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Home Backend Development PHP Tutorial 新手有一個有關(guān)問題很糾結(jié)

新手有一個有關(guān)問題很糾結(jié)

Jun 13, 2016 pm 01:06 PM
echo else quot string

新手有一個問題很糾結(jié),請教各位
為什么下面這段代碼我無論輸出什么都會顯示登錄密碼錯誤呢??用var_dump($num1)發(fā)現(xiàn)無論html那邊num1輸入什么,$num1和$num2都是string類型
$num1=$_REQUEST['num1'];
$num2=$_REQUEST['num2'];
if(is_string($num1))
{
if(is_int($num2))
{
echo"歡迎您的登錄"
}
else
{
echo"登錄密碼錯誤"
}
}
else
{
echo"您的用戶名格式錯誤"
}
?>

------解決方案--------------------
從html那邊提交過來的數(shù)據(jù)一般都是string的.
if(is_int($num2))
{
echo"歡迎您的登錄"
}
else
{
echo"登錄密碼錯誤"
}
$num2是stirng類型,所以顯示"登錄密碼錯誤"
------解決方案--------------------
是的,form 提交過去的都是字符串。你應(yīng)該用 is_numeric() 來檢測更為恰當(dāng)。
------解決方案--------------------
都有,前臺可進行一般驗證。但是為了安全性在后臺驗證也有必要性。
------解決方案--------------------
if(!preg_match('/^\d+$/',$num1)
echo '純字符串';

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