Question

我使用类似下面的代码通过匹配第一个表的id从第二个表中获取数据。代码运行良好，但我知道它会降低性能，我是一个新的蜜蜂。请帮助我以简单正确的方式做同样的事情。

<?php
$result1 = mysql_query("SELECT * FROM table1 ") or die(mysql_error());
while($row1 = mysql_fetch_array( $result1 ))
{
$tab1_id = $row1['tab1_id'];
echo $row['tab1_col1'] . "-";

$result2 = mysql_query("SELECT * FROM table2 WHERE tab2_col1='$tab1_id' ") or die(mysql_error());
while( $row2 = mysql_fetch_array( $result2 ))
{
echo $row2['tab2_col2'] . "-";
echo $row2['tab2_col3'] . "</br>";
}

}
?>

Answer 1

您可以加入这两个表并在单个循环中处理结果。你需要一些额外的逻辑来检查table1的id是否发生了变化，因为你只想在有不同的id时输出这个值：

<?php
// Join the tables and make sure to order by the id of table1.
$result1 = mysql_query("
  SELECT
    *
  FROM 
    table1 t1
    LEFT JOIN table2 t2 ON t2.col1 = t1.id
  ORDER BY
    t1.id") or die(mysql_error());

// A variable to remember the previous id on each iteration.
$previous_tab1_id = null;

while($row = mysql_fetch_array( $result1 ))
{
  $tab1_id = $row['tab1_id'];
  // Only output the 'header' if there is a different id for table1.
  if ($tab1_id !== $previous_tab1_id)
  {
    $previous_tab1_id = $tab1_id;
    echo $row['tab1_col1'] . "-";
  }

  // Only output details if there are details. There will still be a record
  // for table1 if there are no details in table2, because of the LEFT JOIN
  // If you don't want that, you can use INNER JOIN instead, and you won't need
  // the 'if' below.
  if ($row['tab2_col1'] !== null) {
    echo $row['tab2_col2'] . "-";
    echo $row['tab2_col3'] . "</br>";
  }
}

Answer 2

而不是有2个while循环，你可以做的是加入2个表然后迭代结果。

如果您不确定此处的加入方式是什么：https://dev.mysql.com/doc/refman/5.1/de/join.html

此处还有一个使用join编写的相当简单的查询：Join Query Example

Answer 3

你可以用它。与两个表的一个关系：

 <?php
    $result1 = mysql_query("SELECT tab2_col2, tab2_col3 FROM table1, table2 where tab2_col1 = tab1_id ") or die(mysql_error());

    while($row1 = mysql_fetch_array( $result1 ),)
    {

        echo $row2['tab2_col2'] . "-";
        echo $row2['tab2_col3'] . "</br>";

    }
    ?>

Answer 4

像Sushant所说，最好使用一个JOIN或更简单的东西：

 SELECT * FROM table1, table2 WHERE `table1`.`id` = `table2`.`id

在mysql_fetch_array中获取mysql_query的结果

4 个答案: