有條件的準備好的語句不切換

Question

我有以下PHP代碼，這是爲了在用戶名字段爲給定用戶名爲空的情況下插入數據，或者在用戶名存在時更新數據。此刻，插入先前工作正常，但它永遠不會切換到更新子句。

現在，插入子句無法識別我的測試變量，因爲沒有明顯的原因。我得到的錯誤是：

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given 

和

Notice: Undefined variable: checkUsername 

這是相當最近。

if($cmd=="submitinfo"){ 

$usernameQuery = "select username from USERS where username = $username"; 

$xblah = $con->query($usernameQuery); 
    while ($row = mysqli_fetch_assoc($xblah)) 
    { 
    $checkUsername = $row['username']; 

    } 

if ($checkUsername == null) { 

$userQuery = "INSERT INTO USERS VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?)"; 
if ($userInfo = $con->prepare($userQuery)) { 
    $userInfo->bind_param("ssssssssssssssssssss", $username, $firstname, $lastname, $flaggedauctions, $lastauction, $street1, $city1, $postcode1, $street2, $city2, $postcode2, $phone, $mobilephone, $fax, $email, $website, $bank, $banknumber, $accountnumber, $comments); 
    $userInfo->execute(); 
    $userInfo->close(); 
    echo "true"; 
} else { 
echo "false"; 
} 
print_r($con->error); 
} 

else if ($checkUsername == $username) { 

$userQuery = "UPDATE USERS SET firstname = ?, lastname = ?, flaggedauctions = ?, lastauction = ?, street1 = ?, city1 = ?, postcode1 = ?, street2 = ?, city2 = ?, postcode2 = ?, phone = ?, mobilephone = ?, fax = ?, email = ?, website = ?, bank = ?, banknumber = ?, accoutnumber = ? WHERE username = ?"; 
if ($userInfo = $con->prepare($userQuery)) { 
    $userInfo->bind_param("sssssssssssssssssss", $firstname, $lastname, $flaggedauctions, $lastauction, $street1, $city1, $postcode1, $street2, $city2, $postcode2, $phone, $mobilephone, $fax, $email, $website, $bank, $banknumber, $accountnumber, $username); 
    $userInfo->execute(); 
    $userInfo->close(); 
    echo "true"; 
} else { 
echo "false"; 
} 
print_r($con->error); 
} 
} 

什麼是做一個更新，或者根據$用戶名對用戶名字段相匹配的內容插入一個優選的方法是什麼？

Answer 1

Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given 

這是您的錯誤信息？我認爲你的問題是這樣的：

$xblah = $con->query($usernameQuery); 

在$ xblah上做一個var_dump。我懷疑你的查詢失敗。可能是因爲你沒有從你的輸入中提取$username。爲了愛的根，一定要逃脫它！ mysql_real_escape_string()會做你想做的。

Answer 2

如果你正在使用MySQL和用戶名是唯一的，你可以用這個方便的SQL：

INSERT INTO users (?, ?, ...) ON 
DUPLICATE KEY UPDATE firstname=?, 
lastname=?, ...) 

的完整文檔見http://dev.mysql.com/doc/refman/5.0/en/insert-on-duplicate.html。