Php 在单独的表中显示mysql

您好，我只是一个phpmysql的初学者，并试图在单独的表中按行显示结果

我有一个疑问：

$albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citid ASC");
while($row = mysqli_fetch_array($albania))

我知道这会将所有结果输出到一个表中。我想根据$row['citizensity']将结果放在单独的表中

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我在那一行中有大约40个不同的结果，所以我想按那一行将其分开，然后分别输出。

如果您按要在公民身份上拆分的字段排序，并使用$CurrentCitizen跟踪下面的当前值，然后，您应该能够查找该值中的更改，并在下一节开始时启动一个新表

$albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citizenship ASC, citid ASC");
$currentCitizenship = ""

while($row = mysqli_fetch_array($albania)){
   //handle the first time through the loop and start the first table
   if ($currentCitizenship = ""){
       //Start the first table
       echo "<table>";
       $currentCitizenship = $row['citizenship'];
   }elseif ($currentCitizenship != $row['citizenship']){
       //change to next citizenship table
       echo "</table><table>";
       $currentCitizenship = $row['citizenship'];
   }
   /*
      Your current row printing code
   */
}
//close off the last table
echo "</table>"

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你写的东西还不会输出任何东西。它将只从数据库中获取所有行。您需要在PHP中添加代码以在屏幕上显示它

像这样的。需要输入正确的php语法：

    $albania = mysqli_query($con,"SELECT * FROM `data` ORDER BY citid ASC");
    while($row = mysqli_fetch_array($albania))
    {
       if($row['citizenship'] equals 'valuea')
              print in one table
       else
              print in another table
    }