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Question

这是我做的一个搜索功能，它获取搜索词并显示正常。 HM_Jobs 有 3 个 JobStatus 选项： Open、Accepted 或 Complete。

但是，当我运行它时，搜索还会从 Accepted 和 Complete 行中提取结果，为什么 WHERE 语句没有阻止这种情况发生？

function search_jobs($search) //This function checks a search term against the job title and description.
{
    $this->db->select('*');
    $this->db->from('HM_Jobs');
    $this->db->like('JobTitle',$search);
    $this->db->or_like('JobDescription',$search);
    $this->db->where('JobStatus','Open');  

    $query = $this->db->get();
    return $query->result(); 
}

Answer 1

试试这个

 $this->db->select('*');
 $this->db->from('HM_Jobs');
 $this->db->where("(JobTitle LIKE '$search' OR JobDescription LIKE '$search')"  );
 $this->db->where('JobStatus','Open'); 
 $query = $this->db->get();

您可以使用echo $this->db->last_query()查看您的查询执行情况

您的查询创建类似

where JobTitle like 'search' or JobDescription  like 'search' and JobStatus='Open'

但是您需要的查询需要类似

  where (JobTitle like 'search' or JobDescription  like 'search') and JobStatus='Open'

Answer 2

尝试使用以下查询

function search_jobs($search)
{
    $query = $this->db->select('*')
                  ->from('HM_Jobs')
                  ->where('JobStatus','Open')
                  ->where("(`JobTitle` LIKE '%$search%' OR 
                            `JobDescription` LIKE '%$search%')",null,false)
                  ->get();
    return $query->result(); 
}