Php 条件运算符未按预期执行

我目前正在尝试比较用户选择的两组返回值，但是当我比较结果中的数据时，elseif语句会默认，尽管信息明显相反。不管值是什么，程序总是引用elseif。任何帮助都将不胜感激。谢谢大家!

// DB Constant Defines
define('DB_NAME','NurseData'); 
define('DB_USER','root'); 
define('DB_PASSWORD','root'); 
define('DB_HOST','localhost'); 

$state1 = $_REQUEST['state1'];
$state2 = $_REQUEST['state2'];
$city1 = $_REQUEST['city1'];
$city2 = $_REQUEST['city2'];
$jobTitle1 = $_REQUEST['job1'];
$jobTitle2 = $_REQUEST['job2'];

function showerror() {
    die("Error " . mysql_errno() . " : " . mysql_error());
}

$connection = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die(mysql_error()); 
mysql_select_db(DB_NAME, $connection) or die(mysql_error());

$query1 = "SELECT DISTINCT
    TOT_EMP,  
    JOBS_1000,
    A_MEAN,
    A_PCT90
    FROM Nurse_Local 
    WHERE PRIM_STATE='" . $state1 . "' 
    AND AREA_NAME='" . $city1 . "' 
    AND OCC_TITLE='" . $jobTitle1 . "'";

$query2 = "SELECT DISTINCT
    TOT_EMP,  
    JOBS_1000,
    A_MEAN,
    A_PCT90
    FROM Nurse_Local 
    WHERE PRIM_STATE='" . $state2 . "' 
    AND AREA_NAME='" . $city2 . "' 
    AND OCC_TITLE='" . $jobTitle2 . "'";

if (!($getPosts1 = mysql_query ($query1, $connection))) {
    showerror();
}

if (!($getPosts2 = mysql_query ($query2, $connection))) {
    showerror();
}


while($rows1 = mysql_fetch_array($getPosts1)) {     

    while($rows2 = mysql_fetch_array($getPosts2)) {

    //Retrieve array values
        for ($i1 = 0; $i1 < count($rows1); $i1++) {
            for ($i2 = 0; $i2 < count($rows2); $i2++) {
                //Assign array values
                $tot_EMP1 = $rows1['TOT_EMP'];
                $tot_EMP2 = $rows2['TOT_EMP'];
                $jobs_PER1 = $rows1['JOBS_1000'];
                $jobs_PER2 = $rows2['JOBS_1000'];
                $a_MEAN1 = $rows1['A_MEAN'];
                $a_MEAN2 = $rows2['A_MEAN'];
                $A_PCT901 = $rows1['A_PCT90'];
                $A_PCT902 = $rows2['A_PCT90'];

                //Convert array values to numbers
                $tot_EMP1 = 0 + $tot_EMP1;
                $tot_EMP2 = 0 + $tot_EMP2;

                //Functions for calculating differences
                /*
                function compareEMP1($diffEMP1) {
                    $diffEMP1 = $rows1['TOT_EMP'] - $rows2['TOT_EMP'];
                    return $diffEMP1;
                }

                function compareEMP2() {
                    $diffEMP2 = $rows2['TOT_EMP'] - $rows1['TOT_EMP'];
                    return $diffEMP2;
                }
            */
            }
        }

        if($tot_EMP1 > $tot_EMP2 || $tot_EMP2 < $tot_EMP1) {
            echo $tot_EMP1;//"In " . $state1 . " there are " . compareEMP1() . " more jobs than in " . $state2;
        }
        elseif ($tot_EMP2 > $tot_EMP1 || $tot_EMP1 < $tot_EMP2) {
            echo $tot_EMP2;//"In " . $state2 . " there are " . compareEMP2() . " more jobs than in " . $state1;
        }
        else {
            echo "<p>There was a problem comparing the employment numbers.</p>";
        }
    }
}

---

找到了答案，这是因为数字被输出为字符串，而PHP无法识别它们为数字。我首先在$tot_EMP1和$tot_EMP2上执行str_replace来删除逗号，然后在这两个变量上执行intval来解决这个问题。这会将它们转换为整数，逻辑将按需要执行。现在我需要弄清楚如何添加逗号并保留结果的int状态

进行转换的实际代码，但请记住，my DB将信息存储为字符串，因此部分问题是我自己未能正确存储数据：

                //Assign individual array values
                $tot_EMP1 = $rows1['TOT_EMP'];
                $tot_EMP2 = $rows2['TOT_EMP'];
                $jobs_PER1 = $rows1['JOBS_1000'];
                $jobs_PER2 = $rows2['JOBS_1000'];
                $a_MEAN1 = $rows1['A_MEAN'];
                $a_MEAN2 = $rows2['A_MEAN'];
                $a_PCT901 = $rows1['A_PCT90'];
                $a_PCT902 = $rows2['A_PCT90'];


                //Convert individual array values to numbers
                $tot_EMP1 = str_replace(",", "", $tot_EMP1);
                $tot_EMP2 = str_replace(",", "", $tot_EMP2);
                $tot_EMP1 = intval($tot_EMP1);
                $tot_EMP2 = intval($tot_EMP2);
                $jobs_PER1 = floatval($jobs_PER1);
                $jobs_PER2 = floatval($jobs_PER2);
                $a_MEAN1 = str_replace(",", "", $a_MEAN1);
                $a_MEAN2 = str_replace(",", "", $a_MEAN2);
                $a_MEAN1 = intval($a_MEAN1);
                $a_MEAN2 = intval($a_MEAN2);
                $a_PCT901 = str_replace(",", "", $a_PCT901);
                $a_PCT902 = str_replace(",", "", $a_PCT902);
                $a_PCT901 = intval($a_PCT901);
                $a_PCT902 = intval($a_PCT902);

---

请正确缩进代码。现在一团糟。我会试试的，但我都是自学成才，所以我不知道什么是好的缩进。看起来像是用于你的函数的缩进，是不是更好？我试图删除多余的空行，并在彼此之间正确对齐函数。这似乎是多余的$tot_EMP1>$tot_EMP2 | |$tot_EMP2<$tot_EMP1-如果tot1小于tot2或tot2大于tot1，可以说好像tot1不等于2听起来您没有使用正确的数字数据类型开始，但我们一直在存储字符串，这是您应该解决的问题。如果在显示这些数字时需要不同的小数点或千位分隔符，那么这就是格式问题。是的，不幸的是，它们在数据库中存储为字符串。在尝试转换为INT时，这并不是一个特别容易的转换，至少从我尝试将VARCHAR转换为INT/FLOAT的经验来看是这样的，所以这种方法非常有效。