How to know if $_POST works php

Question

I have this code:

if (isset($_POST['food-input']) && $_POST['food-input']!="") {
    echo "RABOTI !!";
}

'food-input' is the name of my Input box with type="text" it, but it doesn't print given echo, what can i do to make it work. And other question will code work with !="" at the end,i wanna tell to server if'food-input' have something in it and IS NOT EMPTY(second $_POST method) to post the given information. Thanks!!

EDIT: There is the HTML .

<!----------------------------------------------------------------
                             HTML
----------------------------------------------------------------->
<p>Търсене на храни: <input type="text" name="food-input" id="food_search"></p>
<div id="food_search_result"></div>

Have you tried `if(!empty( $_POST['food-input'])){ echo "RABOTI !!"; }` — Ray, Jan 13 '17 at 19:32
Then you'll need to show the form where you get this input from. — Qirel, Jan 13 '17 at 19:39
Your form inputs actually have to be in a form and that form has to have its action set to post if your looking to access post variables. See [here](https://developer.mozilla.org/en-US/docs/Web/HTML/Element/form). — cteski, Jan 13 '17 at 19:45
Show full form starting from the form tag to the end maybe you didn't have the form tag and set the method to post — smartnet, Jan 13 '17 at 19:48
Ohh didn't know that guys it displayed error after adding var_dump($_POST['food-input']); and after reading cteski and smartnet comments and adding
it worked <3 thanks to all Thanks for your time <3 — Tsvetomirov Yordan, Jan 13 '17 at 19:53

score 1 · Accepted Answer · edited May 23 '17 at 11:52

1

step by step

try to activate error messages

How do I get PHP errors to display?

and after

 var_dump($_POST['food-input']);

edited May 23 '17 at 11:52

Community

1
1

answered Jan 13 '17 at 19:44

score 1 · Answer 2 · answered Jan 13 '17 at 19:55

Based on the HTML you provided, if that's the full code - you're missing the most important tag - the actual <form> tag. Simply put it inside a form, using the POST method. After submitting it, you can then use the information submitted in the $_POST array.

<form method="POST">
    <p>Търсене на храни: <input type="text" name="food-input" id="food_search"></p>
</form>
<div id="food_search_result"></div>

Also, instead of doing

if (isset($_POST['food-input']) && $_POST['food-input']!="") {

you can replace that entire line by the below, which does exactly the same

if (!empty($_POST['food-input'])) {

Reference

I am really thankfull to guys like you with exhaustive answers,THANKS mate.I already fixed my problem and marked this question for solved,i voted for you if there is anything other i can do(because i am new in this site) feel free to say me <3 Thanks one more time — Tsvetomirov Yordan, Jan 13 '17 at 20:03

score -1 · Answer 3 · answered Jan 13 '17 at 19:49

-1

You can do,

$test = $_POST['name_of_form'];
echo $test;
die ();

Regards

answered Jan 13 '17 at 19:49

juanitourquiza

2,097
1
30
52

I fixed it with roison and comments of smartnetand and cteski.Thanks for time mate <3 – Tsvetomirov Yordan Jan 13 '17 at 19:54

How to know if $_POST works php

3 Answers3

Reference