If语句返回true，而它应该是false。我认为

Question

我有个很奇怪的问题。我正在尝试从WoW军火库导入数据，并在此过程中将此信息添加到我自己的数据库中的一个名为user的表下。

现在，我在添加新成员的同时检查当前成员的所有信息是否仍然正确(请记住，我将在此之后修复所有sql注入问题)：

@$json = file_get_contents("http://$region.battle.net/api/wow/guild/$realm/$guild?fields=members,achievements");
if($json == false)
{
throw new Exception("Failed To load infomation. check setup options");
} 
$decode = json_decode($json, true);

foreach($decode['members'] as $p) {
    $mrank = $p['rank'];
    $mname = $p['character']['name'];
    $mclass = $p['character']['class'];
    $mrace = $p['character']['race'];
    $mlevel = $p['character']['level'];
    $mgender = $p['character']['gender'];

    $check = mysql_query("SELECT * FROM user WHERE charactername='$mname'");
    if($check == false)
    { 
        die("Sql query failed"); 
    }
    if(mysql_num_rows($check) != 0)
    {
        if($mlevel !== $check['level'])
        {
            mysql_query("UPDATE user SET level='$mlevel' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "level $mname<br />"; 
        }
        if($mclass !== $check['class'])
        {
            mysql_query("UPDATE user SET class='$mclass' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "class $mname<br />"; 
        }
        if($mrace !== $check['race'])
        {
            mysql_query("UPDATE user SET race='$mrace' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "race $mname<br />"; 
        }
        if($mgender !== $check['gender'])
        {
            mysql_query("UPDATE user SET gender='$mgender' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "gender $mname<br />"; 
        }
        if($mrank !== $check['rank'])
        {
            mysql_query("UPDATE user SET rank='$mrank' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "rank $mname<br />"; 
        }
    $pCount = $pCount + 1;

    }   
    else
    {
    //add new user otherwise
        $sql="INSERT INTO user (charactername, class, race, level, gender, rank) VALUES ('$mname','$mclass','$mrace','$mlevel','$mgender','$mrank')";
        $nCount = $nCount + 1;
        if (!mysql_query($sql,$con))
            {
            die('Error: ' . mysql_error());
            }
    }
}
    echo "$nCount new members added.<br />";
    echo "$pCount member(s) already existed.<br />";
    echo "$uCount member(s) got updated <br />";

但现在的问题是，使用此代码的更新部分，所以所有的：

if($mlevel !== $check['level'])
        {
            mysql_query("UPDATE user SET level='$mlevel' WHERE charactername='$mname'");
            $uCount = $uCount + 1; 
            echo "level $mname<br />"; 
        }

不管怎么说，我不明白为什么。我比较了两个数据，当回声时两者是相等的。因此，据我所知，使用!==应该只在它们不相同的情况下触发。在当前情况下，我让update脚本运行了800次。这样一来，user表中的所有数据也会更新800次。而据我所知，这不应该是必要的。

任何帮助都将不胜感激。

Answer 1

您应该尝试!= php as !==检查它们是否具有完全相同的值和数据类型，并且由于检索两者的方式不同，php可能会为它们分配不同的数据类型

或者，您可以将$a =(int)$mlevel和$b =(Int)$check‘’level‘转换，然后执行!==

Answer 2

虽然这可能不是最好的方法，但在找到@MatthewBlancarte的查询结果为Null后，我决定修改代码：

foreach($decode['members'] as $p) {
$mrank = $p['rank'];
$mname = $p['character']['name'];
$mclass = $p['character']['class'];
$mrace = $p['character']['race'];
$mlevel = $p['character']['level'];
$mgender = $p['character']['gender'];

$check = mysql_query("SELECT * FROM user WHERE charactername='$mname'");


while($row = mysql_fetch_array($check)){
    $prank = $row['rank'];
    $pname = $row['charactername'];
    $pclass = $row['class'];
    $prace = $row['race'];
    $plevel = $row['level'];
    $pgender = $row['gender'];
}

if($check == false)
{ 
    die("Sql query failed"); 
}

if(mysql_num_rows($check) != 0)
{   
    if($mlevel != $plevel)
    {
        mysql_query("UPDATE user SET level='$mlevel' WHERE charactername='$mname'");
        $uCount = $uCount + 1; 
        //echo "level $mname<br />"; 
    }

使用mysql_fetch_array添加额外的查询while循环：

while($row = mysql_fetch_array($check)){
    $prank = $row['rank'];
    $pname = $row['charactername'];
    $pclass = $row['class'];
    $prace = $row['race'];
    $plevel = $row['level'];
    $pgender = $row['gender'];

是修复它的方法，将其与这些变量进行比较。谢谢大家的帮助。在更多的人评论之前是的..我将修复sql注入..>.>