好吧,我的發佈腳本有一些小問題。除了當我輸入一個像撇號的短語時,一切都很好,例如「這是我的標題」不會發佈,請將「我的標題」添加到後。不知道這是爲什麼。也許條形碼標籤我真的不知道,我該怎麼辦?當它不起作用,我得到了底部錯誤信息PHP表單/發佈錯誤
} else {
if(isset($_POST['what'])&&isset($_POST['when'])&&isset($_POST['where'])&&isset($_POST['details'])&&isset($_POST['sponsored_by'])&&isset($_POST['collegeId'])){
$what = nl2br(htmlspecialchars(strip_tags(stripslashes(trim($_POST['what'])))));
$where = nl2br(htmlspecialchars(strip_tags(stripslashes(trim($_POST['where'])))));
$when = nl2br(htmlspecialchars(strip_tags(stripslashes(trim($_POST['when'])))));
$sponsored_by = nl2br(htmlspecialchars(strip_tags(stripslashes(trim($_POST['sponsored_by'])))));
$details = nl2br(htmlspecialchars(strip_tags(stripslashes(trim($_POST['details'])))));
$collegeId = intval($_POST["collegeId"]);
if(isset($_SESSION['username'])){
$username = htmlspecialchars(strip_tags(stripslashes(trim(($_SESSION['username'])))));
$query = "select id, Name from users where username='$username' and activated = 1";
$doQuery = mysql_query($query);
if(mysql_num_rows($doQuery)>0){
$results = mysql_fetch_array($doQuery);
$userName = $results['Name'];
$email = $username;
$id = $results['id'];
$query = "insert into events values(NULL,$id,$collegeId,'$what','$when','$where','$details','$sponsored_by',NOW())";
if(mysql_query($query)) header("Location: collegeInfo.php?college=$collegeId&message=added");
else echo "Failed to create new Event!".$query;
我認爲你應該開始閱讀一些關於編程的文章,做一些教程 - 也許玩一些框架來感受應用程序的工作方式。 SO不會很好地處理輔導 - 它更適合於更精確的問題。看看http://www.laravel.com/ – Michael