来自查询的 PHP 回显日期和数字

Question

您好，我正在尝试通过编写用于显示销售数据的小型 Web 应用程序来学习 php。我有一个查询，现在我已经对其进行了测试，但我希望它回显匹配的日期以及在该日期找到的行数/结果数。这就是我目前所拥有的

<?php
$con=mysqli_connect("host","user","password","database");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM matched WHERE datematched IN (
    SELECT datematched FROM matched GROUP BY datematched HAVING count(*) > 1");

while($row = mysqli_fetch_array($result))

  {
  echo "['";
 echo "" . $date['datematched'] . "', ";
  echo "" . $num_rows . "],";

   }

mysqli_close($con);
?>

我知道我在这里做错了什么。瑞安

编辑：

<?php
$con=mysqli_connect("host","user","password","database");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$result = mysqli_query($con,"SELECT * FROM matched WHERE datematched IN (
    SELECT datematched FROM matched GROUP BY datematched HAVING count(*) > 1");


 echo "['";
 echo " 16/08/2013 ', ";
  echo "12345}],";


mysqli_close($con);
?>

好的，我刚刚检查了我的回声，他们工作了，我输入了一些数据，所以我只需要找到一种方法来获取已找到的日期匹配的信息，然后找到与之相关的行数.谢谢瑞恩

Answer 1

首先，您需要对查询进行调整，使其具有您期望的行数。

$result = mysqli_query($con,"SELECT datematched, COUNT(*) as num_rows "
       . "FROM matched GROUP BY datematched HAVING num_rows > 0");

那么就可以如下显示数据了

while($row = mysqli_fetch_array($result))
{
 echo $row['datematched'] . ",";
 echo $row['num_rows'];
}

Answer 2

如果你的 sql 查询是完美的，那么你应该这样写

while($row = mysqli_fetch_array($result))

  {
  echo "['";
 echo "" . $row['datematched'] . "', ";
  echo  "" . $row['num_rows'] . "', ";

   }

请按照您在 mysql 查询中获得的列设置。

Answer 3

<?php
$query=mysqli_query($con,"SELECT datematched FROM matched GROUP BY datematched");
$num=mysqli_num_rows($query);
if($num>1)
{ 
$result = mysqli_query($con,"SELECT * FROM matched");
$num_rows=mysqli_num_rows($result);
while($row = mysqli_fetch_array($result))
{
echo '['; echo $row['datematched']; echo $num_rows; echo ']';
}
}