0
需要幫助我的mysqli聲明。我得到的頁面上沒有錯誤,但我的if語句沒有任何打印出任何東西....Mysqli聲明:空白結果
//$dbConnection works.
$extensionAdd_query_domain_extension = mysqli_query($dbConnection, "
SELECT *
FROM domain
WHERE extensions='$extensionAdd_extension'
LIMIT 1");
$extensionAdd_query_domain_row = mysqli_fetch_array($extensionAdd_query_domain_extension, MYSQLI_NUM);
$extensionAdd_query_domain_id = sanitizeInt($extensionAdd_query_domain_row[0],$dbConnection);
$extensionAdd_query_domain_votes = sanitizeInt($extensionAdd_query_domain_row[2],$dbConnection);
//If domain id found in Db update votes otherwise add to list
if($extensionAdd_query_domain_id!='' && $extensionAdd_query_domain_votes!='')
{
echo 'Found in DB. Update Votes.'.$extensionAdd_extension;
}
else
{
echo 'Not Found in DB. Enter in DB'.$extensionAdd_extension;
}
是否在頁面上輸出任何內容?該頁面投擲500? – chris85
如果回聲沒有執行,那麼在代碼達到該點之前,某些內容正在殺死您的腳本。腳本無法輸出任何內容,因爲這兩個代碼路徑都會產生SOME輸出。 –
確保你已啓用[錯誤報告](http://stackoverflow.com/questions/845021/how-to-get-useful-error-messages-in-php)。 – Barmar