Php SQL组内部联接无法获得所有结果
我真的很想让这个SQL正常工作。我不是专家,所以我真的搞不懂Php SQL组内部联接无法获得所有结果,php,sql,Php,Sql,我真的很想让这个SQL正常工作。我不是专家,所以我真的搞不懂 $sqlquery = " SELECT s.searchword AS searchword, s.id AS id, COUNT( c.id ) AS searchresult, s.region AS region FROM search_words AS s INNER JOIN company_data AS c ON c.branch_text LIKE
$sqlquery = " SELECT
s.searchword AS searchword,
s.id AS id,
COUNT( c.id ) AS searchresult,
s.region AS region
FROM search_words AS s
INNER JOIN company_data AS c ON
c.branch_text LIKE CONCAT( '%', s.searchword, '%' )
GROUP BY 1 ORDER BY s.date DESC";
这给了我:
Array
(
[0] => Array
(
[searchword] => WHOLESALE
[searchid] => 427
[searchresult] => 98
[region] => stockholm
)
[1] => Array
(
[searchword] => cars
[searchid] => 426
[searchresult] => 26
[region] =>
)
[2] => Array
(
[searchword] => Retail
[searchid] => 342
[searchresult] => 41
[region] => stockholm
)
[3] => Array
(
[searchword] => Springs
[searchid] => 339
[searchresult] => 4
[region] => stockholm
)
[4] => Array
(
[searchword] => Leasing
[searchid] => 343
[searchresult] => 2
[region] => stockholm
)
[5] => Array
(
[searchword] => Food
[searchid] => 340
[searchresult] => 37
[region] => stockholm
)
)
但是,它没有给我任何其他没有searchhits的结果,会返回类似[searchresult]=>0的结果。这意味着它们没有按照我的意愿进行分组,因为company_数据表中没有此类搜索词
如何修复此问题,请帮助:(
编辑:
以下是完整的代码:
public function getUserSearches()
{
$sqlquery = " SELECT
s.searchword AS searchword,
s.id AS id,
s.userId AS userid,
COUNT( c.id ) AS searchresult,
s.region AS region
FROM search_words AS s
INNER JOIN company_data AS c ON
c.branch_text LIKE CONCAT( '%', s.searchword, '%' )
GROUP BY 1 ORDER BY s.date DESC";
// IS THERE ANYTHING WRONG HERE?? LIKE IT DOES NOT MATCH AGAINST THE USER?
$result = $this->dbh->query($sqlquery, array(":userId" => $this->user_id));
$arr = array();
foreach ($result as $item) {
array_push($arr, array('searchword' => $item['searchword'], 'searchid' => $item['id'],
'searchresult' => $item['searchresult'], 'userid' => $item['userid'],
'region' => $item['region']));
}
return json_encode($arr);
return print_r($arr);
}
使用
LEFT JOIN
,这样任何与连接条件中的搜索条件不匹配的行都将为null,因此count
将为0。类似于以下内容:
SELECT
s.searchword AS searchword,
s.id AS id,
s.region AS region,
COUNT(COALESCE(c.id, 0)) AS searchresult
FROM search_words AS s
LEFT JOIN company_data AS c
ON c.branch_text LIKE CONCAT( '%', s.searchword, '%' )
GROUP BY s.searchword, s.id, s.region
ORDER BY s.date DESC;
有关SQL联接类型的更多详细信息,请参阅。您需要一个
左联接
,而不是内部联接
。但是,您也可以进行其他简化:
SELECT s.searchword, s.id, COUNT( c.id ) AS searchresult, s.region
FROM search_words s INNER JOIN
company_data c
ON c.branch_text LIKE CONCAT('%', s.searchword, '%')
GROUP BY 1
ORDER BY s.date DESC
例如,在不更改列名的情况下,不需要列别名。您仅在searchword
上进行聚合;我假设这在searchwords
中是唯一的。否则,s.id
,s.region
和s.date
将具有不确定的值
对
连接使用LIKE
会影响性能。如果您只有少量数据,这很好。否则,您可能需要考虑其他数据结构。我认为您需要的可能是:
public function getUserSearches()
{
$sqlquery = " SELECT
s.searchword AS searchword,
s.id AS id,
s.userId AS userid,
COUNT( c.id ) AS searchresult,
s.region AS region
FROM search_words AS s
WHERE s.userId='"+$this->userid+"'
LEFT JOIN company_data AS c ON
c.branch_text LIKE CONCAT( '%', s.searchword, '%' )
GROUP BY 1 ORDER BY s.date DESC";
// IS THERE ANYTHING WRONG HERE?? LIKE IT DOES NOT MATCH AGAINST THE USER?
$result = $this->dbh->query($sqlquery);
$arr = array();
foreach ($result as $item) {
array_push($arr, array('searchword' => $item['searchword'], 'searchid' => $item['id'],
'searchresult' => $item['searchresult'], 'userid' => $item['userid'],
'region' => $item['region']));
}
return json_encode($arr);
return print_r($arr);
}
你可以用左外连接代替内连接。告诉我这是否解决了你的问题。几乎:)它给了我更多的结果几乎所有的结果都是32分36秒,奇怪。而且它也没有检查用户id。这就是我获得结果的方式,添加用户id:$result=$This->dbh->query($sqlquery,array(“:userId”=>$This->userId));也许完全错了。对不起,我没有正确理解你的评论。如果你仍然面临一些问题,你能发布你的php代码吗?是的,现在看看。谢谢我不确定为什么要添加数组(“:userId”=>$this->user\u id)
,因为需要的参数是数据库链接。如果希望与userid匹配,那么可以修改查询以包含userid的where子句,并在替换userid之后添加一个参数。在查询
函数调用中,它没有任何意义。谢谢,但这一个不起作用。我得到了所有的结果,但我确实得到了[searchresult]=>0。@AdrianMcCool-你能从两个表search\u words
和company\u data
中发布一些示例记录来知道问题出在哪里吗?我编辑了上面的帖子,包括整个函数,也许更简单。@AdrianMcCool Thaks,但是我需要一些样本记录来运行查询,看看它为什么不起作用,每个表中只有3或4条记录作为样本。好吧,我不能给你company_数据表,它太多了。它只是检查branch_文本列是否匹配。search_words表只包含id、userId、searchword、date和region