Question

这是一个基本问题，我知道，但是我似乎无法使用我的代码破解它。

我将我的登录会话用户名分配给$ username - 这完全插入，但是，我希望为$ id分配位于'users'表中的用户ID。

我目前有2张桌子，用户和旅行。

Users包含以下列：

id (PK,AI), 
username, 
email, 
password, 
trn_date

Trips包含以下列：

id (PK,AI), 
user (FK_users.ID), 
name, 
from, 
to, date, space, email, telephone, comments

这是我的代码：

<?php
session_start();
include("auth.php"); 
include("db.php"); 
?>              
<?php

// SERVER AND DATABASE DETAILS --- WORKING
$servername = "<redacted>";
$username1 = "<redacted>";
$password = "<redacted>";
$dbname = "<redacted>";



//sets session variable username --- WORKING
$username = $_SESSION['username'];

// sets variable to result of sql to determine user_id - NOT WORKING
$id = mysqli_query("SELECT id from `users` where username = '$username'");


// Create connection --- WORKING
$conn = new mysqli($servername, $username1, $password, $dbname);

// Check connection --- WORKING
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
 // Insert script when adding trip --- WORKING
$sql = "INSERT INTO `trips` (`user`,`name`, `from`, `to`, 
`date`, `space`, `email`, `telephone`, `comments`) VALUES   ('$id','$username','".$_POST["from"]."','".$_POST["to"]."','".$_POST["datetime"]."','".$_POST["space"]."','".$_POST["email"]."','".$_POST["telephone"]."','".$_POST["comments"]."')";

// Alert pop-up confirming when insert is successful --- WORKING
if ($conn->query($sql) === TRUE) {
   echo '<script type="text/javascript">'; 
    echo 'alert("Your trip was successfully added!");'; 
    echo 'window.location.href = "/index.php";';
    echo '</script>';

    // Error when sql insert fails
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
    }

$conn->close();
?>

Answer 1

替换此行

$id = mysqli_query("SELECT id from `users` where username = '$username'");

用这个

$conn = new mysqli($servername, $username1, $password, $dbname); //since I am not sure where did you initiate the mysqli connection
$res = mysqli_query($conn, "SELECT id from `users` where username = '$username'");
$id = mysqli_fetch_assoc ( $res ) ['id'] ;

我想这会解决你的问题。

mysqli_query函数不返回您期望的值，而是返回mysqli结果http://php.net/manual/en/mysqli.query.php

因此，您需要先从该对象获取结果，然后从结果中提取值。

在我的示例中，我使用mysqli_fetch_assoc将数据作为关联数组获取，然后使用['id']获取特定的列值。

将PHP变量分配给SQL的结果

1 个答案: