在具有相同输入的另一个表单内形成

Question

Is there a way to send exactly same variables to another page different from action of the form? I tried this structure but the second submit button did not work;

<form name="2" action="page2" method="post" >

<form name="1" action="page1" method="post" >

<input type="radio" name="radio" value="value1" >
<input type="radio" name="radio" value="value2" >
<input type="radio" name="radio" value="value3" >

<input type="submit" value="Submit1">

</form>

<input type="submit" value="Submit2">

</form>

The form 1 shows the information of inputs on the page1 and updates database also.

I want form 2 to show the information of inputs on the page2 only (no update of database).

Is that possible?

Answer 1

You cannot have form inside another form , change your code like this.

<form name="2" action="page2" method="post" >
    <input type="radio" name="radio" value="value1" >
    <input type="radio" name="radio" value="value2" >
    <input type="radio" name="radio" value="value3" >
    <input type="submit" value="Submit1" name="submit1">
    <input type="submit" value="Submit2" name="submit2">
</form>

Give different name for each submit button.

in php

if(isset($_POST['submit1'])){
    // submit1 is pressed
}

if(isset($_POST['submit1'])){
    // submit2 is pressed
}

Changing action dynamically.

add class to submit buttons say, class="submit". ANd add id to form say id="my-form"

$(".submit").change(function() {
      var action = $(this).val() == "submit1" ? "submit1.php : "submit2.php";
     $("#my-form").attr("action",  action);
});