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我已经检查过类似的答案。但我在这里有不同的问题!

要创建 googlce char,我通过从表中获取数据来创建 json 字符串。表只有两行。下面是 json 字符串的构造方式:

 $result = $mysqli->query('SELECT * FROM view_name');

   $p= $mysqli->query('SELECT Index_val FROM view_name where ind_type=pcount');
 //echo "$p";

  $rows = array();
  $table = array();
  $table['cols'] = array(
    array('label' => 'pcount', 'type' => 'string'),
    array('label' => 'ncount', 'type' => 'number')
);
    /* Extract the information from $result */
    foreach($result as $r) {
      $temp = array();
      $temp[] = array('v' => (string) $r['ind_type']); 
      $temp[] = array('v' => (int) $r['Index_val']); 
      $rows[] = array('c' => $temp);
    }

$table['rows'] = $rows;

这里我只需要从数据库pcountncount. 我不想使用上述程序。

我想要的是手动从表中获取值pcountncount存储在变量中并构造 json 字符串。

像这样: $p= $mysqli->query('SELECT Index_val FROM view_name where ind_type=pcount');//如果错误,请更正 get get ncount

现在我怎样才能从这个创建上面的 json 字符串?结果字符串将如下所示:

{
    "cols":[
        {"label":"pcount","type":"string"},
        {"label":"ncount","type":"number"}],
    "rows":[
        {"c":
            [
            {"v":"pcount"},
            {"v":179}  // 179 is pcount
            ]
        },
        {"c":
            [
            {"v":"ncount"},
            {"v":285} //285 is ncount
            ]
        }
        ]
}
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