如何通過PHP觸發表單動作？

Question

我有一個可以觸發paypal或其他鏈接的表單。

如何通過PHP觸發這些操作？

HTML：

<form action="myphp.php" method="POST /> 
    <input type="submit" value="paypal" name="action1" /> 
    <input type="submit" value="other" name="action2"> 
</form> 

myphp.php：

if($_POST["action1"]) { 
//https://www.paypal.com/cgi-bin/webscr should trigger here 
} 
if($_POST["action2"]) { 
//https://www.someotherwebsite.com/pay should trigger here 
} 

Answer 1

用途：

if($_POST["action1"]) { 
header('Location: https://www.paypal.com/cgi-bin/webscr'); 
exit;//it's a good habit to call exit after header function because script don't stop executing after redirection 
} 

Answer 2

只需使用header：

header('Location: http://www.example.com/'); 

Answer 3

在PHP中，你可以做到這一點與header()方法：

if($_POST["action1"]) { 
    header('Location: https://www.paypal.com/cgi-bin/webscr'); 
} 
if($_POST["action2"]) { 
    header('Location: https://www.someotherwebsite.com/pay'); 
} 

Answer 4

使用一個隱藏的輸入。

<form action="myphp.php" name="form1" method="POST" /> 
    <input type="hidden" name="action" /> 
    <input onclick="setHidden(this)" type="button" value="paypal" /> 
    <input onclick="setHidden(this)" type="button" value="other" /> 
</form> 

<script> 
    function setHidden(element) { 
     document.form1.action.value = element.value; 
     document.form1.submit(); 
    } 
</script> 

然後在PHP

if($_POST["action"] == "paypal") { 
    header('Location: https://www.paypal.com/cgi-bin/webscr'); 
} 
else if($_POST["action"] == "other") { 
    header('Location: https://www.someotherwebsite.com/pay'); 
} 

Answer 5

只是檢查是否存在isset：

if(isset($_POST["action1"])) { 
//https://www.paypal.com/cgi-bin/webscr should trigger here 
} 
if(isset($_POST["action2"])) { 
//https://www.someotherwebsite.com/pay should trigger here 
}