unable to get value of form with $_post

Question

i want to get value of input of my form via $_post in my insert page but my problem is that i get error massage:undefined index:track_code and i cant get value of it.the names is same what is problem? this is my problem why stackoverflow want more detail. this is all detail

<form action="insert.php" method="post">
<input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd ; ?>" style="width: auto;height: auto" disabled />
</form>


<form action="insert.php" method="post" class="form-horizontal form-checkout">
 <div class="control-group">
  <label class="control-label" for="firstName">نام<span class="red-clr bold">*</span></label>
  <div class="controls">
  <input name="first_name" type="text" id="firstName" class="span4" >
  </div>
  </div>
  <div class="control-group">
  <label class="control-label" for="lastName">نام خانوادگی<span class="red-clr bold">*</span></label>
  <div class="controls">
  <input name="last_name" type="text" id="lastName" class="span4">
  </div>
  </div>
  <input name="submit" type="submit" class="btn btn-success" value="خرید"     style="width: 66px"/>
  </form>

  <form action="insert.php" method="post">
  <input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd ; ?>" style="width: auto;height: auto" disabled />
  </form>


  <?php
  $track_code = $_POST['track_code'];
  ?>

Possible duplicate of [PHP: "Notice: Undefined variable", "Notice: Undefined index", and "Notice: Undefined offset"](https://stackoverflow.com/questions/4261133/php-notice-undefined-variable-notice-undefined-index-and-notice-undef) — user3942918, Jul 29 '17 at 06:07
@Exprator i have two forms and that form have submit . im confused . whats the solution? — sinak, Jul 29 '17 at 06:25
I have added a sample form with submit button please check that out. @sinak — always-a-learner, Jul 29 '17 at 06:25
can you post the both forms?? it will be easy for us to answer by checking the structure of both form — Exprator, Jul 29 '17 at 06:26
@Exprator yeah 2 forms because the locations are diffrent but i want when i submit both forms submit — sinak, Jul 29 '17 at 06:27

always-a-learner · Answer 1 · 2017-08-04T09:21:54.713

Try Putting the PHP code on the form. as given below.

<?php
  if (isset($_POST['submit'])) {
    $track_code = $_POST['track_code'];
  }
?>
<form action="your_page.php" method="post">
   <input name="track_code" type="text" class="tiny-size" value="<?php echo $track_code_rnd; ?>" style="width: auto;height: auto" readonly />
  <input name="submit" type="submit" />
</form>

It will execute the whole file as PHP. The first time you open it, $_POST['submit'] won't be set because the form has not been sent. Once you click on the submit button, it will print the information.

Disabled controls never post on server. So make it readonly or hidden.

@Sinak Try this way and also put the comma in the value attribute of input. or you can do it like this `= $track_code_rnd ?>` — always-a-learner, Jul 29 '17 at 06:13

score 0 · Answer 2 · answered Jul 29 '17 at 06:09

0

Set this one code in your form

<form action="insert.php" method="post">
  <input name="track_code" type="text" class="tiny-size" value="<?php echo ($track_code_rnd)?$track_code_rnd:'';?>" style="width: auto;height: auto" disabled />
</form>

answered Jul 29 '17 at 06:09

gaurang vyas

68
9

score 0 · Answer 3 · answered Jul 29 '17 at 06:29

0

Remove disable tag from your input code. Disable controls wont be posted on server. for that you can make it readonly or make it hide

answered Jul 29 '17 at 06:29

Tejas Patel

61
1
5

unable to get value of form with $_post

3 Answers3