如何检查按钮是否被推，如果表单是一个PHP文件，但它在另一个检查 - how to check if button is pushed if the form in is one php file but its checked in another

how to check if button is pushed if the form in is one php file but its checked in another

我有两个PHP文件，我们称它们为fileOne.php和fileTwo.php。如果fileOne.php有一个表单与action="fileTwo.php"和按钮button1和button2，我如何检查fileTwo.php如果isset($_POST['button1'])被推，或isset($_POST['button2']被推。

这是"fileOne.php"中"fileTwo.php"的形式:(execute.php = fileTwo.php)

< form action="execute.php" method = "POST">
< input type="hidden" name="edit" value="Edit"/>
< input type="hidden" name="delete" value="Remove"/>
< /form>

这些按钮是在"fileOne.php"中创建的，但形式是action= "fileTwo.php":("button1 = edit, button2 = delete)

echo "<td><input type='submit' name='Edit' value='Edit'>
<input type='submit' name='Delete' value='Remove'></td>'n";

这是我尝试检查"fileTwo.php"的地方，如果其中一个按钮被按下:

    //if edit is clicked
if(isset($_POST['edit'] == 'Edit'))
{
    echo"hello";
}
//if remove is clicked
if(isset($_POST['delete'] == 'Remove'))
{
    echo"good bye";
}

我希望使用PHP解决方案，而不是JavaScript、AJAX等。

这样重写:

fileOne.php

<form action="fileTwo.php" method="post">
   <input type="submit" name="edit" value="edit" />
   <input type="submit" name="delete" value="delete" />
</form>

fileTwo.php

<?php
if (isset($_POST['edit']) && ($_POST['edit'] == 'edit')) {
    echo "Hello";
}
else if (isset($_POST['delete']) && ($_POST['delete'] == 'delete')) {
    echo "Goodbye";
}
?>

好的，这样你就不需要任何额外的代码，没有隐藏的输入，只有2个按钮。两者都将提交带有值的表单，以便在下一页上处理程序正确地处理它们。

制作两种形式:

< form action="execute.php" method = "POST">
< input type="hidden" name="delete" value="Remove"/>
< input type="submit" value="Delete"/>
< /form>
< form action="execute.php" method = "POST">
< input type="hidden" name="edit" value="Edit"/>
< input type="submit" value="Edit"/>
< /form>

在fileTwo.php:

//input names(edit, delete) and values(Edit, Remove) are case sensetive
if( isset($_POST['edit']) && $_POST['edit'] == 'Edit') {
    echo"hello";
}
//if remove is clicked
if( isset($_POST['delete']) && $_POST['delete'] == 'Remove') {
    echo"good bye";
}