pass value from $_GET to $_POST

Question

So, I have a disabled email form field whose placeholder and value are retrieved from $_GET['email']:

<input name="email" type="text" placeholder="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" disabled>

When the user fills out the form, I was hoping that $_POST['email'] would have the email value, but it doesn't (it's empty). What am I missing/forgetting? Is there a clever way to pass this value along? Thanks!

i believe i do? i left the rest out but the above is sandwiched between
*form fields and submit button* — Matt, Apr 21 '16 at 02:59
If your form method is post, change disabled to readonly! http://stackoverflow.com/a/36758818/2209876 — Maxi Schvindt, Apr 21 '16 at 03:06

score 3 · Accepted Answer · answered Apr 21 '16 at 03:00

3

change attribute disabled to readonly because disabled not submit values..

<input name="email" type="text" placeholder="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" readonly>

answered Apr 21 '16 at 03:00

Maxi Schvindt

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more info http://stackoverflow.com/questions/1355728/values-of-disabled-inputs-will-not-be-submited – Maxi Schvindt Apr 21 '16 at 03:02
wonderful! that did the trick. thanks to others who said the same. – Matt Apr 21 '16 at 03:14

score 3 · Answer 2 · edited May 23 '17 at 11:52

3

As Cuchu stated, you could use readonly instead of disabled. Or you could duplicate the field and change the type to hidden.

<form method="post" action="register.php">
    <input type="text" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" disabled>
    <input name="email" type="hidden" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>">
</form>

edited May 23 '17 at 11:52

Community

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answered Apr 21 '16 at 03:01

programmerKev

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score 1 · Answer 3 · answered Apr 21 '16 at 03:04

1

I think it is better to use the readonly attribute instead of disabling the input.

<input name="email" type="text" placeholder="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" readonly="readonly">

answered Apr 21 '16 at 03:04

Aron

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or you could add hidden input and then put the email value in it. – Aron Apr 21 '16 at 03:05

score 0 · Answer 4 · answered Apr 21 '16 at 02:59

0

Placeholder is a prompt, not a value. If you want the textfield to have a value of the email, use the "value" attribute, not the "placeholder".

answered Apr 21 '16 at 02:59

Ethan Brown

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score 0 · Answer 5 · answered Apr 21 '16 at 03:01

You should enclose in a form tag and set the method to post as

<form action="" method="post">
<input name="email" type="text" placeholder="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" value="<?php if(isset($_GET['email'])) echo $_GET['email']; ?>" disabled>
</form>

score 0 · Answer 6 · answered Apr 21 '16 at 03:09

0

  //try this
<form method="post">
<input type="text" name="email"><input type="submit" value="click" name="btnClick" id="btnClick">
<input type="text" name="email1" placeholder="inserted value" value="<?php echo (isset($_POST['email']))? $_POST['email'] : "" ?> ">      
</form>

answered Apr 21 '16 at 03:09

Tosif

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" disabled="disabled"> you can disable it so it can b edited,,inserted value only can be seen or use read only attribute. – Tosif Apr 21 '16 at 03:11

pass value from $_GET to $_POST

6 Answers6