CODEWITHSUNDEEP
phpformsquery-string
user2247178
user2247178

Reputation: 21

PHP Added to query string on submit

I have a text field and a submit button, when I click on the submit button I want to add on to my query string.

example

http://www.fakewebsite.com/?name=mike

But my name appears blank

Here is my code

<?php

if($_POST){


    echo $_POST['name'];




}else{ ?>


    <form action="http://www.fakewebsite.com/test.php?name=<?php $_SERVER['name']; ?>" method="post">
    <label for="name">Name</label><input type="text" name="name" id="name" />
    <input type="submit" name="submit" id="submit" />
    </form>


<?php } ?>

Upvotes: 0

Views: 7441

Answers (2)

Praveen kalal
Praveen kalal

Reputation: 2129

alternate solution

<?php

if(!empty($_POST)){   

    echo $_POST['name'];  

}else{ ?>


    <form action="http://www.fakewebsite.com/test.php" method="post">
    <label for="name">Name</label><input type="text" value="<?=$_SERVER['name']; ?>" name="name" id="name" />
    <input type="submit" name="submit" id="submit" />
    </form>


<?php } ?>

Upvotes: 0

scones
scones

Reputation: 3345

POST request differ from GET requests in terms of how they send data to the server.

with get the data elements are appended to the url (like you tried).

with post the data elements are within the body of the http request.

This is problem 1: You are trying to pass the argument as a get parameter. this can work, but is not guarranteed to.

The other thing is, that you have an input field with the same name name. So even if the first thing would work, the input field would override that data.

what can you do to fox that? you can add a hidden input field with a different name to send this input back to your server:

<form action="foo.php" method="post">
  <input type="hidden" name="other_name" value="<?= $_SERVER['name'] ?>" />
  <input type="text" name="name" value="" />
  <input type="submit" />
</form>

And also keep the name collision in check.

Edit:

A third problem that just appeared to me <? $_SERVER['name'] ?> does not output the data within $_SERVER['name'].

To do That you should use

<?php echo $_SERVER['name'] ?>

// or, works sometimes

<?= $_SERVER['name'] ?>

Upvotes: 2

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