我需要什麼幫助? - 當我將圖像上載到數據庫時,我想將用戶的ID鏈接到我的SQL的正確字段中。不幸的是,當我上傳圖像時,沒有任何內容被輸入到ID字段中,因此似乎是它沒有正確捕獲它。使用MYSQL和PHP將會話ID鏈接到數據庫
所以分解:當用戶登錄時,他有一個唯一的ID,即管理員ID是1.當他在他的用戶面板上時,他點擊上傳第二張圖片:然後他被導向到這個表格。
一旦在窗體上,他將輸入一個描述,圖像,他的ID應該從_SESSION中採取。
如果需要更多信息,我很樂意多寫。
由於提前,
所以...繼承人的代碼:
// // FORM
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<form enctype="multipart/form-data" id="form1" name="form1" method="post" action="secondPic.php">
<p>
<label for="name1">Fav Location Name: </label>
<input type="text" name="name1" id="name1" />
</p>
<p>
<label for="photo1">Fav Location Photo: </label>
<input type="file" name="photo1"><br>
</p>
<p>
<label for="id">ID: <? echo $rows['id']; ?> </label>
<input name="id" type="hidden" id="id" value="<? echo $rows['id']; ?>">
</p>
<p>
<input type="submit" name="submit" id="submit" value="Submit" />
</p>
</form>
</body>
</html>
//輸入到數據庫中//
<?php
include "common.php";
$secondid = $_GET['id'];
DBConnect();
$Link = mysql_connect($Host, $User, $Password);
//This is the directory where images will be saved
$target = "second/";
$target = $target . basename($_FILES['photo1']['name']);
$favname = $_POST["name1"];
$pic2=($_FILES['photo1']['name']);
$id = $_POST["$id"];
$Query ="INSERT into $Table_2 values ('0', '$id', '$favname', '$pic2')";
if (mysql_db_query ($DBName, $Query, $Link)){
print ("A record was created <br><a href=index.php> return to index </a>\n");
// Connects to your Database
//mysql_connect("localhost", "jonathon_admin", "hello123") or die(mysql_error()) ;
//mysql_select_db("jonathon_admin1") or die(mysql_error()) ;
//Writes the photo to the server
if(move_uploaded_file($_FILES['photo1']['tmp_name'], $target))
{
//Tells you if its all ok
echo "The file ". basename($_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory";
}
else {
//Gives and error if its not
echo "Sorry, there was a problem uploading your file.";
}
} else {
print (" - Your Record was not created");
}
mysql_close($Link);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
</body>
</html>
下面是我輸入數據到DB時的表格:
**s_id id favname pic2
1 0 testing the db piccy.png**
你有沒有試過測試過你的任何代碼?你真的應該試着弄清楚代碼到底有多遠,哪裏是..等等,然後發佈它,而不是發佈整個事情。 – Dave 2012-01-27 17:05:03
你在使用會話嗎?將ID存儲在一個會話中,並在您的帖子中使用它 – Drewdin 2012-01-27 17:12:31
我敢打賭,您的問題是在'INSERT'查詢的第一個字段處出現'0'。如果這對應於PK,則不會插入任何後續行。你應該指定一個沒有PK的字段列表,或者提供'NULL'(* not *''NULL'')來解決這個問題。 – DaveRandom 2012-01-27 17:13:53