POST不會在php中插入值

-1

美好的一天！我試圖將一些數據保存到我的數據庫中，但它不會工作，但不會顯示任何類型的錯誤。請告訴我我做錯了什麼。我試圖找到這個錯誤，但我認爲另一雙眼睛和大腦可以真正幫助我。謝謝！ :)POST不會在php中插入值

這是我的代碼

<?php 
$user_name = "root"; 
$password = ""; 
$database = "db_tocode"; 
$server = "127.0.0.1"; 
$content = "content_temp1"; 

$uname = $_GET['username']; 

$db_handle = mysql_connect($server, $user_name, $password); 
$db_found = mysql_select_db($database, $db_handle); 

$header = $_POST['header']; 
$title1 = $_POST['title1']; 
$content1 = $_POST['content1']; 
$title2 = $_POST['title2']; 
$content2 = $_POST['content2']; 
$title3 = $_POST['title3']; 
$content3 = $_POST['content3']; 
$title4 = $_POST['title4']; 
$content4 = $_POST['content4']; 
$title5 = $_POST['title5']; 
$content5 = $_POST['content5']; 
$title6 = $_POST['title6']; 
$content6 = $_POST['content6']; 
$title7 = $_POST['title7']; 
$content7 = $_POST['content7']; 
$title8 = $_POST['title8']; 
$content8 = $_POST['content8']; 
$title9 = $_POST['title9']; 
$content9 = $_POST['content9']; 
$title10 = $_POST['title10']; 
$content10 = $_POST['content10']; 
$content11 = $_POST['content11']; 

if ($db_found) { 

    $SQL="INSERT INTO $content (user_uname, header, title1, content1, title2, content2,     title3,content3,title4, content4, title5, content5, title6, content6, title7, content7, title8, content8, title9, content9, title10, content10, content11) VALUES('$uname',    ' $header', '$title1', '$content1', '$title2', '$content2', '$title3', '$content3', '$title4', '$content4', '$title5', '$content5', '$title6', '$content6', '$title7', '$content7', '$title8', '$content8', '$title9', '$content9', '$title10', '$content10', '$content11')"; 

header('Location:1stTemplate/index.php'); 

} 
else { 

print "Database NOT Found "; 
mysql_close($db_handle); 

} 

?>

來源

2014-01-28 bjmonts

**危險**：您正在使用[an **過時的**數據庫API]（http://stackoverflow.com/q/12859942/19068），並應使用[現代替換]（http：// php。淨/手動/ EN/mysqlinfo.api.choosing.php）。你也**易受[SQL注入攻擊]（http://bobby-tables.com/）**，現代的API會使[防禦]更容易（http://stackoverflow.com/questions/60174/best-way-to-prevent-sql-injection-in-php）自己從。 – Quentin

echo $ sql並在phpmyadmin中手動觸發該查詢並檢查是否有任何錯誤 –

@krishna - 表名是'content_temp1'。 '$'表示PHP中變量的開始。 – Quentin

執行查詢PLZ

$query = mysql_query($sql);

並且使用的是本地主機，以便您的服務器名稱將是

$server = "localhost";

和PLZ使用mysqli不是mysql

來源

2014-01-28 11:42:19

非常感謝。哈哈。我早就知道了！謝謝您的幫助！！ – bjmonts

歡迎您:-) –

POST不會在php中插入值

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