Question

如何使用php回显MySql中某行的某个id？

我有这段代码向我展示了DB中的所有值：

while($row=mysqli_fetch_array($res))
{
// some code goes here 
$HTML.='
    <li>
        <h4 class="">'.$row['product_name'].'</h4>

        <div class="grid">
            '.$like.'
        </div>
    </li>';

最后在页面的某个地方：

<ul class="thumbnail-list">
    <?php echo $HTML; ?>
</ul>

这显示了我所有的product_names（照片），但我想将它们彼此分开。我的数据库看起来像这样：

我如何能够彼此独立地回应Photos？

更新：

我的While循环看起来像这样：

while($row=mysqli_fetch_array($res))
{
// get likes and dislikes of a product
$query = mysqli_query($connection,"select sum(`like`) as `like`,sum(`unlike`) as `unlike` from `likes` where pid = ".$row['id']);
$rowCount = mysqli_fetch_array($query);
if($rowCount['like'] == "")
    $rowCount['like'] = 0;

if($rowCount['unlike'] == "")
    $rowCount['unlike'] = 0;

if($uid == "") // if user not loggedin then show login link on like button click
{
    $like = '
        <input onclick="location.href = \'login.php\';" type="button" value="'.$rowCount['like'].'" rel="'.$row['id'].'" data-toggle="tooltip"  data-placement="top" title="Login to Like" class="button_like" />
        <input onclick="location.href = \'login.php\';" type="button" value="'.$rowCount['unlike'].'" rel="'.$row['id'].'" data-toggle="tooltip" data-placement="top" title="Login to Unlike" class="button_unlike" />';
}
else
{
    $query = mysqli_query($connection,"SELECT * from `likes` WHERE pid='".$row['id']."' and uid='".$uid."'");
    if(mysqli_num_rows($query)>0){ //if already liked od disliked a product
        $likeORunlike = mysqli_fetch_array($query);
        // clear values of variables
        $liked = '';
        $unliked = '';
        $disable_like = '';
        $disable_unlike = '';

        if($likeORunlike['like'] == 1) // if alredy liked then disable like button
        {
            $liked = 'disabled="disabled"';
            $disable_unlike = "button_disable";
        }
        elseif($likeORunlike['unlike'] == 1) // if alredy dislike the disable unlike button
        {
            $unliked = 'disabled="disabled"';
            $disable_like = "button_disable";
        }

        $like = '
        <input '.$liked.' type="button" value="'.$rowCount['like'].'" rel="'.$row['id'].'" data-toggle="tooltip"  data-placement="top" title="Like" class="button_like '.$disable_like.'" id="linkeBtn_'.$row['id'].'" />
        <input '.$unliked.' type="button" value="'.$rowCount['unlike'].'" rel="'.$row['id'].'" data-toggle="tooltip" data-placement="top" title="Un-Like" class="button_unlike '.$disable_unlike.'" id="unlinkeBtn_'.$row['id'].'" />
        ';
    }
    else{ //not liked and disliked product
        $like = '
        <input  type="button" value="'.$rowCount['like'].'" rel="'.$row['id'].'" data-toggle="tooltip"  data-placement="top" title="Like" class="button_like" id="linkeBtn_'.$row['id'].'" />
        <input  type="button" value="'.$rowCount['unlike'].'" rel="'.$row['id'].'" data-toggle="tooltip" data-placement="top" title="Un-Like" class="button_unlike" id="unlinkeBtn_'.$row['id'].'" />
        ';
    }
}

$HTML.='
    <li>
        <h4 class="">'.$row['product_name'].'</h4>

        <div class="grid">
            '.$like.'
        </div>
    </li>';

}

Answer 1

我建议您在变量$HTML的字符串上使用数组。因为您的字符串只是将所有结果连接在一起，这就是代码应该如何工作。

请参阅PHP ARRAYS。

所以在你的while循环中这样做：

$HTML[]='
    <li>
    <h4 class="">'.$row['product_name'].'</h4>

    <div class="grid">
        <p>Photo Name : '.$row['product_name'].' <br>
           File Name : '.$row['image'].' </p>
    </div>
</li>';
//because I know nothing about how you are retrieving your images, I am only 
//display the text value of the code as it would be in the database

当您回显$HTML时，您需要在数组中指定所需的索引。

<ul class="thumbnail-list">
    <?php echo $HTML[0]; ?>
</ul>
<ul class="thumbnail-list">
    <?php echo $HTML[1]; ?>
</ul>

或者如果你想遍历每个元素

foreach($HTML as $item){
    <ul class="thumbnail-list">
        <?php echo $item; ?>
    </ul>
}

如何用php从MySql数据库中选择某个ID

1 个答案: