Question

我一直在这里寻找一些答案，但我碰壁了。我找到了一堆代码非常相似的帖子，显然有效。代码是通过我正在寻找的电子邮件中的硬编码进行测试的，并且工作正常。当我通过POST功能发送信息时，由于某种原因它似乎没有翻译。

$Email = $_POST["Email"];

$sql = 'SELECT * FROM DewSweepers WHERE Email = "$Email"';
$result = mysqli_query($conn,$sql) or trigger_error("Query Failed! SQL: $sql - Error: ".mysqli_error(), E_USER_ERROR);


while($row = mysqli_fetch_row($result))
{
    echo "<tr>";

    echo "<td>$row[1]</td>";


    echo "</tr>\n";
}

Answer 1

在单引号内部，所有内容都被视为字符串。因此需要正确转义或连接

$sql = 'SELECT * FROM DewSweepers WHERE Email = "'.$Email.'"';

更新1：

您的代码输出将是这样的

SELECT * FROM DewSweepers WHERE Email = "$Email"

警告mysql_query，mysql_fetch_array，mysql_connect等..扩展在PHP 5.5.0中已弃用，并且已在PHP 7.0.0中删除。相反，应该使用MySQLi或PDO_MySQL扩展。

//db connection
$servername = "localhost";  //host name

$username = "username"; //username

$password = "password"; //password

$mysql_database = "dbname"; //database name

// mysqli准备好的陈述

$conn = mysqli_connect($servername, $username, $password) or die("Connection failed: " . mysqli_connect_error());

mysqli_select_db($conn,$mysql_database) or die("Opps some thing went wrong");

$Email = $_POST["Email"];

$stmt = $conn->prepare("SELECT * FROM DewSweepers WHERE Email =?");

$stmt->bind_param('s',$Email);
$stmt->execute();
$get_result =$stmt->get_result();
$row_count= $get_result->num_rows;

if($row_count>0)
{
   echo "row exists";

}
else
{
   echo "row not exists";

}

$ Email = $ POST _ [＆＃34; Email＆＃34;]在尝试在数据库中查找电子邮件时，不会申请我的Mysqli代码

1 个答案: