我正在使用下面这样的代码,通过匹配第一个表的id来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,我是一个新手。请帮我用一种简单正确的方法做同样的事情。
<?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>";
}
}
?>
您可以连接这两个表,并在单个循环中处理结果。您将需要一些额外的逻辑来检查表1的id是否发生了更改,因为您只想在有不同的id时输出此值:
<?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>";
}
}
不需要2个while循环,您可以连接2个表,然后迭代结果。
如果您不确定什么是join,请查看此处:https://dev.mysql.com/doc/refman/5.1/de/join.html
这里还有一个使用join编写的相当简单的查询:join query Example
您可以使用它。具有两个表的一个关系:
<?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>";
}
?>
正如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