Question

我正在尝试获取一个区域中的物品清单。

我希望将地区名称作为标题然后，下表中该区域中的所有项目即

HEADING-Region1

表格-Region1项

HEADING-Region2

表格-Region2项

HEADING-Region3

表格-Region3项

HEADING-Region4

表格-Region4项

我可以输出：

HEADING

如果使用“实际区域名称”之类的代码，则可以在表格中输出区域详细信息

我想使用LIKE $ region进行输出，因此不需要为每个“实际区域名称”编写新的语句。

2个SELECT查询：

$region = mysqli_query($conn,"SELECT DISTINCT region FROM country") or die($conn->error);
$result = mysqli_query($conn,"SELECT * FROM country WHERE region LIKE '$region'") or die($conn->error);

输出1

这会将每个区域输出为

区域名称

$i = 0;
while($row = $region->fetch_assoc())
{
 if ($i == 0) {
 foreach ($row as $value) {
 echo "<p>" . $value . "</p>";
                      }

             }
}

输出2

这将输出所有区域数据并将其放在表格中。

echo "This is table Build";
echo "<table border='1'>";

$i = 0;
while($row = $result->fetch_assoc())
{
    if ($i == 0) {
      $i++;
      echo "<tr>";
      foreach ($row as $key => $value) {
        echo "<th>" . $key . "</th>";
      }
      echo "</tr>";
    }
    echo "<tr>";
    foreach ($row as $value) {
      echo "<td>" . $value . "</td>";
    }
    echo "</tr>";
}
echo "</table>";

mysqli_close($conn);

?>

OUTPUT 1和OUTPUT 2都分别工作。我不能让他们一起工作

Answer 1

您可以从查询中获取所有region的名称，然后将其传递到next query并打印与该查询相关的所有record，即：

<?php

$region = mysqli_query($conn,"SELECT DISTINCT region FROM country") or die($conn->error);

if($region->num_rows > 0) 
    {                               {
       while($row = $region->fetch_assoc())
         {
         //printing region name
         echo "<p>" . $row['region'] . "</p>";
         //only those row will be retrieve which belong to current region name 
         $result = mysqli_query($conn,"SELECT * FROM country WHERE region LIKE '$row['region']'") or die($conn->error);
//printing table 
      echo "This is table Build";
      echo "<table border='1'>"; 
              $i = 0;
    while($row1 = $result->fetch_assoc())
    {
        if ($i == 0) {
          $i++;
          echo "<tr>";
          foreach ($row1 as $key => $value) {
            echo "<th>" . $key . "</th>";
          }
          echo "</tr>";
        }
        echo "<tr>";
        foreach ($row1 as $value) {
          echo "<td>" . $value . "</td>";
        }
        echo "</tr>";
    }
    echo "</table>";
             }

}
  mysqli_close($conn);

        ?>

在mysqli WHERE LIKE语句中使用$ var

1 个答案: