duandanxiu6965 2017-04-23 16:27
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我如何解码json数组

I am using ajax to send data to a php page that queries mysql and returns the results. I have in my page, a variable that holds the dept value. However, if I var_dump($_POST) I see that the var is an array.

If I manually enter a value in the query, then data is returned. However, just not using my $dept var.

How do I decode this array to make the variable available for my query. Thanks

ajax code

$.ajax({
   url: 'deptdata.php',
   type: "POST",
   contentType: "application/json; charset=utf-8",
   data: {dept: depts},
   dataType: "json",
      success: function (data) {
        console.log(data);

   },
      error: function (data) {

        alert('error');

        }
      });

post in firebug tab

dept=DEMOBILL

deptdata.php

<?php

    $dept = $_POST['dept']; <--- array?
    //open connection to mysql db
    $connection = mysqli_connect("localhost","root","","sample") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select custref from boxes where department = '".$dept."' and status = 1";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row = mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }
    echo json_encode($emparray);

    //close the db connection
    mysqli_close($connection);
?>
  • 写回答

1条回答 默认 最新

  • dongyoufo5672 2017-04-23 16:31
    关注
    $myArray = json_decode($data, true);
    
    echo $myArray[0]['id']; // Fetches the first ID
    echo $myArray[0]['c_name']; // Fetches the first c_name
    // ...
    echo $myArray[2]['id']; // Fetches the third ID
    // etc..
    

    If you do NOT pass true as the second parameter to json_decode it would instead return it as an object:

    echo $myArray[0]->id;
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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