Question

我正在对mysql表进行查询：

// Perform queries
$sth = mysqli_query($con,"SELECT * FROM statistics");

$rows = array();

while($r = mysqli_fetch_assoc($sth)) {
    $rows['cols'] = $r;
}

print json_encode（$ rows）;

但这是这样的：

[{"id":"3","session_id":"22052015","user_id":"2","house":"3","cars":"30","bikes":"40","code":"22"}]

我需要这种格式：

{
  "cols": [
        {"id":"","label":"Topping","pattern":"","type":"string"}
      ],
  "rows": [
        {"c":[{"v":"Mushrooms","f":null},{"v":3,"f":null}]}
      ]
}

我该怎么做？

Answer 1

您需要自己创建两个单独的数组。一个包含键，一个包含值，并使用JSON将它们合并数组（＆＃34; cols＆＃34; =＆gt; $ arrayCols，＆＃34; rows＆＃34; =＆gt; $ arrayRows）;

查看php中的array_keys和array_values函数以完成此任务。

Php从SQL查询输出中获取Json格式

1 个答案: