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我正在尝试编写一个函数，该函数返回一个多维数组，其中键是日期，并且日期还有两个项目成本和价值。如果当天股市关闭（即周末或节假日），我正在使用的查询结果将为零行。我想将那一天的成本和价值设置为等于前一天。显然我在这里遗漏了一些东西，但我不明白。

function getNetWorth()
{
    // Query for First and Last date
    $query = $this->db->query("SELECT min(date) as start from Transactions ORDER BY date ASC;");
    $date = $query->row_array();
    $current = $date['start'];
    $networth = array();
    while(strtotime($current)<= strtotime(date('Y-m-d')))
    {
        $cost = 0;
        $value = 0;
        $run = date('Y-m-d', $current);
        // Query for net worth for given day
        echo $current;
        $sql = "SELECT Transactions.symbol, sum(shares) AS shares, sum(shares * price) AS cost, History.close as `close`, (sum(shares) * History.close) as value ".
                "FROM Transactions INNER JOIN History ON (History.symbol = Transactions.symbol AND History.date ='".$run."') ".
                "WHERE (action <>5) AND Transactions.date <= '".$run."' GROUP BY Transactions.symbol HAVING sum(shares) > 0";
        $query = $this->db->query($sql);
        if($query->num_rows() < 1)
        {

            $date = strtotime("-1 day", $current);
            $last = date('Y-m-d', $date);
        //  $networth[$current][] = $networth[$last];
        }
        else
        {
            $result = $query->result_array();
            $cost += $result['cost'];
            $value += $result['value'];
            $networth[$current]= array("cost" => $cost, "value" => $value);

        }


        $current = strtotime("+1 day", strtotime($current));    
    }
    return $networth;
}
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1 回答 1

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看起来您的格式$last与 不同$current，因此$networth[$last]不会匹配任何内容。$last并且$run格式相同，因此您可能需要更改该行：

$networth[$run]= array("cost" => $cost, "value" => $value);
于 2013-02-19T02:06:26.847 回答