PHP：以特定顺序和特定条件显示2D数组？

我有一个数据库，它有四个表，为了简单起见，我将它们称为表1，表2，表3，表4

每个表都有不同的信息，但它们都共享一个唯一的id，sessionid

我已经通过PHP数组中的PHPMyAdmin导出了所有四个表。这给了我一个名为mydomain.php的文件

其中，数据结构如下所示：

$table1 = array(
  array('id'=>1,'sessionid'=>'12','field1_1'=>'data','field2_1'=>'data','field3_1'=>'data, 'done'=>1),
  array('id'=>2,'sessionid'=>'13','field1_1'=>'data','field2_1'=>'data','field3_1'=>'data, 'done'=>0)
);

$table2 = array(
  array('id'=>4,'sessionid'=>'12','field1_2'=>'data','field2_2'=>'data','field3_2'=>'data),
  array('id'=>6,'sessionid'=>'13','field1_2'=>'data','field2_2'=>'data','field3_2'=>'data)
);

$table3 = array(
  array('id'=>2,'sessionid'=>'12','field1_3'=>'data','field2_3'=>'data','field3_3'=>'data),
  array('id'=>5,'sessionid'=>'13','field1_3'=>'data','field2_3'=>'data','field3_3'=>'data)
);

$table4 = array(
  array('id'=>6,'sessionid'=>'12','field1_4'=>'data','field2_4'=>'data','field_43'=>'data),
  array('id'=>1,'sessionid'=>'13','field1_4'=>'data','field2_4'=>'data','field3_4'=>'data)
);

我希望根据会话id显示它们（如您所见，每个表的id可能不同），我希望检查第一个表中的字段“done”是否设置为1或0，并跳过其中为空或包含“

NULL

”的字段

我尝试过一些不同的方法，比如相互嵌套的foreach循环，但似乎不太管用

希望你能帮助解决这个问题

真诚的

---

-Mestika

为什么不使用SQL？这种语言就是为了做这样的事情而设计的

• 使用JOIN语句合并四个表的内容
• 使用SELECT语句检索字段[done]
• 使用WHERE筛选空字符串和空值

最后，使用PHP和mySQL特性直接检索表值，而不是使用phpMyAdmin导出数据

示例SQL语句：

SELECT t1.sessionid, t1.done
FROM table1 AS t1
JOIN table2 AS t2 ON t1.sessionid = t2.sessionid
JOIN table3 AS t3 ON t2.sessionid = t3.sessionid
JOIN table4 AS t4 ON t3.sessionid = t4.sessionid
WHERE t1.done IS NOT NULL

PHP资源：

SELECT t1.sessionid, t1.done
FROM table1 AS t1
JOIN table2 AS t2 ON t1.sessionid = t2.sessionid
JOIN table3 AS t3 ON t2.sessionid = t3.sessionid
JOIN table4 AS t4 ON t3.sessionid = t4.sessionid
WHERE t1.done IS NOT NULL

---

为什么不使用SQL？这种语言就是为了做这样的事情而设计的
• 使用JOIN语句合并四个表的内容
• 使用SELECT语句检索字段[done]
• 使用WHERE筛选空字符串和空值
最后，使用PHP和mySQL特性直接检索表值，而不是使用phpMyAdmin导出数据
示例SQL语句：

SELECT t1.sessionid, t1.done
FROM table1 AS t1
JOIN table2 AS t2 ON t1.sessionid = t2.sessionid
JOIN table3 AS t3 ON t2.sessionid = t3.sessionid
JOIN table4 AS t4 ON t3.sessionid = t4.sessionid
WHERE t1.done IS NOT NULL

PHP资源：

SELECT t1.sessionid, t1.done
FROM table1 AS t1
JOIN table2 AS t2 ON t1.sessionid = t2.sessionid
JOIN table3 AS t3 ON t2.sessionid = t3.sessionid
JOIN table4 AS t4 ON t3.sessionid = t4.sessionid
WHERE t1.done IS NOT NULL

---

这不是最有效的解决方案，但它是直截了当的

foreach ($table1 as $data) {
   if ($data['done']===0 || $data['done']===1) {
      print_r($data);
      echo "Data table 2:";
      foreach ($table2 as $data2) {
         if ($data2['sessionid']===$data['sessionid']){
            print_r($data2);
         }
      }
      echo "Data table 3:";
      foreach ($table3 as $data3) {
         if ($data3['sessionid']===$data['sessionid']){
            print_r($data3);
         }
      }
      echo "Data table 4:";
      foreach ($table4 as $data4) {
         if ($data3['sessionid']===$data['sessionid']){
            print_r($data4);
         }
      }
   }
}

---

我没有测试它，我只是简单地写了下来，希望没有打字错误。
这不是最有效的解决方案，但它是直截了当的

foreach ($table1 as $data) {
   if ($data['done']===0 || $data['done']===1) {
      print_r($data);
      echo "Data table 2:";
      foreach ($table2 as $data2) {
         if ($data2['sessionid']===$data['sessionid']){
            print_r($data2);
         }
      }
      echo "Data table 3:";
      foreach ($table3 as $data3) {
         if ($data3['sessionid']===$data['sessionid']){
            print_r($data3);
         }
      }
      echo "Data table 4:";
      foreach ($table4 as $data4) {
         if ($data3['sessionid']===$data['sessionid']){
            print_r($data4);
         }
      }
   }
}

我没有测试，我只是简单地写了下来，希望没有打字错误