當我的形式提交(通過Ajax)提交,我得到了以下錯誤消息:漸漸空虛查詢PHP錯誤時表單通過Ajax
[17-Oct-2012 11:46:29] PHP Warning: mysqli_query() [<a href='function.mysqli-query'>function.mysqli-query</a>]: Empty query in /home1/xenongro/public_html/testing/enrolment/thanks.php on line 32
我有一個懷疑,它的東西做if/else語句,但不確定實際問題是什麼。
任何人都可以幫忙嗎?
<?php
$firstname = htmlspecialchars(trim($_POST['fname']));
$lastname = htmlspecialchars(trim($_POST['lname']));
$worktel = htmlspecialchars(trim($_POST['worktel']));
$dbc = mysqli_connect('localhost', 'xxxxx', '<xxxx>', 'xxxx')
or die ('Could not connect to MySQL server.');
if ($level != "IOSH Managing Safely"){
if ($funding == "Self Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
else if ($funding == "Employer Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
}
else if ($level == "IOSH Managing Safely"){
if ($funding == "Self Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
else if ($funding == "Employer Funding"){
$query = "INSERT INTO enrolments (fname, lname, worktel)" .
"VALUES ('$firstname', '$lastname', '$worktel')";
}
}
$result = mysqli_query($dbc, $query)
or die ('error querying database');
mysqli_close($dbc);
?>
意味着你的程序沒有進入任何條件塊! – Deepak
那麼$ level和$ funding的定義在哪裏? –
是否設置了$ _POST?您的AJAX請求是否實際使用'POST'方法? $資金和$等級來自哪裏(?),它們在您的示例代碼中無處定義(請始終打擾顯示所有相關片段)。爲什麼你不使用正確的數據庫轉義? – mario