Php 为什么我的函数不返回返回值？

好的，这是我的函数：

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;
        }
    }
}
}

代码执行得非常完美，从数据库中提取值并进行比较，然后按照我希望的方式执行所有操作。它在我可以测试的每种情况下都能得到正确的值，但它完全拒绝将该值返回给调用脚本。有人看到我掩饰过的愚蠢错误吗？在return语句之前立即使用debug echo似乎可以消除在return语句之前出错的可能性，但我现在还不知道

编辑：需要澄清的是，3005是我目前期望从数据库中得到的值。这是为了在工作中设置工作编号，该编号始终为Zxxx，其中Z是一年中的最后一位数字。这些都是按顺序创建的，但是对于跨越一年以上的作业，我们只更改Z，因此这是我用来解决以下事实的代码：在创建3000个作业之前，3030可以（并且确实）存在。

当您打电话时

getNewJobNumber（$jobPrefix，$addition+1）

您实际上并没有返回值

换成

返回getNewJobNumber（$jobPrefix，$addition+1）
您正在递归调用函数，但没有对返回值执行任何操作：

    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;
    }

if（$addition）另一个小提示：在开始时递增addition变量，然后传递（再次递增）对self的值，看起来可能是个错误。谢谢，这确实是个问题，我现在也明白为什么它有意义了。关于双增量，你也是对的。为了测试它可以跳过多个增量，我将数据库设置为跳过两个，但没有意识到它同时跳过了两个。这是一个错误lso现在也修复了，非常感谢您的帮助。谢谢，这个解决方案是绝对正确的。只是因为更详细的信息而检查了另一个人，同时也指出了我的其他问题。非常感谢您的回复，我非常感谢。
return:nextJobNumber-3005
:}{:

    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;
    }

    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?
    }