我有两个使用INNER JOIN的表出现在用户信息页面的位置。现在我需要创建另一个表,但我无法将 3 个表放在一起进行功能。
遵循旧代码(包含两个表),并在当前代码下方出现错误(包含三个表)。
旧代码:
// Pega subdomínio
$urlExplode = explode('.', $_SERVER['HTTP_HOST']);
if (count($urlExplode) > 2 && $urlExplode[0] !== 'www') {
$subdomain = $urlExplode[0];
// echo $subdomain;
}
// Diz que o usuário é igual ao subdomínio
$usuario = $subdomain;
// Select DB da Tabela TEXTOS
$sql = "SELECT * FROM vms_textos i INNER JOIN vms_users u on u.id = i.id where u.usuario='$usuario'";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error());
// TODO: better error handling
}
else {
$row = mysql_fetch_array($result);
// Tabela Textos
$userKeywords = $row['userKeywords'];
$userDesc = $row['userDesc'];
$userTitleSite = $row['userTitleSite'];
$userTextSobre = $row['userTextSobre'];
$userTextContatos = $row['userTextContatos'];
$userTextMaisInfos = $row['userTextMaisInfos'];
}
当前代码
// Pega subdomínio
$urlExplode = explode('.', $_SERVER['HTTP_HOST']);
if (count($urlExplode) > 2 && $urlExplode[0] !== 'www') {
$subdomain = $urlExplode[0];
// echo $subdomain;
}
// Diz que o usuário é igual ao subdomínio
$usuario = $subdomain;
// Select DB da Tabela TEXTOS
$sql = "SELECT * FROM (vms_textos t INNER JOIN vms_users u ON u.id = t.id) INNER JOIN vms_cores c ON u.id = c.id where u.usuario='$usuario'";
$result = mysql_query($sql);
if($result === FALSE) {
die(mysql_error());
// TODO: better error handling
}
else {
$row = mysql_fetch_array($result);
// Tabela Textos
$userKeywords = $row['userKeywords'];
$userDesc = $row['userDesc'];
$userTitleSite = $row['userTitleSite'];
$userTextSobre = $row['userTextSobre'];
$userTextContatos = $row['userTextContatos'];
$userTextMaisInfos = $row['userTextMaisInfos'];
}
提前感谢您的帮助。
听起来你需要使用外部连接而不是(来自评论)
SELECT
*
FROM
vms_textos t
INNER JOIN vms_users u
ON u.id = t.id
left outer JOIN vms_cores c
ON u.id = c.id
where
u.usuario='$usuario'
基本上,使用内部联接,如果数据不存在于所有表中,则不会返回。使用内部/外部连接的组合,您可以确定需要带回的内容。
根据下面的注释,查询现在是:
SELECT
*
FROM
vms_cores c
INNER JOIN vms_users u
ON u.id = t.id
left outer JOIN vms_textos t
ON u.id = c.id
where
u.usuario='$usuario'
阅读我写的这个问答也可能是一个好主意,它更详细地详细介绍了这个答案的细节。
我从不喜欢这种 sql 92 模式,我觉得很难阅读,所以,这是一种更简单的方法:
$sql = "SELECT * FROM vms_textos t ,
vms_users u ,
vms_cores c
where u.id = t.id
and u.id = c.id
and u.usuario='$usuario'";