Question

我这里有问题。我试过到处寻找，但似乎无法找到它。我知道有关于此的帖子，也许我无法正确搜索我的搜索查询。

我发现它类似于这个2： Applying different CSS to top 10 results of a query .addClass with IF statement based on value

目前这是我的代码的简短示例：

<?php
$select=mysql_query("select * from products_list ORDER BY id ASC");
while($row=mysql_fetch_array($select)){
echo "<tr>
<td>".$row['id']."</td>
<td>".$row['title']."</td>
<td>".$row['new_arrival']."</td>";
?>

数据库中new_arrival的数据是yes还是no。

我的问题是如果它是“是”，我如何在TD上为“new_arrival”添加一个类，如果是“否”则不添加类

希望你们明白我想说的话。感谢。

此致杰夫

Answer 1

如果new_arrival == yes添加了其他类，则使用简单的if语句将其清空

$select=mysql_query("select * from products_list ORDER BY id ASC");
while($row=mysql_fetch_array($select)){
    $class = ($row['new_arrival'] == 'yes') ? 'class="new_arrival"' : '';
    echo "<tr>
    <td ".$class.">".$row['id']."</td>
    <td ".$class.">".$row['title']."</td>
    <td ".$class.">".$row['new_arrival']."</td></tr>";
}

Answer 2

<?php
$select=mysql_query("select * from products_list ORDER BY id ASC");
while($row=mysql_fetch_array($select)){
echo "<tr>
<td>".$row['id']."</td>
<td>".$row['title']."</td>";
  if(if($row['new_arrival']=='yes')
  {
   echo "<td class='classname'>".$row['new_arrival']."</td>";
  }
  else
  {
    echo "<td>".$row['new_arrival']."</td>";
  }
?>

Answer 3

试试这个简单的IF

echo "<tr>
<td>".$row['id']."</td>
<td>".$row['title']."</td>";
if($row["new_arrival"] == "yes"){
 echo "<td class='positive'>".$row["new_arrival"]."</td>"; //yes
}else{
 echo "<td>".$row['new_arrival']."</td>"; // no
}

Answer 4

<?php

    $select = mysql_query( "SELECT * FROM products_list ORDER BY id ASC" );
    while( $row = mysql_fetch_array( $select ) ) {

        $class = '';
        if ( $row[ 'new_arrival' ] == 'yes' )
            $class = ' class="your_class" ';

        $result = "<tr>
                    <td>" . $row['id'] . "</td>
                    <td>" . $row['title'] . "</td>
                    <td" . $class. ">" . $row['new_arrival']  ."</td>
                  </tr>";

        echo $result;
    }

?>

使用mysql查询结果将CSS类添加到元素

4 个答案: