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我试图生成一个数组，但不知道如何去做。

我目前正在像这样获取我的数据：

$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test@test.com'");
$row = mysql_fetch_array($query);

$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){

    $query2 = mysql_query("SELECT * FROM usersettings WHERE userId = ".$row['userId']." AND usersettingCategory".$row1['categoryId']." LIKE 'y'");
    $isyes = mysql_num_rows($query2);

    if($isyes > 0){

        $cat1 = mysql_query("SELECT * FROM shops WHERE shopstateId = 1 AND (categoryId1 = ".$row1['categoryId']." OR categoryId2 = ".$row1['categoryId']." OR categoryId3 = ".$row1['categoryId'].")");
        $cat1match = mysql_num_rows($cat1);

        if($cat1match > 0){
            while($cat1shop = mysql_fetch_array($cat1)){
                $cat1msg = mysql_query("SELECT * FROM messages WHERE shopId = ".$cat1shop['shopId']." and messagestateId = 1");
                while($cat1msgrow = mysql_fetch_array($cat1msg)){

                    echo $cat1msgrow['messageContent']." - ".$cat1msgrow['messageCode'];
                    $cat1img = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$cat1shop['shopimageId']);
                    $imgpath = mysql_fetch_array($cat1img);
                    echo " - ".$imgpath['shopimagePath']."<br/>";
                }
            }
        }
     }
}

但是，当用户在他们的偏好中选择了所有 3 个商店类别时，这可能会导致重复。我正在尝试找到一种方法来提取消息 ID 而不是整个内容并将其放入给我的数组中，例如：

1,3,5,7,1,3,5,2,4,7,8

然后我可以运行一个单独的查询来告诉我ID在数组中的所有消息，但我不确定构建这样一个数组的最有建设性的方法，我见过的while循环中的数组示例似乎并不成为我正在寻找的东西。

有没有人可以把我推向正确的方向？

4

2 回答 2

0

无法帮助使用此代码。但是，如果您想从查询中获得一个没有重复结果的数组，您可以在查询中使用“select DISTINCT (id)”或更简单的解决方案：

$id_arr = array();
$sql = mysql_query("select id from id_table"); 
while ($id_result = mysql_fetch_array($sql) {
  $id  = $id_result['id'];
  if (!in_array($id, $id_arr)) {
    $id_arr[] = $id;
  }
}
于 2012-05-01T08:27:08.030 回答
0

我找到了一种更简单的方法来创建所需的结果。我想在早上 6 点经过一夜艰苦的编码后，我的大脑被炸了，我让事情变得比我需要的要复杂得多。我的问题的一个简单解决方案如下：

$query = mysql_query("SELECT * FROM users WHERE userEmail LIKE 'test2@test2.com'");
$row = mysql_fetch_array($query);

$categories = "(";

$query1 = mysql_query("SELECT * FROM categories");
while($row1 = mysql_fetch_array($query1)){

    $query2 = mysql_query("SELECT usersettingCategory".$row1['categoryId']." FROM usersettings WHERE userId = ".$row['userId']);
    $row2 = mysql_fetch_array($query2);

    if($row2['usersettingCategory'.$row1['categoryId']] == y){

        $categories .= $row1['categoryId'].",";
    }

}

$categories = substr_replace($categories ,")",-1);

echo $categories."<br />";

$query3 = mysql_query("SELECT * FROM shops,messages WHERE shops.shopId = messages.shopId AND messages.messagestateId = 1 AND (shops.categoryId1 IN $categories OR shops.categoryId2 IN $categories OR shops.categoryId3 IN $categories)");

while($row3 = mysql_fetch_array($query3)){

    $query4 = mysql_query("SELECT shopimagePath FROM shopimages WHERE shopimageId = ".$row3['shopimageId']);
    $row4 = mysql_fetch_array($query4);

    echo $row3['messageContent']." - ".$row3['messageCode']." - ".$row4['shopimagePath']."<br />";
}
于 2012-05-01T20:16:09.957 回答