如何使用php在json中转换mysql数据库表数据-Howtoconvertmysqldatabasetabledatainjsonusingphp

13  

Try like this:

试试这样：

$query = mysql_query("SELECT * FROM table");
$rows = array();
while($row = mysql_fetch_assoc($query)) {
    $rows[] = $row;
}
print json_encode($rows);

If you don't have json_encode add this before the code above:

如果你没有json_encode，请在上面的代码之前添加：

if (!function_exists('json_encode'))
{
  function json_encode($a=false)
  {
    if (is_null($a)) return 'null';
    if ($a === false) return 'false';
    if ($a === true) return 'true';
    if (is_scalar($a))
    {
      if (is_float($a))
      {
        // Always use "." for floats.
        return floatval(str_replace(",", ".", strval($a)));
      }

      if (is_string($a))
      {
        static $jsOnReplaces= array(array("\\", "/", "\n", "\t", "\r", "\b", "\f", '"'), array('\\\\', '\\/', '\\n', '\\t', '\\r', '\\b', '\\f', '\"'));
        return '"' . str_replace($jsonReplaces[0], $jsonReplaces[1], $a) . '"';
      }
      else
        return $a;
    }
    $isList = true;
    for ($i = 0, reset($a); $i  $v) $result[] = json_encode($k).':'.json_encode($v);
      return '{' . join(',', $result) . '}';
    }
  }
}

0  

I guess you mean the data in a MySQL-database table right?

我猜你的意思是MySQL数据库表中的数据对吗？

In that case, have a look at PHP's json_encode().

在这种情况下，看看PHP的json_encode（）。

You can fetch the data from the db into an array and then covert it to JSON.

您可以将数据从db中提取到数组中，然后将其转换为JSON。

0  

put the resultset of the query into an array and then use json_encode

将查询的结果集放入数组中，然后使用json_encode

$sql="Select * from table";
$l= array();
$result = mysql_query($sql);
while ($row = mysql_fetch_assoc($result)) {
  $l[] = $row;
}
$j = json_encode($l);
echo $j;

you may use id table as index of the array.

您可以使用id表作为数组的索引。

0  

This code converts your mysql data in phpmyadmin to json. Works perfect

此代码将phpmyadmin中的mysql数据转换为json。工作完美

0  

using this code :

使用此代码：

json_encode($array)

for example :

例如 ：

public function SELECT($tableName,$conditions){

      $cOnnection= mysqli_connect($hostname, $userName, $password,$dbName);
      try {

        if (!$connection)
            die("Connection failed: " . $connection->connect_error);
        else
        {
            $qry = "";
            if(!$this->IsNullOrEmptyString($conditions))
               $qry = "SELECT * FROM `".$tableName."` WHERE ".$conditions;
            else
               $qry = "SELECT * FROM `".$tableName."`";

            $result = mysqli_query( $connection, $qry);
            if($result) {
                $emparray = array();
                while($row =mysqli_fetch_assoc($result))
                    $emparray[] = $row;

                echo(json_encode($emparray));           
            }
            else
                echo(mysqli_error($connection));       
            } 
            mysqli_close($connection); 
      } catch(Exception $ex) {
          mysqli_close($connection);
          echo($ex->getMessage());
      }  
 }

0  

As long as you are using MySQL server 5.7 or later, you can produce JSON data by using just SQL and nothing more, that is, you need PHP just to pass the SQL and get the JSON result. For example:

只要您使用的是MySQL服务器5.7或更高版本，您就可以仅使用SQL生成JSON数据，也就是说，您只需传递SQL并获取JSON结果即可。例如：

SELECT JSON_OBJECT( 'key1', column1,
                    'key2', JSON_OBJECT('key3', column2)) as fοο;

There is more that JSON_OBJECT! Please check this page: https://dev.mysql.com/doc/refman/5.7/en/json-function-reference.html

JSON_OBJECT还有更多！请查看此页面：https：//dev.mysql.com/doc/refman/5.7/en/json-function-reference.html