Php 连接两个表以显示数据
我试着把两张桌子连接起来。一个是类别,另一个是子类别。类别表中的id将成为子类别表中的cat id 类别Php 连接两个表以显示数据,php,mysql,sql,database,phpmyadmin,Php,Mysql,Sql,Database,Phpmyadmin,我试着把两张桌子连接起来。一个是类别,另一个是子类别。类别表中的id将成为子类别表中的cat id 类别 id name catimage 2 cat1 image1 子类别 id name subcatimage catid 1 subcat1 image2 2 代码是 <?php ob_start(); require_once('config.php'); $selectsubcategory = mysql_query(
id name catimage
2 cat1 image1
子类别
id name subcatimage catid
1 subcat1 image2 2
代码是
<?php
ob_start();
require_once('config.php');
$selectsubcategory = mysql_query("SELECT category.name, subcategory.name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");
$posts = array();
if(mysql_num_rows($selectsubcategory))
{
while($post = mysql_fetch_assoc($selectsubcategory))
{
$posts[] = $post;
}
header('Content-type: application/json');
echo stripslashes(json_encode(array('subcategorylist'=>$posts)));
}
else
{
header('Content-type: application/json');
echo stripslashes(json_encode(array('subcategorylist'=>'No subcategory')));
}
?>
我正在获取所有详细信息,但问题是结果中没有获取category.name。有人能帮忙吗
另外,我已经使用了sql,稍后将对其进行更改。我关心的是功能部分子类别名称将覆盖您的类别名称。 您必须为子类别名称使用别名,如下所示:
$selectsubcategory = mysql_query("SELECT category.name, subcategory.name AS sub_name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");
subcategory.name将覆盖您的类别名称。 您必须为子类别名称使用别名,如下所示:
$selectsubcategory = mysql_query("SELECT category.name, subcategory.name AS sub_name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");
subcategory.name将覆盖您的类别名称。 您必须为子类别名称使用别名,如下所示:
$selectsubcategory = mysql_query("SELECT category.name, subcategory.name AS sub_name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");
subcategory.name将覆盖您的类别名称。 您必须为子类别名称使用别名,如下所示:
$selectsubcategory = mysql_query("SELECT category.name, subcategory.name AS sub_name, category.catimage, subcategory.id, subcategory.catid FROM category INNER JOIN subcategory ON category.id=subcategory.catid ");
希望这能帮助你 此处MainCatName将显示类别名称字段数据。。。和 子类别名称将显示为子类别名称字段数据
SELECT
`category`.`name` AS MainCatName
, `subcategory`.`name` AS SubCatName
FROM
`category`
INNER JOIN `subcategory`
ON (`category`.`id` = `subcategory`.`catid`);
希望这能帮助你 此处MainCatName将显示类别名称字段数据。。。和 子类别名称将显示为子类别名称字段数据
SELECT
`category`.`name` AS MainCatName
, `subcategory`.`name` AS SubCatName
FROM
`category`
INNER JOIN `subcategory`
ON (`category`.`id` = `subcategory`.`catid`);
希望这能帮助你 此处MainCatName将显示类别名称字段数据。。。和 子类别名称将显示为子类别名称字段数据
SELECT
`category`.`name` AS MainCatName
, `subcategory`.`name` AS SubCatName
FROM
`category`
INNER JOIN `subcategory`
ON (`category`.`id` = `subcategory`.`catid`);
希望这能帮助你 此处MainCatName将显示类别名称字段数据。。。和 子类别名称将显示为子类别名称字段数据
SELECT
`category`.`name` AS MainCatName
, `subcategory`.`name` AS SubCatName
FROM
`category`
INNER JOIN `subcategory`
ON (`category`.`id` = `subcategory`.`catid`);
创建别名并选中“选择category.name作为catname,选择subcategory.name作为subcatname,选择category.catimage,选择subcategory.id,选择subcategory.catid作为category.id上的类别内部联接子类别”
创建别名并选中“选择category.name作为catname,subcategory.name作为catname,category.catimage,subcategory.id,subcategory.catid从category内部连接category上的子category.id=subcategory.catid”
创建别名并选中“选择category.name作为catname,subcategory.name作为subcategory,category.catimage,subcategory.id,subcategory.catid来自category.id=subcategory.catid上的category内部联接子类别“
创建别名并选中”选择category.name作为catname,subcategory.name作为subcatname,category.catimage,subcategory.id,subcategory.catid来自category.id=subcategory.catid上的category内部联接子类别“