如何检查 mysql 表的布尔字段并将 0 或 1 值转换为 php 中的真/假

Question

在我的查询中，我有 3 个 tinyint 字段，通过获取它们会给出 0/1 值。但是在获取所有数据后，我想将该结果编码为 JSON 格式，该格式应该得到 true/false 而不是 0/1。 在我的代码中，我已经完成了类型转换，这给出了正确的结果。但我应该为所有我不想要的 tinyint 字段写。 有什么方法可以检查 tinyint 字段并进行转换。

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);

请帮忙！！ 提前致谢。

Answer 1

在 PHP 中没有真正的方法来检查 tinyint

你可以做的是在你的变量前面加上一个 b_ 并使用 regex/strpos/substr 过滤它以将其类型转换为布尔值。这样，您只需调整您的查询，脚本就会为您进行转换。

$result= mysql_query("select incident_main.inc_id as incident_id,
                      ward_master.ward_id as 'ward_id', 
                      GROUP_CONCAT(incident_log.inclog_date) as revisit_dates ,
                      inc_GISlat as lat,
                      inc_GISlon as lon,
    --->              inc_closed as  b_is_closed,
    --->              inc_reviewed as b_is_reviewed,
                      inc_closeDate as closed_date,
                      incident_statuscd.inc_status_desc as status, 
                      inc_date as 'incident_date', 
                      inc_patientName as 'patient_name',
                      inc_patientAge as 'patient_age',
                      inc_patientGender as 'patient_gender',
                      inc_patientMobile as'patient_mobile' ,
                      inc_patientAddress as 'patient_address',
                      inc_type as type,
                      inc_typeOther as 'type_other',
                      inc_diagnosis as diagnosis,
                      inc_recurrence as recurrence,
    --->              inc_hospitalized as b_hospitalized, 
                      inc_treatment as treatment ,
                      inc_treatmentOther as 'treatment_other' 
                  from 
                      incident_main,
                      incident_statuscd ,
                      ward_master,
                      incident_log 
                  where 
                      inc_status=incident_statuscd.inc_status_id 
                     and 
                      incident_main.inc_patientWard=ward_master.ward_id 
                     and incident_log.inc_id=incident_main.inc_id ") 
                      or die(mysql_error);
while($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);
}

编辑

添加了一种将布尔标记的键恢复为正常代码执行的原始值的方法。 是的，它是重复的，但可能有助于进一步实现/代码执行，而无需将所有内容重命名为 b_

Answer 2

很遗憾，在 PHP 中无法区分 TINYINT 和 integer。因此，您需要列出要转换的字段。其缩写版本如下所示：

$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}

当且仅key 列在$bools 数组中时，后者会将integer 转换为bool。更多详情请参考array_map docs。

希望对你有帮助。

Answer 3

尝试编写另一个函数：

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

当使用 0 或 1 调用时返回文本“true”或“false”。然后为每个要编码的 tinyint 添加调用：

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

Answer 4

JS代码：

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