Question

我从mysql中获取数据，如...

$query = "SELECT legal_country, count(*) FROM accounts GROUP BY legal_country ORDER BY count(*) DESC";
$result = mysql_query( $query );

$data = array();
    while ( $row = mysql_fetch_assoc( $result ) ) {
        $data[] = $row;
    }

return json_encode( $data );

目前的结果是：

[{"legal_country":"US","count(*)":"107865"},{"legal_country":"DE","count(*)":"44711"},{"legal_country":"SL","count(*)":"1"}]

然而，我的目标是让json字符串像：

{"map": "worldLow", "areas": [{"id": "US","value": "107865"},{"id": "DE","value": "44711"},{"id": "SL","value": "1"}]}

如何在数组之前添加{“map”：“worldLow”，“areas”：....}以及如何更换标签legal_country - ＆gt; id和count（*） - ＆gt;价值？

此外，这是一个带有'echo'的工作示例......但是我需要'return'功能......

$prefix = '';
echo '{"map": "worldLow", "areas": [';
while ( $row = mysql_fetch_assoc( $result ) ) {
    echo $prefix . " {\n";
    echo '  "id": "' . $row['legal_country'] . '",' . "\n";
    echo '  "value": "' . $row['count(*)'] . '"' . "\n";
    echo " }";
$prefix = ",\n";
}

echo "\n]}";

echo的工作示例如下所示：

{"map": "worldLow", "areas": [ {
  "id": "US",
  "value": "107865"
 },
 {
  "id": "DE",
  "value": "44711"
 },
 {
  "id": "SL",
  "value": "1"
 }
]}

Answer 1

在使用$data

之前，您必须稍微更改一下您的查询和json_encode

 $query = "SELECT legal_country as `id`, count(*) as `value` FROM accounts GROUP BY legal_country ORDER BY count(*) DESC";
    $result = mysql_query( $query );
    $json= array();
    $area = array();
        while ( $row = mysql_fetch_assoc( $result ) ) {
            $area[] = $row;
        }
    $data = array(
        'map' => 'worldlow',
        'area' => $area,
      );
    return json_encode( $data );

Answer 2

根据您的要求创建一个包含结果的数组。

$query = "SELECT legal_country AS `id`, count(*) AS `value` 
          FROM accounts 
          GROUP BY legal_country 
          ORDER BY count(*) DESC";
$dataArr = $areasArr = array();
if ($result = mysql_query( $query )) {
    while ( $row = mysql_fetch_assoc( $result ) ) {
        $areasArr[] = $row;
    }
    $dataArr = array(
        "map"   => "worldLow",
        "areas" => $areasArr
    );
}
return json_encode( $dataArr, true );

还使用mysqli_*个函数代替mysql_*

如何替换json数组的标签并在数组之前添加信息

2 个答案: