我有兩個表使用INNER JOIN出現在哪裏用戶信息頁面。 現在我需要創建另一張桌子,我無法將3張桌子一起做功能車。選擇3個表格並在其中
遵循舊代碼(有兩個表)和當前錯誤代碼(有三個表)。
OLD CODE:
// 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'];
}
當前代碼
// 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'];
}
預先感謝您的幫助。
看起來你可能需要以某種方式連接表'c'和't'?除此之外,在查詢中沒有出現任何問題。你得到什麼錯誤? – Fluffeh
找不到用戶 –
如果未找到用戶可能是因爲您正在加入連接子句的所有表上不存在的數據。 –