Question

我正在尝试使用XSwitch jquery脚本并动态创建＆＃34; section id＆＃34;自动但被卡住了。

如何为每个部分提供一个＆＃34; id＆＃34;从0开始？有些项目有2或3个图像，有些有30 +。

$result = mysqli_query($conn,"SELECT * FROM Images WHERE stock=".$_GET['stock']." ORDER BY orderIndex");
while($row = mysqli_fetch_array($result)){
  $count = $result->num_rows;
  if ($count) {
    $photos .= '<div class="section" id="section__NEED___NUMBER" style="background-image: url(inventory/'.$category.'/large/'.$stock.'_'.$row['id'].'.jpg)"></div>';
  } else {
    $photos = 'No Photos are currently available';
  }
}

Answer 1

你可以使用一个简单的计数器变量（$counter = 0;），你在每个循环中增加（$ counter ++;`） -

$counter = 0;
$result = mysqli_query($conn,"SELECT * FROM Images WHERE stock=".$_GET['stock']." ORDER BY orderIndex");
$count = $result->num_rows;
if ($count) {
    $photos = "";
    while($row = mysqli_fetch_array($result)){
        $photos .= '<div class="section" id="section__'.$counter.'" style="background-image: url(inventory/'.$category.'/large/'.$stock.'_'.$row['id'].'.jpg)"></div>';
        $counter++;
    }
} else {
    $photos = 'No Photos are currently available';
}

请注意，您的$count = $result->num_rows;if ($count) {...}圈内有while()，而它应该在外面。

如何获取返回的每个项目的行号

1 个答案: