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Home Backend Development PHP Tutorial 為什么小弟我在php上寫的這個代碼,在瀏覽器上什么都不顯示

為什么小弟我在php上寫的這個代碼,在瀏覽器上什么都不顯示

Jun 13, 2016 am 10:24 AM
post pass submit quot user

為什么我在php上寫的這個代碼,在瀏覽器上什么都不顯示啊
if(isset($_POST['Submit'])&& $_POST['Submit']=="登陸"){
$user=$_POST['user'];
$pass=$_POST['pass'];
if(empty($user)||empty($pass)){
echo"<script>alert('用戶名和密碼不能為空');</script>";
}
else{
echo"輸入的用戶名為:$user 密碼為:$pass";
}
}
?>





在瀏覽器上運行的結(jié)果空白的


------解決方案--------------------
說明你這個條件if(isset($_POST['Submit'])&& $_POST['Submit']=="登陸") 沒有成立,所以是空白的,你在表里看下的按鈕的name是不是 name="登陸",
如果不是 說明$_POST['submit']=="登陸"條件不成立,把它們改成一致的就行了,
另外:form 里的 method 是要用post 方法來提交的,即method="post"
可以這樣
在表單中
php中這樣寫:
if(isset($_POST['login'])&& $_POST['login']=="登陸"){
$user=$_POST['user'];
$pass=$_POST['pass'];
if(empty($user)||empty($pass)){
echo"<script>alert('用戶名和密碼不能為空');</script>";
?}
?else{
? echo"輸入的用戶名為:$user 密碼為:$pass";
?}
}
else{
?echo "表單沒有提交過來";
}
??>

------解決方案--------------------

探討

說明你這個條件if(isset($_POST['Submit'])&& $_POST['Submit']=="登陸") 沒有成立,所以是空白的,你在表里看下的按鈕的name是不是 name="登陸",
如果不是 說明$_POST['submit']=="登陸"條件不成立,把它們改成一致的就行了,
另外:form 里的 method……


------解決方案--------------------
參考PHP手冊:語言參考->變量->來自PHP之外的變量

  • HTML code
<?php// 自 PHP 4.1.0 起可用   echo $_POST[&#39;username&#39;];   echo $_REQUEST[&#39;username&#39;];   import_request_variables(&#39;p&#39;, &#39;p_&#39;);   echo $p_username;// 自 PHP 3 起可用。自 PHP 5.0.0 起,這些較長的預(yù)定義變量// 可用 register_long_arrays 指令關(guān)閉。   echo $HTTP_POST_VARS[&#39;username&#39;];// 如果 PHP 指令 register_globals = on 時可用。不過自// PHP 4.2.0 起默認值為 register_globals = off。// 不提倡使用/依賴此種方法。   echo $username;?><form action="foo.php" method="POST">????Name:??<input type="text" name="username"><br />????Email:?<input type="text" name="email"><br />????<input type="submit" name="submit" value="Submit me!" /></form>

$_POST['Submit']=="登陸"

表單傳遞過來的值(中文)是編碼過的。當然驗證不通過!
如果你換成英文就通過了

或者你把這個去掉也一樣,如下面的那個。這個鍵值存在時就執(zhí)行
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