Question

我需要一个按钮来替换另一个按钮，具体取决于多个PHP if statements。它的工作方式是，用户只能favourite另一个用户一次，如果用户已经受宠，它将显示另一个按钮，而不是默认的一组按钮，这些按钮可供非受欢迎的用户使用。

我的数据库中有一个名为favourites的表。我已插入一行进行测试。

id - 16
favourited_who - Freddy
favourited_by - AliceP

仅供参考，我的网站目前有三位用户：AliceP，Freddy和Fred。只是为了解释一下，我以[{1}}身份登录，并且您可以看到AliceP不在我的Fred中，在这种情况下，应该为favourites显示两个按钮是Fred和Send Message。另一方面，Add to Favourites在我的收藏夹中，在这种情况下，我想显示两个按钮，Freddy和Send Message。

使用下面的代码，Remove from Favourites和Remove from Favourites同时显示Fred按钮。

我需要在我的代码中涵盖这三个条件，我相信我有：

Freddy

以下是我的尝试：

// 1. Con 1: Dont display any buttons for the logged in user as they cannot fav or msg themselves.
// 2. Con 2: Display sendmsg and addfriend btn if the user is not in the logged in users favs
// 3. Con 3: Display sendmsg and remfriend btn is the user is in the logged in users favs.

我已经涵盖了所有三个条件，但我无法理解为什么代码没有为每个场景显示正确的按钮？

修改：

找到解决方案，结果我必须在if (isset($_POST['addfriend'])) { $fav_request = $_POST['addfriend']; $favourited_who = $user; // u variable $favourited_by = $username; // logged in user // Get all the favourites of the logged in user... $q = mysqli_query ($connect, "SELECT * FROM favourites WHERE favourited_by='$username'"); $num_of_favs = mysqli_num_rows($q); $r_query = mysqli_fetch_array($q); $db_fav_who = $r_query['favourited_who']; $db_fav_by = $r_query['favourited_by']; // Check: See if $user isn't already in there favourites. /*if ($db_fav_by == $username){ // if logged in user has anyone favourited // if the user already exists in the logged in users favourites, then display remove from favourites button. if ($db_fav_who == $user){ echo "<div class='edit_profile'> <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> </div>"; } }*/ //}// while loop closed if ($user != $username) { // Check: See user isnt favouriting themself. $favourite_user = mysqli_query($connect, "INSERT INTO favourites VALUES ('', '$favourited_who', '$favourited_by')"); } } // if isset closed // Condition 1: if ($user == $username){ // dont display buttons } // Condition 2: if ($user != $username && $favourited_who != $db_fav_who){ // if $user is not equal to db fav_who value echo "<form method='post'> <input type='submit' class='btn btn-info' name='sendmsg' value='Send Message'/> <input type='submit' class='btn btn-info' name='addfriend' value='Add to Favourites'> </form>"; } // Condition 3: if ($user != $username && $favourited_who == $db_fav_who) { echo "<form method='post'> <input type='submit' class='btn btn-info' name='sendmsg' value='Send Message'/> <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> </form>"; }闭括号外添加以下内容：

if isset

Answer 1

你需要一个

public static double value(char[][] array, int x, int y) {
    for (int i = 0; i < array.length; i++) {
        for (int j = 0; j < array[0].length; j++) {
            if (array[i][j] == (char) 120) {
                int x_cord = i;
                int y_cord = j;
                int width = (x_cord - x);
                int height = (y_cord - y);
                Math.abs(width);
                Math.abs(height);
                double distance = Math.sqrt(Math.pow(width, 2) + Math.pow(height, 2));
                return distance;
            }

        }
    }

    return 0; // Or some other default
}

以下一行？

<form method='post'>

另一个有他们。

Answer 2

你写 while loop closed 但是在哪里循环???

可能是

while($r_query = mysqli_fetch_array($q)) { ... };

另外，您在此代码中评论一个{和两个}

  /*  if ($db_fav_who == $user){
        echo "<div class='edit_profile'> 
            <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'>
        </div>";
        }
    }*/

按钮替换另一个按钮取决于PHP查询

2 个答案: