php查詢alwaysfalse不管是什麼

Question

我試圖閱讀和做（合併）我在這裏找到的任何東西。但我還沒有找到解決方案。我使用的是wamp服務器，我有一個用戶表，其中有兩個用戶，一個用email和password作爲test和test1，不管我怎麼試，if語句總是返回false。

<?php 

$user = "root"; 
$pass = ""; 
$db = "testdb"; 

$db = new mysqli("localhost", $user, $pass, $db) or die("did not work"); 

echo "it connected"; 

$email = "test"; 
$pass1 = "test1"; 
$qry = 'SELECT * FROM user WHERE email = " '. $email .' " AND password = " '.$pass1.' " '; 

$result = mysqli_query($db, $qry) or die(" did not query"); 

$count = mysqli_num_rows($result); 

if($count > 0) 
    echo "  found user "; 
else 
    echo " did not find user or password"; 

?> 

我試圖以增加mysqli_num_rows但隨後它出來總是如此

Answer 1

我需要消除的空格中的變量

<?php 

$user = "root"; 
$pass = ""; 
$db = "testdb"; 

$db = new mysqli("localhost", $user, $pass, $db) or die("did not work"); 

echo "it connected"; 

$email = "test"; 
$pass1 = "test1"; 
$qry = 'SELECT * FROM user WHERE email = "'. $email .'" AND password = "'.$pass1.'"'; 

$result = mysqli_query($db, $qry) or die(" did not query"); 

$count = mysqli_num_rows($result); 

if($count > 0) 
    echo "  found user "; 
else 
    echo " did not find user or password"; 

?> 

Answer 2

您的查詢周圍的變量空間：

" '. $email .' " 

變化：

"'. $email .'" 

MySQL在搜索匹配時會逐字地佔用這些空格。

Answer 3

封裝如果您使用的mysqli類版本，那麼你應該象下面這樣使用：

<?php 

$user = "root"; 
$pass = ""; 
$db = "testdb"; 

$mysqli = new mysqli("localhost", $user, $pass, $db); 

$email = "test"; 
$pass1 = "test1"; 

$qry = sprintf('SELECT * FROM user WHERE email = "%s" AND password = "%s"',$email,$pass1); 
$result = $mysqli->query($qry); 
$count = $result->num_rows; 

if($count > 0) 
    echo "  found user "; 
else 
    echo " did not find user or password"; 

$mysqli->close(); 
?>