAdmin()) {header("Location: login.php");
exit();
}
$host = $_SERVER['HTTP_HOST'];
$host_upper = strtoupper($host);
$login_path = @ereg_replace('admin','',dirname($_SERVER['PHP_SELF']));
$path = rtrim($login_path, '/\\');
foreach($_GET as $key => $value) {
$get[$key] = filter($value);
}
foreach($_POST as $key => $value) {
$post[$key] = filter($value);
}
?>
<?php
$target = "images/test/";
$target = $target . basename($_FILES['photo']['name']);
$title = mysql_real_escape_string($_POST['title']);
$desc = mysql_real_escape_string($_POST['desc']);
$pic = "images/test/" .(mysql_real_escape_string($_FILES['photo']['name']));
mysql_query("INSERT INTO `test` (`title`, `desc`, `photo`) VALUES ('$title', '$desc', '$pic')") ;
if(move_uploaded_file($_FILES['photo']['tmp_name'], $target))
{
echo "The file ". basename($_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
echo "Sorry, there was a problem uploading your file.";
}
?>
<form enctype="multipart/form-data" action="uploader.php" method="POST">
Title: <input type="text" name="title"><br>
Description: <input type="text" name = "desc"><br>
Photo: <input type="file" name="photo"><br>
<input type="submit" value="Add">
</form>
出於某種原因,當該行進入MySQL是插入一個重複的空行,以便該表的樣子:
ID Title Desc Photo
15 images/test/
16 test title test description images/test/test1.jpg
沒有任何理由,這是從發生上面的代碼 - 它相當簡陋,但考慮到它的痛苦和鬥爭,以獲得這個工作,我真的不能再次面對開始!
在此先感謝您的幫助。
JD
嗨 - 試過,但現在它似乎並沒有做任何事情 - 無論是張貼文件或進入數據庫表......? – JD2011
我的不好,你需要檢查是否上傳文件沒有得到錯誤信息。如果文件尚未上傳,我添加了一些調試信息。如果這段代碼無法正常工作 - 請告訴我輸出問題,我會盡力解決問題出在哪裏 – Greenisha
嗨,謝謝你回到我身邊。輸出是 抱歉,上傳您的file.array(1){[「photo」] => array(5){[「name」] => string(12)「IMG_0161.jpg」[「type 「] => string(10)」image/jpeg「[」tmp_name「] => string(14)」/ tmp/phpM9Qa9k「[」error「] => int(0)[」size「] => int 35516)}} – JD2011