Php:While循环组输出 - Php: While loop group output

Php: While loop group output

本文关键字：输出循环 While Php | 更新日期: 2023-09-27

我的数据库中有两列，年份和内容我想按年份对内容进行分组（我有很多内容和很多年，不仅仅是2015年和2016年为例）

所以我会在html 中得到这个输出

<div class="all">
  <div class="2016">
    2016
      <div class="content_2016">
          All the content of the column that is in the same ligne as 2016 in the database
    </div>
  </div>
  
  <div class="2015">
    2015
      <div class="content_2015">
          All the content of the column that is in the same ligne as 2015 in the database
    </div>
  </div>
</div>

<?php
	$query = "SELECT * FROM publi where pub_type=0 order by pub_year DESC, pub_publi ";
	$result = mysqli_query($connection, $query);
	$previous =0;
	while ($val = mysqli_fetch_array($result))
	{
		if ($previous <> $val['pub_year'])
		{
			$previous = $val['pub_year'];
			$year = $previous;
			echo $year;
			echo  '<br>';
			$Temp = highlight("person1",$val['pub_publi'],"0000FF");
			$Temp = highlight("person2",$Temp,"0000FF");
			$Temp = highlight("person3",$Temp,"0000FF");
			$Temp = highlight("person4",$Temp,"0000FF");
			$Temp = highlight("person5",$Temp,"0000FF");
			$Temp = highlight("person6",$Temp,"0000FF");
			$Temp = highlight("person7",$Temp,"0000FF");
			$Temp = highlight("person8",$Temp,"0000FF");
			$Temp = highlight("person9",$Temp,"0000FF");
			$Temp = highlight("person10",$Temp,"0000FF");
			echo '<a target=blank href="http://www.test.com/query.f?term=' . $val['pub_pubmed'] . '";)><img border="0" src="img/test.gif" align=MIDDLE alt="Search in  for ' . $val['pub_publi'] . '"></a>';
   		echo $Temp;
			echo  '<br>';
			
					}
			
			
				}
?>

它正在输出正确的年份

2016年

2015年

2014年

等等。。。

但它只显示了每年的第一行，而不是每年的所有内容。

运行以下查询：

Select Content From table order by year

类似回声的下列（或concat）：

while rs.hasNext(){
if (currentYear != lastyear){ 
    echo div #with year As Class;
} 
echo row # a single entry the way you want it dispalyed;       
lastYear = currentYear; 
}

您所要求的是要编写的巨大数据块。以下是步骤：

使用下面的SQL语句创建数组，并在PHP 中获取assoc

从表中选择不同年份*
现在在数组上使用foreach（$years作为$year）并获取内容

从表中选择内容，其中年份=$year
要将SQL中的代码输出夹在中间，请像一样保持连接

$html = "";
$html .= "<div class='"all'">";
foreach($years as $year){
$html .= "<div class='"content_$year'">";
//then content fetched from second sql query
$html .= "</div>";
}
$html .= "</div>";

如何在sql|PHP.NET MANUAL 中获取数据

如何在php|php.NET MANUAL 中加入/concat

稍后添加的内容，我自己的代码片段：

$letter = $_POST['letter'];
$query_out = "SELECT link_id,singer_url FROM letter_mapping WHERE letter = '$letter'";
if($result_out = $conn_out->query($query_out)){
    if($result_out->num_rows > 0){
        while($row = $result_out->fetch_assoc()) {
            $link_id = $row['link_id'];
            $singer_url = $row['singer_url'];
            /*
            foreach($row as $name=>$value){
            }
            */
        }
    }
$result_out->free();
}
if($conn_out->close()){
    // Message for disconnect
    //echo "Status : DISCONNECTED<br/>";
}