PHP 脚本，获取挂起的请求列表 - PHP script, getting a pending request list

PHP script, getting a pending request list

我才刚刚开始编写PHP脚本来为我的安卓应用程序的后端提供支持。我目前正在尝试做的是运行一个名为 send_friendreq 的 PHP 脚本和名为 pending_req 的表，并获取行 toUser 并将其添加到数组中。我目前遇到的唯一问题是我无法让 PHP 脚本正常运行。任何帮助将不胜感激。这是我目前拥有的 PHP 脚本的代码。非常感谢您的帮助！

if (isset($_POST['Username']) && isset($_POST['FriendReq']))
{
    $username = $_POST['Username'];
    $usernamebeingreq = $_POST['FriendReq'];
    $i=0;
    //$sqlCheck = "SELECT Username FROM Users WHERE Username = '" . $usernamebeingreq . "'";
    //$resultCheck = mysqli_query($con, $sqlCheck);
    //if(!$resultCheck)
    //{
    //echo "Invalid Username";
    //}
    //else
    //{
    $sql="SELECT fromUser FROM pending_req WHERE toUser='&username'";
    $result = mysqli_query($con, $sql);
    $array = array();
    while ($row = pg_fetch_array($result))
    {
         $i++;
    }
    for($x=0;$x<$i;$x++)
    {
        echo $array[$x];
    }
    if(!$result)
    {
        echo 'Failed';
    }
    else
    {
        echo json_encode($array[$x]);
        echo "<br>";
    }

如果您对更好/

更有效/更安全的东西有建议，请告诉我！

如果我没记错的话，您正在尝试从查询中获取结果的 json。试试这段代码。

if (isset($_POST['Username']) && isset($_POST['FriendReq']))
{
$username = $_POST['Username'];
$usernamebeingreq = $_POST['FriendReq'];
$i=0;
//$sqlCheck = "SELECT Username FROM Users WHERE Username = '" . $usernamebeingreq . "'";
//$resultCheck = mysqli_query($con, $sqlCheck);
//if(!$resultCheck)
//{
//echo "Invalid Username";
//}
//else
//{
$sql = 'SELECT fromUser FROM pending_req WHERE toUser='. $username ;
$result = mysqli_query($con, $sql);
if(!$result) {
    echo 'Failed';
}elseif($result){
    $myArray = array();
    while($row = $result->fetch_array(MYSQL_ASSOC)) {
        $myArray[] = $row;
    }
    echo json_encode($myArray);
}

您查询语法似乎不正确，请修改为：

 $sql="SELECT fromUser FROM pending_req WHERE toUser='$username'";