Question

先谢谢，我在表格列中有三个薪资数据。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}。如何将循环的三个薪水值分成三个变量

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

使用json_encode创建所有薪水和回声的数组，数组将根据需要转换为json，

$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>

请试试这个。我检查过这个。并且工作正常。

如何将数据库中的值传递给php中的json

2 个答案: