PHP MySQL的：添加朋友

我有MySQL數據庫具有下面的表：PHP MySQL的：添加朋友

用戶表：由ID（int）和用戶名的（VARCHAR）
朋友表：由ID（int）和Friends_ID（varchar）

我在我的本地主機上測試了我的腳本，它工作正常，但只要我將腳本上傳到我選擇的免費虛擬主機，它不能正常工作。例如，FriendID被添加到所有用戶的Friendlist而不是僅指定的一個。

這裏是我的PHP腳本：

<?php 
$servername= "********"; 
$serverusername="********"; 
$password="**********"; 
$dbName= "***********"; 

//Make Connection 
$connection = new mysqli($servername, $serverusername,$password, $dbName); 
//Check Connection 
if(!$connection){ 
    die("Connection Failed. ". mysqli_connect_error()); 
} 

$friend_username = 'Steven';//$_POST["friendusernamePost"]; 
$username ='Hans'; //$_POST["usernamePost"] 

$getOwnID_q = "SELECT ID FROM Users WHERE Username = '$username' "; 
$getFriendID_q = "SELECT ID FROM Users WHERE Username = '$friend_username' "; 

$ownID_result = mysqli_query($connection, $getOwnID_q); 
$ownID = mysqli_fetch_assoc($ownID_result)["ID"]; 

$friendID_result = mysqli_query($connection, $getFriendID_q); 
$friendID = mysqli_fetch_assoc($friendID_result)["ID"]; 

if($ownID<1) 
    echo "ERROR: couldn't get correct Own ID"; 
else if($friendID<1) 
    echo "ERROR: couldn't get correct friend ID"; 
else{ 
    $getFriendlist_q= "SELECT Friends_ID From Friends WHERE ID = '$ownID' "; 
    $oldFriendlist_result = mysqli_query($connection, $getFriendlist_q); 
    $oldFriendlist = mysqli_fetch_assoc($oldFriendlist_result)["Friends_ID"]; 
    //check if user isnt already friends with other user 
    $splitted = explode(";", $oldFriendlist); 


    if(in_array($friendID, $splitted)){ 
     echo "You are already Friends."; 
    } 
    else{ 
     //append ';' at the end of friendID 
     $friendID .= ';' ; 
     $updatedFriendlist = $oldFriendlist . $friendID; 
     $addID_query = "UPDATE Friends SET Friends_ID = $updatedFriendlist WHERE ID = $ownID "; 

     if ($connection->query($addID_query) === TRUE) { 
      echo "Friendlist updated successfully"; 
     } 
     else { 
      echo "Error updating Friendlist: " . $conn->error; 
     } 
    } 

    $connection->close(); 
} 

?>

我有PHP或MySQL這使你的幫助，將不勝感激之前沒有任何經驗。

來源

2017-07-19 shawn

第一你有一個大問題來連接一個SQL請求與可變

$getOwnID_q = "SELECT ID FROM Users WHERE Username = '$username' ";

成爲

$getOwnID_q = "SELECT ID FROM Users WHERE Username like ' ".$username." ' ";

$getFriendID_q = "SELECT ID FROM Users WHERE Username = '$friend_username' ";

成了

$getFriendID_q = "SELECT ID FROM Users WHERE Username like ' ".$friend_username." ' ";

$getFriendlist_q= "SELECT Friends_ID From Friends WHERE ID = '$ownID' ";

成了

$getFriendlist_q= "SELECT Friends_ID From Friends WHERE ID =".$ownID;

$addID_query = "UPDATE Friends SET Friends_ID = $updatedFriendlist WHERE ID = $ownID ";

成了

$addID_query = "UPDATE Friends SET Friends_ID = ".$updatedFriendlist." WHERE ID = ".$ownID;

來源

2017-07-19 23:56:13

PHP MySQL的：添加朋友

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