Question

我的代码如下

$sql = <<<SQL
 SELECT p . * , s . * 
 FROM am_user p
 INNER JOIN am_user_status s
 USING ( user_id ) 
 WHERE product_id =4
 AND partner_logo =  '1'
 ORDER BY RAND( ) 
 LIMIT 6
SQL;

$array = Array();

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

echo $array;

但是我收到错误，我只是想把结果放到一个数组中。有谁知道我怎么能做到这一点？

警告：mysql_fetch_array（）要求参数1为资源，在第26行的/var/sites/c/xxxxxxx/public_html/index.php中给出null

谢谢！

Answer 1

    $sql = mysql_query("
     SELECT p . * , s . * 
     FROM am_user p
     INNER JOIN am_user_status s
     USING ( user_id ) 
     WHERE product_id =4
     AND partner_logo =  '1'
     ORDER BY RAND( ) 
     LIMIT 6";

    $array = Array();

    while ($row = mysql_fetch_array(  $sql )) {
      $array[] = $row;
    }

echo "<pre>";
    print_r( $array );

Answer 2

你有sql语句，忘了把它发送到mysql。

...
$result = mysql_query($SQL); // you forgot this

$array = Array();

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

var_dump($array) ; // not echo $array

将结果放入php中的数组中

2 个答案: