“从表中选择电子邮件WHERE name = $ name”;用PHP

时间:2016-05-11 19:41:54

标签: php mysql

我在从MYSQL打印用户名的详细信息时遇到问题。 WHERE name = "xxx"中的引号是原因。 这是代码:     

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
}

$sql = "SELECT name, email FROM MyTable WHERE name=$name";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        echo "id: " . $row["name"]. " - email: " . $row["email"]. "<br>";
    }
} else {
    echo "0 results";
}
$conn->close();
?> 

如何替换WHERE name = $name

2 个答案:

答案 0 :(得分:0)

你错过了'' 应该: WHERE name = '$name' 注意引号

答案 1 :(得分:-1)

我认为你应该使用这种表示法(以前确保正确清理$ name以防止潜在的SQL注入)

 "SELECT name, email FROM MyTable WHERE name='$name'";
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