如何在php中檢查mysql表的布爾字段並將0或1值轉換為true / false

Question

在我的查詢中，我有3個tinyint字段，通過提取給出0/1值。 但是在獲取所有數據之后我想將該結果編碼為JSON格式，該格式應該為真/假而不是0/1。 在我的代碼中，我完成了類型轉換，從而得到了正確的結果。 但我應該為所有我不想要的tinyint字段寫。 是否有任何方法可以檢查tinyint字段和轉換。

PHP CODE: 


$result= mysql_query("select incident_main.inc_id as incident_id,
                      ward_master.ward_id as 'ward_id', 
                      GROUP_CONCAT(incident_log.inclog_date) as revisit_dates ,
                      inc_GISlat as lat,
                      inc_GISlon as lon,
            inc_closed as  b_is_closed,
              inc_closeDate as closed_date,
                      incident_statuscd.inc_status_desc as status, 
                      inc_date as 'incident_date', 
                      inc_patientName as 'patient_name',
                      inc_patientAge as 'patient_age',
                      inc_patientGender as 'patient_gender',
                      inc_patientMobile as'patient_mobile' ,
                      inc_patientAddress as 'patient_address',
                      inc_type as type,
                      inc_typeOther as 'type_other',
                      inc_diagnosis as diagnosis,
                      inc_recurrence as recurrence,
              inc_hospitalized as b_hospitalized, 
                      inc_treatment as treatment ,
                      inc_treatmentOther as 'treatment_other' 
                  from 
                      incident_main,
                      incident_statuscd ,
                      ward_master,
                      incident_log 
                  where 
                      inc_status=incident_statuscd.inc_status_id 
                     and 
                      incident_main.inc_patientWard=ward_master.ward_id 
                     and incident_log.inc_id=incident_main.inc_id ") 
                      or die(mysql_error);
while($res = mysql_fetch_assoc($result)) {
 foreach($res as $key => $value) {
     if(substr( $key, 0, 2 ) === "b_") {
         // assign new value
         $res[substr($key,2,strlen($key))] = !!$value;
         //clean up old value
         unset($res[$key]);
         };

  }
  $data=array_filter($res);

echo json_encode($data, true);

請幫忙！！ 提前致謝。

Answer 1

不幸的是，在PHP中無法區分TINYINT和integer 。 因此，您需要列出要轉換的字段。 縮短版本如下所示：

$a = ['a' => 1, 'b' => 100, 'c' => 0 ];
$bools = ['a', 'c'];   // list of fields to transform

$result = array_map(function($item) use($bools,&$a) { 
  $key=key($a); next($a);  // we need a key to check it’s name
  return in_array($key, $bools) ? (bool)$item : $item; 
}, $a);

print_r(json_encode($result, true));
//⇒ {"a":true,"b":100,"c":false}

后者將integer為bool如果且只有key在$bools數組中列出。 有關詳細信息，請參閱array_map文檔 。

希望能幫助到你。

Answer 2

在PHP中沒有真正的方法來檢查tinyint

您可以做的是使用b_前置變量並使用regex / strpos / substr來過濾它以將其轉換為布爾值。 這樣您只需調整查詢，腳本就會為您進行投射。

$result= mysql_query("select incident_main.inc_id as incident_id,
                      ward_master.ward_id as 'ward_id', 
                      GROUP_CONCAT(incident_log.inclog_date) as revisit_dates ,
                      inc_GISlat as lat,
                      inc_GISlon as lon,
    --->              inc_closed as  b_is_closed,
    --->              inc_reviewed as b_is_reviewed,
                      inc_closeDate as closed_date,
                      incident_statuscd.inc_status_desc as status, 
                      inc_date as 'incident_date', 
                      inc_patientName as 'patient_name',
                      inc_patientAge as 'patient_age',
                      inc_patientGender as 'patient_gender',
                      inc_patientMobile as'patient_mobile' ,
                      inc_patientAddress as 'patient_address',
                      inc_type as type,
                      inc_typeOther as 'type_other',
                      inc_diagnosis as diagnosis,
                      inc_recurrence as recurrence,
    --->              inc_hospitalized as b_hospitalized, 
                      inc_treatment as treatment ,
                      inc_treatmentOther as 'treatment_other' 
                  from 
                      incident_main,
                      incident_statuscd ,
                      ward_master,
                      incident_log 
                  where 
                      inc_status=incident_statuscd.inc_status_id 
                     and 
                      incident_main.inc_patientWard=ward_master.ward_id 
                     and incident_log.inc_id=incident_main.inc_id ") 
                      or die(mysql_error);
while($res = mysql_fetch_assoc($result)) {
 foreach($res as $key => $value) {
     if(substr( $key, 0, 2 ) === "b_") {
         // assign new value
         $res[substr($key,2,strlen($key))] = !!$value;
         //clean up old value
         unset($res[$key]);
     }
  }
  $data=array_filter($res,'not_null');
}


function not_null($test) {
   return !($test === null);
}

編輯

添加了一種方法，將布爾標記的鍵恢復為正常代碼執行的原始值。 是的，它是重復的，但可能有助於進一步實現/代碼執行，而無需將所有內容重命名為b_

Answer 3

嘗試編寫另一個函數：

private function isTrue($input){
     if ((bool)$input){
          return "true";
     }else{
          return "false";
     }         
}

當使用0或1調用時，返回文本“true”或“false”。 然后為每個要編碼的tinyint添加調用：

...
$row['hospitalized'] = isTrue($row['hospitalized']);
...

Answer 4

JS代碼：

if(1) 
    alert('true');
if(0)
    alert('false');