Question

我之前刚问过一个问题，但现在我已将其格式化为一个不会返回任何值的函数。

这是我的代码：

echo GetHours($UID, $DAY, $MONTH, $YEAR);

function GetHours($UserId, $Day, $Month, $Year){
//filter queries:


    //YEAR:
    if($Year==FALSE){
        $Y = "";
    } else {
        $Y = " AND Year = '$Year'";
    }

    //MONTH:
    if($Month==FALSE){
        $M = "";
    } else {
        $M = " AND Month = '$Month'";
    }

    //DAY:
    if($Day==FALSE){
        $D = "";
    } else {
        $D = " AND Day = '$Day'";
    }

    $Query = mysql_query("SELECT SUM(TotalHrs) 
                          FROM WorkLog 
                          WHERE UserId = '$UserId'$D$M$Y");

    $Data = mysql_fetch_array($Query);

    return $Data;

}

现在，我知道mysql_函数已被折旧，但目前此应用程序需要它。

我目前的问题是，在使用GET参数进行测试后，此函数不会返回任何内容。

对此有何解决方案？

修改

我已将最后一行更改为：return json_encode($Data);，现在屏幕显示：{"0":"8","SUM(TotalHrs)":"8"}

Answer 1

始终检查mysql_query()是否未返回错误！特别是，如果您从可能存在或可能不存在的输入生成的部分构建查询。

这至少应该识别SQL中的任何错误。

function GetHours($UserId, $Day, $Month, $Year){
//filter queries:


    //YEAR:
    if($Year==FALSE){
        $Y = "";
    } else {
        $Y = " AND Year = '$Year'";
    }

    //MONTH:
    if($Month==FALSE){
        $M = "";
    } else {
        $M = " AND Month = '$Month'";
    }

    //DAY:
    if($Day==FALSE){
        $D = "";
    } else {
        $D = " AND Day = '$Day'";
    }

    $Query = mysql_query("SELECT SUM(TotalHrs) 
                          FROM WorkLog 
                          WHERE UserId = '$UserId'$D$M$Y");

    if ( ! $Query ) {
        // just debug code, should be amended for a live site situation
        echo mysql_error();
        return 'Its broken';
    }

    $Data = mysql_fetch_array($Query);

    return $Data;

}

Answer 2

我发现通过echo json_encode(GetHours($UID, $DAY, $MONTH, $YEAR));

我返回的数据是一个数组

我完成的代码有效：

<?php
include 'assets/db_connect.php';
//Function to get total from WorkLog:


$DAY = $_GET['day'];
$MONTH = $_GET['month'];
$YEAR = $_GET['year'];
$UID = $_GET['id'];


echo GetHours($UID, $DAY, $MONTH, $YEAR);



function GetHours($UserId, $Day, $Month, $Year){
//filter queries:


//YEAR:
if($Year==FALSE){
    $Y = "";
}
else{
    $Y = " AND Year = '$Year'";
}

//MONTH:
if($Month==FALSE){
    $M = "";
}
else{
    $M = " AND Month = '$Month'";
}

//DAY:
if($Day==FALSE){
    $D = "";
}
else{
    $D = " AND Day = '$Day'";
}




$Query = mysql_query("SELECT SUM(TotalHrs) FROM WorkLog WHERE UserId = '$UserId'$D$M$Y");

$Data = mysql_fetch_array($Query);

return $Data['SUM(TotalHrs)'];

}

功能不返回值

2 个答案: