[英]a field from table A IN a field on table B + mysql + php
I have two tables with the following structure: 我有两个表具有以下结构:
tblA: TBLA:
+----------+---------------+ | id (int) | name (string) | +----------+---------------+ | 1 | a | | 2 | b | | 3 | c | +----------+---------------+
tblB: TBLB:
+----------+---------------+-------------+ | id (int) | name (string) | aid(string) | +----------+---------------+-------------+ | 1 | x | '1,2' | | 2 | y | '2,' | | 3 | z | '1,3' | +----------+---------------+-------------+
$a = $this::$db->prepare('SELECT * FROM tblB WHERE id= :id LIMIT 1');
$a->bindValue(':id', $ID, PDO::PARAM_INT);
$a->execute();
$r = $a->fetch(pdo::FETCH_ASSOC);
if ($a->rowCount() > 0){
$bInf = $r['id'] . '|*|' .
$r['name'] . '|*|' .
$r['aid'] . '|**|';
$b = $this::$db->prepare('SELECT id,name FROM tblA WHERE FIND_IN_SET(id,:ids)');
$b->bindValue(':ids', $r['aid']);
$b->execute();
$rs = $b->fetchAll(pdo::FETCH_ASSOC);
if ($b->rowCount() > 0)
{
foreach ($rs as $srow => $srval)
$aInf .= $srval['id'] . '[|]' .
$srval['name'] . '[#]' ;
} else
$aInf = ' ';
$aInf.= '|***|' . $bInf;
}
}
i need to query from tblA and tblB as above sample, but second query dont return any records. 我需要从tblA和tblB查询上面的示例,但第二个查询不返回任何记录。
i also tried 'IN' operator but dont worked too... 我也试过'IN'运算符但是也没用过......
pls help me... 请帮助我...
Try something like this: 尝试这样的事情:
SELECT * FROM tblb,tbla
WHERE tblb.id= 1 AND FIND_IN_SET(tbla.id, tblb.aid)
for details, refer to: 有关详细信息,请参阅:
You can use a different approach by extracting each of the ids from aid
column separately 您可以通过分别从
aid
列中提取每个id来使用不同的方法
$a = $this::$db->prepare('SELECT * FROM tblB WHERE id= :id LIMIT 1');
$a->bindValue(':id', $ID, PDO::PARAM_INT);
$a->execute();
$r = $a->fetch(pdo::FETCH_ASSOC);
if ($a->rowCount() > 0)
{
$bInf = $r['id'] . '|*|' .
$r['name'] . '|*|' .
$r['aid'] . '|**|';
//extract each of the ids in the variable 'aid'
$tbla_ids = explode(',',$r['aid']);
foreach($tbla_ids as $tbla_id){
//case for the record where aid = '2,'
if(strlen($tbla_id)==0){
continue;
}
$b = $this::$db->prepare('SELECT id,name FROM tblA WHERE id= :ids');
$b->bindValue(':ids', $tbla_id);
$b->execute();
//do what you need to do here. The query returns the single record
//from tbla that matches the id $tbla_id
}
}
我有一个有趣的错误,通过删除'字符从'援助'字段,问题解决了,FIND_IN_SET现在正常工作。
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