使用CakePHP從MySQL存儲過程中檢索輸出變量

Question

我試圖使用CakePHP調用存儲過程。

當前，返回值由存儲過程中第一條SQL select語句的第一條記錄集組成。

儘管在存儲過程（即select @project_id into project_id）中設置了輸出變量，但它並未顯示在查詢結果的var_dump中。

存儲過程：

CREATE DEFINER = 'admin'@'%' 
PROCEDURE thebuggenie.cmdb_project_team_init(
    IN project_name VARCHAR(200), 
    IN project_key VARCHAR(200), 
    IN project_homepage VARCHAR(200), 
    IN team_name VARCHAR(200), 
    OUT project_id INT(10)) 
BEGIN 
    -- start transaction 
    start transaction; 

    -- init variables 
    set @project_id = 0; 
    set @team_id = 0; 
    set @assoc_count = 0; 
    set @scope_id = 1; 

    -- select team and set variable 
    select @team_id := id 
    from tbg3_teams 
    where name = team_name; 

    -- if team_id = 0, insert team and set variable 
    if @team_id is NULL or @team_id = '' or @team_id = 0 then 
    -- insert new project 
    insert into tbg3_teams(ondemand, name, scope) values(0, team_name, @scope_id); 
    -- set team_id variable 
    set @team_id = LAST_INSERT_ID(); 
    end if; 

    -- select project and set variable 
    select @project_id := id 
    from tbg3_projects 
    where name = project_name; 

    -- if project_id = 0, insert project and set variable 
    if @project_id is NULL or @project_id = '' or @project_id = 0 then 
    -- insert project 
    insert into tbg3_projects (name, locked, use_scrum, `key`, homepage, deleted, owner_team, scope, workflow_scheme_id, issuetype_scheme_id) values(project_name, 0, 1, project_key, project_homepage, 0, @team_id, @scope_id, 1, 1); 
    -- set project_id variable 
    set @project_id = LAST_INSERT_ID(); 
    end if; 

    select @assoc_count := count(*) 
    from tbg3_projectassignedteams 
    where uid = @team_id 
    and project_id = @project_id; 

    if(@assoc_count = 0 and @project_id > 0 and @team_id > 0) then 
    insert into tbg3_projectassignedteams (project_id, role_id, uid, scope) values(@project_id, 35, @team_id, @scope_id); 
    end if; 

    -- setup default views 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (101, 0, 0, @project_id, 2, 1); 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (102, 0, 0, @project_id, 2, 1); 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (110, 0, 0, @project_id, 2, 1); 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (105, 0, 0, @project_id, 2, 1); 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (106, 0, 0, @project_id, 2, 1); 
    INSERT INTO tbg3_dashboard_views (name, view, pid, tid, target_type, scope) VALUES (111, 0, 0, @project_id, 2, 1); 

    commit; 

    -- return values 
    select @project_id INTO project_id; 
END 

PHP代碼：

$sql = "call thebuggenie.cmdb_project_team_init("; 
$sql .= '\''.$results[0]['Asset']['project_name'].'\''; 
$sql .= ',\''.$results[0]['Asset']['project_name'].'\''; 
$sql .= ',\'\''; 
$sql .= ',\''.$results[0]['Repository']['team_name'].'\''; 
$sql .= ',@project_id'; 
$sql .= ');'; 
$sql .= 'select @project_id as project_id'; 

var_dump($sql); 

$results = $this->Asset->query($sql); 

print_r($results); 

PHP代碼輸出：

string 'call thebuggenie.cmdb_project_team_init('CMDB','CMDB','','team-app-platforms',@project_id);select @project_id as project_id;' 

Array ([0] => Array ([0] => Array ([@team_id := id] => 6))) 

注：我還沒有最終確定錯誤陷阱呢。

Answer 1

我結束了使用此：

$sql = "call thebuggenie.cmdb_project_team_init("; 
$sql .= '\''.$results[0]['Asset']['project_name'].'\''; 
$sql .= ',\''.$results[0]['Asset']['project_name'].'\''; 
$sql .= ',\'\''; 
$sql .= ',\''.$results[0]['Repository']['team_name'].'\''; 
$sql .= ',@project_id'; 
$sql .= '); select @project_id as project_id'; 

var_dump($sql); 

$mysqli = new mysqli("DB_HOST", "DB_USER", "DB_PWD", "DATABASE"); 
if ($mysqli->connect_errno) { 
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error; 
} 

$out = array(); 

if($mysqli->multi_query($sql)) 
{ 
    do 
    { 

     if($result=$mysqli->store_result()) 
     { 
      while($row=$result->fetch_row()) 
      { 
       array_push($out, $row); 
      } 
      $result->free(); 
     } 

    }while($mysqli->more_results() && $mysqli->next_result()); 
} 


$mysqli->close(); 

$出[3] [0]包含我一直在尋找的價值。