如何将数据库中的值传递给php中的json
问题描述：
Thanks in advance, I have a three salary data in the table column. I am using while loop to display three salary values in the view page. But I have to pass these three values to three variables in json like {salary1:$sal1,salary2:$sal2,salary3:$sal3}. How to seperate looped three salary values into three variables
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>
答
create an array of all salary and echo with json_encode, array will be convert in to json as you wants,
$sql = mysql_query("select * from register");
$salary =array();
while ($row = mysql_fetch_array($sql)) {
$salary[$row['name']] = $row['salary'];
}
echo json_encode($salary);
答
<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>
Please have a try on this. I have checked this. and working fine.