Question

我意识到我使用的是mysql而不是mysqli。我需要在更新现有代码之前解决此问题。我目前通过php查询数据以及数据库中的表行名称。我无法弄清楚为什么我在运行json_encode时没有收到响应。有人可以帮帮我吗？

我想要的JSON格式输出如下..

{ "projects" : [
  {
    "title":"title 1",
    "image":"Image1",
    "subtitle":"subtitle 1",
    "about":"Blah Blah Blah"
  },
  {
    "title":"title 2",
    "image":"Image2",
    "subtitle":"subtitle 2",
    "about":"Blah Blah Blah"
  }
]}

现在我的代码看起来像这样..

// 1) Connect to mysql database
    $con = mysql_connect($host,$user,$pass);
    $dbs = mysql_select_db($databaseName, $con);

  // 2) Query database for data
  $result = mysql_query("SELECT * FROM $tableName");   
    if (mysql_num_rows($result) > 0) 
    {
      $numFields = mysql_num_fields($result);
      $fields = array();
      $i = 0;
      while ($i < $numFields) {
        $fieldInfo = mysql_fetch_field($result, $i);
        $fields[] = $fieldInfo->name;
        $i++;
      }

      $userInfo = array();
      $userInfo["projects"] = array();
      while($row = mysql_fetch_row($result)) {

          $project = array();
          $i = 0;
          while ($i < mysql_num_fields($result)) {
            $project[$fields[$i]] = $row[$i];
            $i++;
          }
          $userInfo["projects"][] = $project;
      }

      //echo '<pre>';
      //    print_r($userInfo);
      //echo '</pre>';

      echo json_encode($userInfo["projects"]);
    }
    else 
    { 
        echo ( "No rows found!" );
    }

当我执行上面的echo '<pre>'（当前已注释掉）时，我得到以下结果。但是当我删除它并尝试将数据作为JSON返回时，我得不到任何结果。有人可以帮帮我。

Array
(
    [projects] => Array
        (
            [0] => Array
                (
                    [title] => title
                    [subtitle] => subtitle 1
                    [about] => Blah Blah Blah
                    [image] => Image1
                )

            [1] => Array
                (
                    [title] => title 2

                    [subtitle] => subtitle 2

                    [about] => Blah Blah Blah
                    [image] => Image2
                )
         )
)

json_encode不适用于行名数组

0 个答案: