C9 - 如何獲取查詢的結果以及如何獲取查詢中的行數？

Question

我寫了這個函數獲取用戶名並需要檢查用戶是否已經在數據庫中。

這是我的代碼：

$conn = new mysqli($servername, $username, $password, $database, $dbport); 
mysql_select_db("myDB",$conn); 
//$sql="CALL checkIfExsist(".$name.")"; 
$sql = "select * from Users where userName='".$name."' LIMIT 1;"; 
$myData = $conn-> query($sql); 
$rowc = mysqli_num_rows($myData); 
printf("Result set has %d rows.\n", $rowc); 
    if($rowc > 0) 
    { 
     echo "User ".$name." checked"; 

    } 
    else 
    echo "not exsist"; 

當我運行在終端這個查詢，我得到1個的結果，但在康壽我得到：「結果集有0行。」

什麼問題？

Answer 1

您應該使用mysqli_而不是mysql_由於mysql_已被棄用。

首先，我看到一個問題：（你混合mysql_和mysqli_）

你連接MySQL的：

mysql_select_db("myDB",$conn); 

更改爲：

$conn=mysqli_connect("localhost","my_user","my_password","my_db"); 
// Check connection 
if (mysqli_connect_errno()) 
    { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 

你應該a LSO使用預處理語句：

$stmt = $mysqli->prepare('SELECT * FROM Users WHERE userName = ?') 
$stmt->bind_param('s', $username); 
if(!$stmt->execute()){ 
trigger_error("there was an error....".$mysqli->error, E_USER_WARNING); 
} 
$stmt->store_result(); 

那麼當然你的if語句：

 if ($stmt->num_rows > 0){ 
     echo "User ".$name." checked"; 
    }else 
     echo"User doesn't exist" 

Answer 2

我認爲你需要改變你的連接代碼。請看看，並通過這種方式

<?php 

define("DB_HOST", "db_host"); 
define("DB_USER", "db_user"); 
define("DB_PASSWORD", "db_pass"); 
define("DB_DATABASE", "db_name"); 
$conn= mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD,DB_DATABASE); 

$sql = "select * from Users where userName='".$name."' LIMIT 1;"; 
$myData = $conn->query($sql); 
$rowc = mysqli_num_rows($myData); 
printf("Result set has %d rows.\n", $rowc); 
    if($rowc > 0) 
    { 
     echo "User ".$name." checked"; 

    } 
    else 
    echo "not exsist"; 

Answer 3

，您是否試圖$rowc = $myData->num_rows;