Question

我刚刚意识到自己的错误！我两次使用表jokecategory ...学习PHP是一种令人羞愧的经历......

感谢您的帮助，到目前为止一切顺利！

抱歉，我没有粘贴SQL，但你的建议仍有帮助。我修改了SQL查询以使用“AS”，但我认为我将某些内容与“类别”表混合在一起。无法弄清楚是什么，但它现在不允许我根据类别进行搜索。有任何想法吗？我收到以下错误：

错误：不唯一的表/别名：'jokecategory'

SQL查询：

    $select = 'SELECT DISTINCT joke.id, joke.joketext, author.id AS author_name, author.name AS author_name, jokecategory.jokeid AS cat_jokeid, jokecategory.categoryid AS joke_catid, category.id AS cat_id, category.name as cat_name'; 
$from   = ' FROM joke, author, jokecategory, category'; 
$where = ' WHERE joke.authorid = author.id AND joke.id = jokecategory.jokeid AND jokecategory.categoryid = category.id';


$aid = $_POST['aid']; 
if ($aid != '') { // An author is selected
    $where .= " AND authorid='$aid'";
}

$cid = $_POST['cid']; 
if ($cid != '') { // A category is selected
    $from .= ', jokecategory'; 
    $where .= " AND joke.id=jokecategory.jokeid AND categoryid='$cid'";
}

$tid = $_POST['tid'];
if ($tid != '') { // A theme is selected
    $from .= ', joketheme';
    $where .= " AND joke.id=joketheme.jokeid AND themeid='$tid'";
}

$gfid = $_POST['gfid'];
if ($gfid != '') { // A region is selected
    $from .= ', jokegeofocus';
    $where .= " AND joke.id=jokegeofocus.jokeid AND geofocusid='$gfid'";
}

$searchtext = $_POST['searchtext']; 
if ($searchtext != '') { // Some search text was specified
    $where .= " AND joketext LIKE '%$searchtext%'";
    }

?>

Answer 1

我们没有看到您的SQL查询，但您应该使用AS关键字为不明确的列提供列别名，如：

SELECT 
  t1.name AS t1name,
  t2.name AS t2name
FROM
  t1 
  JOIN t2 ON t1.id = t2.id

然后可以通过别名在提取的行中访问它们：

while ($joke = mysql_fetch_array($jokes)) { 
  echo "Table 1 name: {$joke['t1name']}\n";
  echo "Table 2 name: {$joke['t2name']}\n";
}

进一步的建议（尽管我们没有看到它可能没有实际意义，因为它可能是没有意义的），是明确你选择哪些列。

而不是做

SELECT t1.*, t2.*

更好的做法是明确列出您需要的列：

SELECT t1.name, t2.id, t2.category .... t2.whatever

Answer 2

这可以通过对$ jokes中的sql稍作修改来完成

SELECT j.name as jokename, a.name as authorname 
FROM joke j, author a 
WHERE j.author_id = j.id

然后，您可以通过$ joke ['jokename']和作者姓名列通过$ joke ['authorname']

引用笑话的名称列

PHP回显多个表中的模糊名称

2 个答案: