如何将数据库中的值传递给php中的json

Question

在此先感谢，我在表列中有三个薪水数据。我正在使用while循环在视图页面中显示三个薪水值。但我必须将这三个值传递给json中的三个变量，如{salary1：$ sal1，salary2：$ sal2，salary3：$ sal3}。如何单独循环三个薪金值到三个变量

My code as below: 

<table border="1"> 
     <caption><h2>View Registration</h2></caption> 
     <tr> 
      <th>Name</th> 
      <th>Designation</th> 
      <th>Email</th> 
      <th>Salary</th> 
     <tr> 
      <?php 

      include('common.php'); 
      $sql = mysql_query("select * from register"); 
      while($row = mysql_fetch_array($sql)) 
      { 

      ?> 
     <tr> 
      <td><?php echo $row['name']?></td> 
      <td><?php echo $row['designation']?></td> 
      <td><?php echo $row['email']?></td> 
      <td><?php 

      $salary = $row['salary'];  
      echo $salary; 


      ?></td> 
     <tr> 
      <?php 
      }   


      ?> 
     </table> 

Answer 1

创建所有工资的数组，并json_encode呼应，阵列将在转换成JSON作为你想，

$sql = mysql_query("select * from register"); 

$salary =array(); 
while ($row = mysql_fetch_array($sql)) { 
    $salary[$row['name']] = $row['salary']; 
} 
echo json_encode($salary); 

Answer 2

<table border="1"> 
     <caption><h2>View Registration</h2></caption> 
     <tr> 
      <th>Name</th> 
      <th>Designation</th> 
      <th>Email</th> 
      <th>Salary</th> 
     <tr> 
      <?php 

      include('common.php'); 
      $sql = mysql_query("select * from register"); 
      $arraySalary=array();// declaring an array for json array 
      $loopCounter=1; 
      while($row = mysql_fetch_array($sql)) 
      { 

      ?> 
     <tr> 
      <td><?php echo $row['name']?></td> 
      <td><?php echo $row['designation']?></td> 
      <td><?php echo $row['email']?></td> 
      <td><?php 

      $salary = $row['salary']; 
      $arraySalary['salary'.$loopCounter]= $row['salary']; // making an array os diffrent index as salry1, salary2, salary3. 
      $loopCounter++; 
      ?></td> 
     <tr> 
      <?php 
      } 
      echo json_encode($arraySalary); // here you will get the salary json.  


      ?> 
     </table> 

请尝试一下。我已经检查过了。并且工作正常。