php - 为什么我的函数不返回返回值？

Question

好的，这是我的功能:

function getNewJobNumber($jobPrefix, $addition = "0") {
$addition = $addition + 1;
//echo $addition . "<br />";    
$yearDate = date("Y");
$firstDigit = $yearDate[strlen($yearDate) - 1];
$db = DatabaseHelpers::getDatabaseConnection();
$jobQuery = 'SELECT jobID, jobNumber, jobPrefix FROM tblJobNumbers WHERE jobPrefix = "' . $jobPrefix . '" AND jobNumber LIKE "' . $firstDigit . '___" ORDER BY jobID DESC LIMIT 1';
//echo $jobQuery . "<br />";
$stmt1 = $db->query($jobQuery);
$stmt1->setFetchMode(PDO::FETCH_OBJ);
$firstResult = $stmt1->fetch();
//above should select the latest created job number with selected prefix
//print_r($firstResult);
$jobNumber = $firstResult->jobNumber; //top row, will be last job number
//echo "jobNumberFromDB:" . $jobNumber . "<br />";
if (!$jobNumber) {
    //no job number exists yet, create one
    //will be last digit of year followed by "000" ie in 2013 first
    //new job number is "3000"
    $newJobNumber = str_pad($firstDigit, 4, "0");
    return $newJobNumber;
} else {
    //job number already exists, try next one
    $nextJobNumber = $jobNumber + $addition;
    $nextJobQuery = 'SELECT jobID, jobNumber, jobPrefix FROM tblJobNumbers WHERE jobPrefix = "' . $jobPrefix . '" AND jobNumber = "' . $nextJobNumber . '" ORDER BY jobID DESC LIMIT 1';
    $stmt2 = $db->query($nextJobQuery);
    $stmt2->setFetchMode(PDO::FETCH_OBJ);
    $nextResult = $stmt2->fetch();
    $dbNextJobNumber = $nextResult->jobNumber;      
    if (!$dbNextJobNumber) {
        //new job number is unique, return value
        echo "return:nextJobNumber-" . $nextJobNumber . "<br />";
        return($nextJobNumber);
    } else {
        //new job number is not unique, and therefore we need another one
        if ($addition <= 99) { //don't let this recurse more than 99 times, it should never need to
            //in order to loop this programatically call function again, adding one to addition factor
            getNewJobNumber($jobPrefix, $addition+1);
        } else {
            return;
        }
    }
}
}

这是我的电话:

        $ourNewJobNumber = getNewJobNumber($_POST['txtJobPrefix'], 0);
        echo ":}" . $ourNewJobNumber . "{:<br />";

这是我的结果:

return:nextJobNumber-3005
:}{:

代码执行完美，从数据库中提取值并比较它们，并按照我想要的方式执行所有操作。它在我可以测试的每种情况下都获得了正确的值，但它完全拒绝将该值返回给调用脚本。有没有人看到我掩盖的任何愚蠢错误？在我的 return 语句之前立即进行调试回显似乎消除了它在 return 语句之前出错的任何可能性，但我现在还不知道。

编辑:需要说明的是，此时 3005 是我期望从我的数据库中获取的值。这是为了在工作中设置工作编号，该编号始终为 Zxxx，其中 Z 是年份的最后一位。这些总是按顺序创建的，但对于跨度超过一年的作业，我们只更改 Z，所以这是我用来解决 3030 可以(并且确实)在创建 3000 之前存在这一事实的代码。

Answer 1

你正在递归地调用你的函数，但你没有对返回值做任何事情:

    if ($addition <= 99) { //don't let this recurse more than 99 times, it should never need to
        //in order to loop this programatically call function again, adding one to addition factor
        getNewJobNumber($jobPrefix, $addition+1);
    } else {
        return;
    }

应该是这样的:

    if ($addition <= 99) { //don't let this recurse more than 99 times, it should never need to
        //in order to loop this programatically call function again, adding one to addition factor
        return getNewJobNumber($jobPrefix, $addition+1);
        ^^^^^^
    } else {
        return -1;    // some kind of error message?
    }