從我的sql查詢中檢查表

Question

您好，我正在檢查表中的重複數據。我有一個問題，從哪裏選擇數據。我的代碼是：

$sub_cat = array(); 
$select = array("core_network","daisy_chain", "rf_bts", "rf_power", "rf_transmission"); 
$d='0'; 
for ($i=0;$i<=4;$i++){ 
    $SQL = "SELECT sub_cat FROM (".$select[$i].") WHERE location=('".$id."')"; 
    $result = mysql_query($SQL); 

    $cs=$d; 

    if ($result) { 
     while ($db_field = mysql_fetch_array($result)) { 
      if(!in_array($db_field['sub_cat'],$sub_cat)) { 
       $sub_cat[]= $db_field['sub_cat']; 
       $cs++; 
       $d=$cs; 
       $d--; 
      } 
     } 
    } 

我需要知道從哪個$select[i]選擇sub_cat。如何找到它？

Answer 1

要得到的值，這樣做：

$sub_cat = array(); 
$select = array("core_network","daisy_chain", "rf_bts", "rf_power", "rf_transmission"); 
$d='0'; 
for ($i=0;$i<=4;$i++){ 
    $SQL = "SELECT sub_cat FROM (" . $select[$i] . ") WHERE location=('".$id."')"; 
    $result = mysql_query($SQL); // deprecated - use PDO 

    $cs = $d; 

    if ($result) { 
     while ($db_field = mysql_fetch_array($result)) { 
      if(!in_array($db_field['sub_cat'], $sub_cat)) { 
       $table = $select[$i]; 
       $sub_cat[$table][] = $db_field['sub_cat']; 

       // I have no clue what's going on here in your example: 
       $cs++; 
       $d=$cs; 
       $d--; 
      } 
     } 
    } 
} 

然後，找回它：！

foreach ($sub_cat as $table_name => $values) { 
    foreach ($values as $row) { 
     // output values here 
    } 
}