Php 如何显示两个不同表中具有相同名称的列?
我从两个表中进行了SQL查询。一切正常,但问题是这两个表具有相同的字段名,在我不知道如何正确显示它们之后,如何区分$data['aaa']来自表1,而$data['aaa']来自表2 以下是我的SQL查询:Php 如何显示两个不同表中具有相同名称的列?,php,sql,Php,Sql,我从两个表中进行了SQL查询。一切正常,但问题是这两个表具有相同的字段名,在我不知道如何正确显示它们之后,如何区分$data['aaa']来自表1,而$data['aaa']来自表2 以下是我的SQL查询: $query_str = "SELECT cm.id, cm.global_category_id, cm.num, cm
$query_str = "SELECT
cm.id,
cm.global_category_id,
cm.num,
cm.menu_lv,
cm.menu_ru,
cm.menu_en,
u.id,
u.menu_lv,
u.menu_ru,
u.menu_en
FROM products_category cm, products_global_category u
WHERE cm.global_category_id = u.id
";
并显示数据
<? foreach ($sub_category_list as $line) : ?>
<tr>
<td><?=$line['menu_lv']?></td> <---- here I want to display data from products_global_category u
<td><?=$line['sub_menu_lv']?></td> <---- products_category cm
<td><?=$line['sub_menu_ru']?></td> <---- products_category cm
<td><?=$line['sub_menu_en']?></td> <---- products_category cm
</tr>
<? endforeach; ?>
作为一种解决方案,您可以更改SQL查询,为具有相同名称的字段提供别名 例如:
SELECT somefield AS othername FROM table.
SELECT
table1.pid AS page_id,
table2.pid AS product_id
FROM
table1
LEFT JOIN
table2
ON
table1.id = table2.id
在这种情况下,somefield
字段将通过别名othername
可用
就你而言:
$query_str = "SELECT
cm.id AS cm_id,
cm.global_category_id,
cm.num,
cm.menu_lv AS cm_menulv,
cm.menu_ru AS cm_menuru,
cm.menu_en AS cm.menuen,
u.id as u_id,
u.menu_lv AS u_menulv,
u.menu_ru AS u_menuru,
u.menu_en AS u_menuen
FROM products_category cm, products_global_category u
WHERE cm.global_category_id = u.id
";
然后在PHP中:
$line['u_menulv'] //Access field menu_lv from products_global_category table
$line['cm_menulv'] //Access field menu_lv from products_category table
编辑:在mysql\u fetch\u数组中
:
如果结果的两列或多列具有相同的字段名,则
最后一列将优先。要访问
如果名称相同,则必须使用列的数字索引或
列的别名。对于别名列,您无法访问
具有原始列名的内容
换句话说,可以创建如上所示的别名,也可以通过数组的数字索引访问字段。只需为要打印的相互冲突的记录添加别名,例如:
SELECT somefield AS othername FROM table.
SELECT
table1.pid AS page_id,
table2.pid AS product_id
FROM
table1
LEFT JOIN
table2
ON
table1.id = table2.id
然后在您的PHP中,您可以按如下方式对其进行回显:
while ($row = mysql_fetch_assoc($res)) {
echo $row['page_id'];
echo $row['product_id'];
}
希望这有帮助