php - 从表中选择值并按时显示？

Question

我想通过'while'选择值但是有问题，这是代码：

<?php 
$following_select_article = mysql_query("SELECT * FROM follow WHERE user_follower_id='$user_id'");

while( $following_select_article_row = mysql_fetch_array($following_select_article) ) {
    $article_following_user_id = $following_select_article_row['user_following_id'].",";
    $mmnnss = substr_replace($article_following_user_id, "", -1);
    $echo $mmnss;
}

笔记$user_id = 1

所需的输出是

2,3,4

但我得到的是

234

数据库是这样的：

下表：

编号 | user_follower_id | user_following_id

1 | 1 | 2

2 | 1 | 4

3 | 1 | 3

谢谢

Answer 1

做这个

<?php 

$following_select_article = mysql_query("SELECT * FROM follow WHERE user_follower_id='$user_id'");
$article_following_user_id = "";
while($following_select_article_row = mysql_fetch_array($following_select_article)){

    $article_following_user_id .= $following_select_article_row['user_following_id'].",";

}
$mmnnss = substr_replace($article_following_user_id, "", -1);
echo $mmnnss;

因为你substr_replace在循环中，所以每次$article_following_user_id用,它创建后，每次都替换最后一个字符

编辑

正如Glavić建议的那样，如果您可以更换您的

substr_replace($article_following_user_id, "", -1);

和

substr($article_following_user_id, 0, -1);