爲陣列中的每個ID運行單獨的查詢

Question

嘗試運行數組中的項目的第二個查詢。 首先，我創建了一組用戶信息，然後爲每個用戶查詢他們在答案表中找到的答案。

這個想法是爲數組提供來自第一個查詢的數據，然後簡單地將答案中的數據添加到相關的用戶元素。下面的代碼只列出了第一個系統。

我一直都很不願意在這裏發表我的問題，但是我的整個一天都被這個消耗掉了，我沒有得到快速的地方。對於我明顯有缺陷的方法，事先抱歉。

Array 
(
    [0] => Array 
     (
      [fname] => asdf 
      [lname] => asdf 
      [minitial] => a 
      [rank] => MAJ 
      [uniq] => !s5$qn 

      [sysName] = System 1 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 

      [sysName] = System 2 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 

      [sysName] = System 3 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 


    [1] => Array 
     (
      [fname] => asdf 
      [lname] => lkjlkj 
      [minitial] => i 
      [rank] => oiuoi 
      [uniq] => @z26dr 

      [sysName] = System 1 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 

      [sysName] = System 2 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 

      [sysName] = System 3 Name 
      [choice] = The choice for This named system 
      [priority] = The priority 
      [termcom] = The termcom 


     ) 

// CODE

$sql = "SELECT fname, lname, minitial, rank, uniq FROM `user` join answers on answers.uniqid = user.uniq"; 
$data = mysqli_query($con, $sql) or die("MySQL ERROR: ". mysqli_error($con)); 

$users = array(); 
$i = 0; 

while ($row = mysqli_fetch_array($data, MYSQL_ASSOC)) 
{ 
    $users['answers'][$i] = array (
     "fname" => $row['fname'], 
     "lname" => $row['lname'], 
     "minitial" => $row['minitial'], 
     "rank" => $row['rank'], 
     "uniq" => $row['uniq'] 
    ); 

    $query2 = "SELECT a.sysid, s.sysName, uniqid, choice, priority, termcom FROM answers a LEFT JOIN systems s ON s.sysID = a.sysid WHERE a.uniqid = '" . $row['uniq'] . "'"; 
    $data2 = mysqli_query($con, $query2); 

    while ($row2 = mysqli_fetch_array($data2, MYSQL_NUM)) 
    {  
     $users['answers'][$i]['sysName'] = $row2[1]; 
     $users['answers'][$i]['choice'] = $row2[3];  
    } 

    $i++; 
} 

預先感謝您爲您可以共享任何見解。

編輯：這是數組回來，並且只有第一個系統正在爲每個用戶列出。

[2] => Array 
     (
      [fname] => asdf 
      [lname] => lkjlkj 
      [minitial] => i 
      [rank] => oiuoi 
      [uniq] => @z26dr 
      [sysName] => Super Terminate System 
     ) 

    [3] => Array 
     (
      [fname] => Juuu 
      [lname] => kjuuu 
      [minitial] => k 
      [rank] => LTC 
      [uniq] => gthdz% 
      [sysName] => Super Terminate System 
     ) 

Answer 1

O.K.你不完全知道當你運行腳本時會發生什麼（問題是什麼）。但是當我看着你的角色時，我可以假設出了什麼問題。

首先，我想你也可以用做它的查詢：

$sql = "SELECT * FROM `user` 
     join answers on answers.uniqid = user.uniq 
     LEFT JOIN systems s ON s.sysID = a.sysid"; 

希望這有助於。否則：

在你的第二個查詢中有一個問題。我想你應該改變：

$query2 = "SELECT a.sysid, s.sysName, uniqid, choice, priority, termcom FROM answers a LEFT JOIN systems s ON s.sysID = a.sysid WHERE a.uniqid = '" . $row['uniq'] . "'"; 

到

$query2 = "SELECT a.sysid, s.sysName, s.uniqid, s.choice, s.priority, s.termcom FROM answers a LEFT JOIN systems s ON s.sysID = a.sysid WHERE a.uniqid = '" . $row['uniq'] . "'"; 

在S。 before choice = s.choice