连接 PHP 个变量

Question

我正在尝试为灯箱画廊显示中的图片信息形成一个字符串。本来我只是用了下面的

<td><div style='height:175px;'></div></td><td><a href='images/memwork/".$cell1['folder']."/".$cell1['filename']."' data-lightbox='image-".$cell1['id']."' data-title='".$cell1['title']." by  ".$cell1['artist']."<br />".$cell1['medium']." &nbsp; &nbsp; ".$cell1['width']." x ".$cell1['height']." cm.  <br />Price: £".$cell1['priceu']."'><img class='tn' src='images/memwork/".$cell1['folder']."/".$cell1['tn']."' /></a></td>

但这变成了一场噩梦，试图用三元运算符捕获零价格、尺寸或 medai 类型，所以我现在尝试用 IF 语句格式化数据并将结果添加到数据数组中。但是，我似乎错过了一些东西，因为结果总是只是“by”。我的代码是：

        $query = "SELECT * FROM gallery WHERE sig=1 LIMIT $start,6";
}
$result = mysqli_query($db, $query);
$i=1;
while($row = mysqli_fetch_assoc($result)) {
    ${'cell'.$i} = $row;
    $gal_artist = $row['artist'];
    $query = "SELECT COUNT(*) AS num FROM gallery WHERE artist='$gal_artist'";
    $total = mysqli_fetch_array(mysqli_query($db,$query));
    ${'cell'.$i}['num_pics'] = $total['num'];

    $picdata = ${'cell'.Si}['title']." by ".${'cell'.Si}['artist'];
    if (${'cell'.Si}['medium'].${'cell'.Si}['width'] !='') $picdata = $picdata."<br />";    
    if (${'cell'.Si}['width'] >0) $picdata = $picdata." &nbsp; &nbsp; ".${'cell'.Si}['medium']." &nbsp; &nbsp; ".${'cell'.Si}['width']." x ".${'cell'.Si}['height']." cm.";
    if (${'cell'.Si}['price'] >0) $picdata = $picdata."<br />Price: £".${'cell'.Si}['price'];
    ${'cell'.Si}['picdata'] = $picdata;

$i++;
}

和 table 数据语句应简化为

<td><div style='height:175px;'></div></td><td><a href='images/memwork/".$cell1['folder']."/".$cell1['filename']."' data-lightbox='image-".$cell1['id']."' data-title='".$cell1['picdata']."'><img class='tn' src='images/memwork/".$cell1['folder']."/".$cell1['tn']."' /></a></td>

Answer 1

您在代码中使用 Si 而不是 $i。另外，我建议你使用数组而不是动态变量名，这通常是一种不好的做法。

所以 :

$result = mysqli_query($db, $query);
$i=1;
while($row = mysqli_fetch_assoc($result)) {
    $cells[$i] = $row;
    $gal_artist = $row['artist'];
    $query = "SELECT COUNT(*) AS num FROM gallery WHERE artist='$gal_artist'";
    $total = mysqli_fetch_array(mysqli_query($db,$query));
    $cells[$i]['num_pics'] = $total['num'];

    $picdata = $cells[$i]['title']." by ".$cells[$i]['artist'];
    if ($cells[$i]['medium'].$cells[$i]['width'] !='') $picdata = $picdata."<br />";    
    if ($cells[$i]['width'] >0) $picdata = $picdata." &nbsp; &nbsp; ".$cells[$i]['medium']." &nbsp; &nbsp; ".$cells[$i]['width']." x ".$cells[$i]['height']." cm.";
    if ($cells[$i]['price'] >0) $picdata = $picdata."<br />Price: £".$cells[$i]['price'];
    $cells[$i]['picdata'] = $picdata;

$i++;
}

然后像

一样打印出来

<td><div style='height:175px;'></div></td><td><a href='images/memwork/".$cells[1]['folder']."/".$cells[1]['filename']."' data-lightbox='image-".$cells[1]['id']."' data-title='".$cells[1]['picdata']."'><img class='tn' src='images/memwork/".$cells[1]['folder']."/".$cells[1]['tn']."' /></a></td>

Answer 2

首先您需要使用以下查询更新您的 sql 查询。

    SELECT t.*, (Select count(*) from test AS t1 where t.artist= t1.artist) AS album_count 
FROM test AS t

希望对您有所帮助。

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