dousuize3082 2017-04-04 15:14
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将php数组转换为js selectpicker数组格式

This is how I build sql data and send back to ajax call:

    (...)
    $sql = "SELECT id_option FROM options WHERE id_win = '{$id_win}'";
    $result = mysqli_query($conn, $sql);
    $rows =  array();
    while ($row = mysqli_fetch_array($result)) {
      $rows[] =  $row;
    }

    $data["win_data"] = $rows;
    echo json_encode($data);

This is my ajax function to get options for selectpicker from DB:

$.ajax({
  type: "POST",
  url: ...,
  data: ...,
  dataType: "json",
  success: function(data)   {       
    $("#id_win").selectpicker("val", data.win_data);
   // $("#id_win").selectpicker("val", [1,3];
  }
});

data.win_data variable should be: [1,3]

but if I do: console.log(JSON.stringify(data.win_data)); I get:

[{"0":"1","id_option":"1"},{"0":"3","id_option":"3"}]

What is the simplest way to get proper format array for selectpicker?

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2条回答 默认 最新

  • duai4379 2017-04-04 15:23
    关注

    It is because $row is an array. Look: https://secure.php.net/manual/pt_BR/mysqli-result.fetch-array.php

    You have to use like this:

        (...)
        $sql = "SELECT id_option FROM options WHERE id_win = '{$id_win}'";
        $result = mysqli_query($conn, $sql);
        $rows =  array();
        while ($row = mysqli_fetch_array($result)) {
          $rows[] =  $row['id_option'];
        }
    
        $data["win_data"] = $rows;
        echo json_encode($data);
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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