我有一个更新个人资料页面,您可以从文件夹中选择一个图像并将其上传到我的MySQL数据库。看看这段代码:
$image ="";
$image_size;
$image_name;
$uploadable = false;
if(isset($_FILES['image']['tmp_name'])){
$image= file_get_contents($_FILES['image']['tmp_name']);
$image_name=$_FILES['image']['name'];
$image_size=getimagesize($_FILES['image']['tmp_name']);
if($image_size==false){
$_SESSION["FLAW"] = '4';
header("Location:editprofile.php");
echo "nope";
$uploadable = false;
}else{
$uploadable = true;
}
}else{}
AND:
if($row[6] == $id && $uploadable == true){
if(($fetchedemail == $emailentered) || ($fetchedemail != $emailentered && ($new == true)) ){
$query = mysql_query("UPDATE `allaccounts` SET `realname` = '$realname' WHERE `allaccounts`.`id` = $id;");
$query2 = mysql_query("UPDATE `allaccounts` SET `twitter` = '$twitter' WHERE `allaccounts`.`id` = $id;");
$query1 = mysql_query("UPDATE `allaccounts` SET `email` = '$emailentered' WHERE `allaccounts`.`id` = $id;");
$query3 = mysql_query("UPDATE `allaccounts` SET `facebook` = '$facebook' WHERE `allaccounts`.`id` = $id;");
$query4 = mysql_query("UPDATE `allaccounts` SET `description` = '$description' WHERE `allaccounts`.`id` = $id;");
$query5 = mysql_query("UPDATE `allaccounts` SET `image` = '$image' WHERE `allaccounts`.`id` = $id;");
$query6 = mysql_query("UPDATE `allaccounts` SET `filename` = '$image_name' WHERE `allaccounts`.`id` = $id;");
// header("Location:Myprofile.php");
$_SESSION["FLAW"] = '';
break;
}else{
//header("Location:editprofile.php");
$_SESSION["FLAW"] = '3';
break;
}
这是我用来从我的HTML页面上的表单中获取图像然后更新我的MYSQL数据库的图像列的PHP代码,这是一个BLOB类型。我去检查数据库,但数据不存在,我不知道为什么。如果我在记事本中打开我的图像,看起来它的所有数据都带有奇怪的符号,这很好。但如果我回显$ image它看起来不同,并且有很多带有问号符号的钻石。这可能表明出现了问题。任何人都可以看到我做错了什么不能正确获取代码或正确上传