Question

我有一个搜索表单，用户可以输入任何字母单词并进行搜索这个搜索在两个表中表1是tbl_project 此表包含

db_projectname db_location db_client  
xxxx              vvv      ccc

dddd              ggg      tttt

tbl_activities包含

db_projectname  db_cost  db_name   
xxxx             500     ttt
xxxx             500     hhh
dddd             200     llll

我使用这个PHP代码进行搜索

$sq = ""; 
$qq=array();
    if(isset($_POST['search']) && !empty($_POST['search'])){  
    $search =  mysqli_real_escape_string($conn,$_POST['search']); 
    $qq[] = "tbl_project.db_projectname like '%".$search."%' ";  
    $qq[] = "tbl_project.db_location like '%".$search."%' ";
    $qq[] = "tbl_project.db_client like '%".$search."%' ";
    $qq[] = "tbl_project.db_offer like '%".$search."%' ";    
    $qq[] = "tbl_project.db_sheet like '%".$search."%' ";
    $qq[] = "tbl_project.db_invoice like '%".$search."%' ";
    $qq[] = "tbl_project.db_po like '%".$search."%' ";
    $qq[] = "tbl_project.db_id like '%".$search."%' ";    
    } 
$second=true;
       foreach($qq as $que){  
        if($second){  
        $sq .= " where ".$que;      
        $second = false;  
        }else{  
        $sq .= " or ".$que;          
        }}

如果用户输入x或v或c或xxxx或vvv或ccc

，结果应如下所示

xxxx 1000 vvv cccc

查询应如何获得此结果我已尝试不同的查询但我没有收到此结果

SELECT  tbl_project.* 
      , (SELECT  sum(tbl_activities.db_totalcost) AS total_cost FROM tbl_activities {$sq}
FROM tbl_project

Answer 1

尝试这种方法：

$cond = "";
if(isset($_POST['search']) && !empty($_POST['search'])){  
    $search =  mysqli_real_escape_string($conn,$_POST['search']); 
    $cond .= " and pa.db_projectname like '%".$search."%' ";  
    $cond .= " and pa.db_location like '%".$search."%' ";
    $cond .= " and pa.db_client like '%".$search."%' ";
    $cond .= " and pa.db_offer like '%".$search."%' ";    
    $cond .= " and pa.db_sheet like '%".$search."%' ";
    $cond .= " and pa.db_invoice like '%".$search."%' ";
    $cond .= " and pa.db_po like '%".$search."%' ";
    $cond .= " and pa.db_id like '%".$search."%' ";    
}

查询将是：

select p.*,sum(pa.db_totalcost) from tbl_project p left join 
tbl_activities pa on p.db_projectname=pa.db_projectname 
where 1 $cond group by p.db_projectname

Answer 2

我想你想要这样的东西：

SELECT p.db_projectname, IFNULL(a.total, 0) as total, p.db_location, p.db_client FROM tbl_project as p
LEFT JOIN (select db_projectname, sum(db_cost) as total from tbl_activities group by db_projectname) as a
    ON p.db_projectname = a.db_projectname

从第二张表中搜索第一张表总和

2 个答案: