PHP搜索查询

我试图搜索一个CVs应用程序，它存储在许多表中，我试图找到一个包含特定信息的简历，如学历和教育

因此，我已通过POST发送了搜索条件：

但这些声明带来了所有具有19个或5个或13个学历的简历

这是我的代码：

$keys = array_keys($_POST);

$found = array();
foreach ($keys as $key) {
    $found[$key] = DataBase::query("SELECT * FROM cv_" . $key . " WHERE " . $key . "_id IN (?)", array($_POST[$key]));
}
echo count($found)."\n";
print_r($found)."\n";

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尝试按单个查询

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假设两个表都使用一个公共的resume_id，您可以使用该id将表连接到一个查询中，该查询将：

SELECT * FROM cv_qualifications q 
JOIN cv_educations e ON e.resume_id = q.resume_id
where q.qualifications_id IN (5,19)
and e.educations_id IN (13)

本例中的SELECT*仅从cv_资格中检索，如果您需要两个表中的字段，可以用SELECT q.*和e.*替换…

查看以下内容：-

SELECT * FROM cv_qualifications q 
JOIN cv_educations e ON e.resume_id = q.resume_id
where (q.qualifications_id = '5'
       or q.qualifications_id = '9')
and e.educations_id = '13'"

连接这两个表，然后根据需要执行where子句。您也不应该执行多次do1execution并动态构建查询。

$keys = array_keys($_POST);

$found = array();
foreach ($keys as $key) {
    $found[$key] = DataBase::query("SELECT * FROM cv_" . $key . " WHERE " . $key . "_id IN (?)", array($_POST[$key]));
}
echo count($found)."\n";
print_r($found)."\n";

select * FROM cv_qualifications as cq
 INNER JOIN cv_educations as ce ON cq.resume_id=ce.resume_id 
WHERE cq.qualifications_id IN (19,5) and ce.educations_id IN (13);

SELECT * FROM cv_qualifications q 
JOIN cv_educations e ON e.resume_id = q.resume_id
where q.qualifications_id IN (5,19)
and e.educations_id IN (13)

SELECT * FROM cv_qualifications q 
JOIN cv_educations e ON e.resume_id = q.resume_id
where (q.qualifications_id = '5'
       or q.qualifications_id = '9')
and e.educations_id = '13'"