php - 如何在 php 中获取当前用户的用户名并在 mysql 查询中使用它

Question

我是使用 php 的初学者，我想使用当前登录用户的用户名发送查询。

我应该从哪里开始 session ？在登录php文件中？我怎样才能将变量传递给其他 php 文件？

<?php

session_start();

//importing required script
require_once '../includes/DbOperation.php';

// Create connection
$con=mysqli_connect(" ","root"," ","myiosapp");

// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$sql = "SELECT * FROM ITEM WHERE username = $_SESSION['username']"; //?

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
    // If so, then create a results array and a temporary one
    // to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    {
        // Add each row into our results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    }

    // Finally, encode the array to JSON and output the results
    echo json_encode($resultArray);
}

// Close connections
mysqli_close($con);
?>

Answer 1

First u need to fetch data off logged-in user
then u have to set those data in session variable
and then u will be able to use that session variable in your query.

you need to create api 

http://example.com/testfile.php?username='abc';

then in php

<?php

//session_start();

//importing required script
require_once '../includes/DbOperation.php';

// Create connection
$con=mysqli_connect(" ","root"," ","myiosapp");

// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$username = $_GET['username'];

$sql = "SELECT * FROM ITEM WHERE username = $username"; //?

// Check if there are results
if ($result = mysqli_query($con, $sql))
{
    // If so, then create a results array and a temporary one
    // to hold the data
    $resultArray = array();
    $tempArray = array();

    // Loop through each row in the result set
    while($row = $result->fetch_object())
    {
        // Add each row into our results array
        $tempArray = $row;
        array_push($resultArray, $tempArray);
    }

    // Finally, encode the array to JSON and output the results
    echo json_encode($resultArray);
}

// Close connections
mysqli_close($con);
?>