Table of Contents
".$row['times']."
回复讨论(解决方案)
".$row['title']."
Home Backend Development PHP Tutorial 求一段根据日期隔行显示内容列表的PHP代码

求一段根据日期隔行显示内容列表的PHP代码

Jun 23, 2016 pm 01:42 PM

<?php  $sql = mysql_query("select title,times from table order by times desc limit 0,50");   while($row=mysql_fetch_array($sql)){   //每个日期只显示一次    echo "<h2 id="row-times">".$row['times']."</h2>";	//该日期发布的全部内容	echo "<li>".$row['title']."</li>";  }?>
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显示方式如下
12-30
12-30发布的内容1
12-30发布的内容2
12-30发布的内容3
12-30发布的内容4
12-30发布的内容5
12-29
12-29发布的内容1
12-29发布的内容2
12-29发布的内容3
12-29发布的内容4
12-29发布的内容5
12-28
12-29发布的内容1
12-29发布的内容2
12-29发布的内容3
12-29发布的内容4
12-29发布的内容5


回复讨论(解决方案)

<?php  $sql = mysql_query("select title,times from table order by times desc limit 0,50");   $init_time="";  while($row=mysql_fetch_array($sql)){ 	if($init_time==''){ //第一次循环 显示时间和第一条信息类容		$init_time=$row['times'];		echo "<h2 id="row-times">".$row['times']."</h2>";		echo "<li>".$row['title']."</li>";	}else{   //不是第一条判断 时间与上一条是否相等		if($init_time==$row['times']){  //相等的话说明是同一时间段的 直接打印内容			echo "<h2 id="row-title">".$row['title']."</h2>";		}else{	//不一样  打印时间 和 内容			echo "<h2 id="row-times">".$row['times']."</h2>";			//该日期发布的全部内容			echo "<li>".$row['title']."</li>";		}	}  }?>
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在你的代码基础上改了下,可以试试。

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