Php:表单返回表单
Php:表单返回表单
提问于 2013-12-02 14:30:07
我有一个单选按钮表单，它为每个特定的选项返回某种类型的表单。我从第二个开始获得的数据将用于在数据库中搜索。我的问题是，如果我在第二个表单中提交数据时，对这两个表单都使用post方法，那么就会注意到，对于第一个表单，我没有这样做，因为每次加载页面时，我的代码都会验证变量是否为空。我可以在PHP中做些什么，这样服务器就可以区分表单，而不会在每次通过post提交数据时进行检查？我试着命名提交按钮，但它似乎不起作用。在数据库中搜索的代码，它是在另一个文件中编写的，并且正在工作。提前谢谢。
<!doctype html>
<html>
<body>
<?php
// define variables and set to empty values
$optiuneErr="";
$optiune="";
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["optiune"]))
{$optiuneErr= "Selectati optiunea";}
else
{$optiune = test_input($_POST["optiune"]);}
}
?>
<?php
$numeErr=$anErr=$regiuneErr=$succesorErr=$predecesorErr="";
$nume=$an=$regiune=$succesor=$predecesor="";
if ($_SERVER["REQUEST_METHOD"] == "POST")
{
if (empty($_POST["nume"]))
{$numeErr= "Introduceti numele";}
else
{$nume = test_input($_POST["nume"]);}
if (empty($_POST["an"]))
{$anErr = "Introduceti anul";}
else
{$an = test_input($_POST["an"]);}
if (empty($_POST["regiune"]))
{$regiuneErr = "Introduceti regiunea";}
else
{$regiune = test_input($_POST["regiune"]);}
if (empty($_POST["succesor"]))
{$succesorErr = "Introduceti succesorul";}
else
{$succesor = test_input($_POST["succesor"]);}
if (empty($_POST["predecesor"]))
{$predecesorErr = "Introduceti predecesorul";}
else
{$predecesor = test_input($_POST["predecesor"]);}
}
function test_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = htmlspecialchars($data);
return $data;
}
?>
<p><span class="error">* Selectati un camp.</span></p>
<form method="post" >
<input type="radio" name="optiune" value="1" />Cauta dupa nume <br />
<input type="radio" name="optiune" value="2" />Cauta dupa an<br />
<input type="radio" name="optiune" value="3" />Cauta dupa regiune<br />
<input type="radio" name="optiune" value="4" />Cauta dupa succesor<br />
<input type="radio" name="optiune" value="5" />Cauta dupa predecesor<br />
<input type="submit" name="submit1" value="Alege" />
<span class="error"> <?php echo $optiuneErr;?></span>
</form>
<?php
switch($optiune)
{
case 1:
echo '
<p><span class="error">* Introduceti o valoare.</span></p>
<form method="POST">
Nume: <input type="text" name="nume"><br>
<input type="submit" value="Cauta" />
<span class="error"> <?php echo $numeErr;?></span>
</form>';
break;
case 2:
echo '
<p><span class="error">* Introduceti o valoare.</span></p>
<form method="POST" >
An: <input type="text" name="an"><br>
<input type="submit" value="Cauta" />
<span class="error"> <?php echo $anErr;?></span>
</form>';
break;
case 3:
echo '
<p><span class="error">* Introduceti o valoare.</span></p>
<form method="POST">
Regiune: <input type="text" name="regiune"><br>
<input type="submit" value="Cauta" />
<span class="error"> <?php echo $regiuneErr;?></span>
</form>';
break;
case 4:
echo '
<p><span class="error">* Introduceti o valoare.</span></p>
<form method="POST">
Succesor: <input type="text" name="succesor"><br>
<input type="submit" value="Cauta" />
<span class="error"> <?php echo $succesorErr;?></span>
</form>';
break;
case 5:
echo '
<p><span class="error">* Introduceti o valoare.</span></p>
<form method="POST">
Predecesor: <input type="text" name="predecesor"><br>
<input type="submit" value="Cauta" />
<span class="error"> <?php echo $predecesorErr;?></span>
</form>';
break;
}
?>
</body>
</html>回答 1
Stack Overflow用户
回答已采纳
发布于 2013-12-02 14:36:28
您应该使用isset()函数。
if (isset($_POST['submit1']))
{
if (empty($_POST["optiune"]))
{$optiuneErr= "Selectati optiunea";}
else
{
if(isset($_PoST['nume']))
{
$nume = test_input($_POST["nume"]);
}
else if($_POST['an'])
{
///////some checking code/////
}
////and so on for all individual forms///
}
}第二阶段就像这样。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/20330701
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