Php 在组中显示要配置的帖子

这是我从数据库获取帖子的代码。我把我的帖子称为快照

$getuser = mysql_query("SELECT userID FROM user WHERE username= '$user' ");
$get_row = mysql_fetch_assoc($getuser);
$userID = $get_row['userID'];

$getposts = mysql_query("SELECT * FROM snaps WHERE userID='$userID' ORDER BY snapID DESC ") or die(mysql_error());
while ($row = mysql_fetch_assoc($getposts)) {
                    $title = $row['title']; 
                    $date_added = $row['date_added'];
                    $time = $row['time'];
                    $photo = "userdata/user_snaps/".$row['photos'];

用于显示快照的html格式为每行3个快照

<!-- Projects Row -->


echo ("
        <div class="row">
        <div class="col-md-4 portfolio-item">

        <div class="thumbnail">
        <a href=''>
        <img src="$photo" class="portrait" alt="Image" />
        </div>

        <h3>
        '$title'</a>
        </h3>
        <p>$content</p>
        </div>
        <div class="col-md-4 portfolio-item">
        <div class="thumbnail">
        <a href=''>
        <img src="$photo" class="portrait" alt="Image" />
        </div>


        <h3>
        '$title'</a>
        </h3>
        <p>'$content'</p>
        </div>
        <div class="col-md-4 portfolio-item">
        <div class="thumbnail">
        <a href="">
        <img src="$photo" class="portrait" alt="Image" />
        </div>

        <h3>
        '$title'</a>
        </h3>
        <p>$content</p>
        </div>
        </div>
");

回声（“
$content

“$content”

$content

");

---

我如何以每行有三篇文章的方式显示它？就像Instagram一样。

在while循环中使用计数器

$counter = 0;
while ($row = mysql_fetch_assoc($getposts)) {
    $counter++;
    if($counter  > 3) { //if your counter goes to 4 then it will add a new line in your html
        $counter = 0;
        echo '<div class="row"></div>';
    }
    echo '<div class="col-md-4 portfolio-item">
            <div class="thumbnail">
            <a href="">
            <img src="$photo" class="portrait" alt="Image" />
            </div>


            <h3>
                $title</a>
            </h3>
            <p>$content</p>
        </div>';
}

$counter=0；
while（$row=mysql\u fetch\u assoc（$getposts））{
$counter++；
如果（$counter>3）{//如果计数器变为4，那么它将在html中添加新行
$counter=0；
回声'；
}
回声'
$content

';
}

目前发生了什么？请注意，mysql_uuu函数被认为是贬值的，与通过mysqli_uuuu函数和PDO类准备的语句相比，它们被视为高风险的。