about $_POST warning

Question

I have a problem with $_POST. I am adding data to my mysql database with this code and it is working.;

if (isset($_POST["d_kayit"])){
     $denetci=$dbpdo->prepare("INSERT INTO denetciler(name,pass) VALUES(:name, :pass)");
     $denetci->bindParam(":name",$_POST["denad"],PDO::PARAM_STR);
     $denetci->bindParam(":pass",$_POST["sif"],PDO::PARAM_STR);
     $denetci->execute();
}

But in the same form i want to use $_POST["denad"] for another insert. It is giving me "Notice: Undefined index: denad in" error. Sample code that giving error is;

if (isset($_POST["add"]))
{
    echo "Person: ".$_POST["denad"];
}

Can you help me please?

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) — iainn, Feb 04 '18 at 13:13
it is not duplicated question. I have looked before ask. If I dismiss something please show me. — Hüseyin Çiftçi, Feb 04 '18 at 13:16
You are checking whether `add` is in the post array and read data from `denad`. So why do you check for one key and then read another? — Nico Haase, Feb 04 '18 at 13:33
i have a button. i am checking if button is pushed. if button pushed i am tring to read the text. — Hüseyin Çiftçi, Feb 04 '18 at 13:37

Herco · Answer 1 · 2018-02-04T14:07:23.923

0

If this:

if (isset($_POST["add"]))
{
    echo "Person: ".$_POST["denad"];
}

is yielding notice: Undefined index: denad then you probably don't have inputs "add" and "denad" on the same form. Or it is an unchecked checkbox.

Edit based on the code of your HTML form. You'll need something like this, you can't end your form with </form> like you did until you included all needed input fields:

<form action="" method="post">
    <!-- content here -->
    <input name="denad" id="denad" type="text" style="margin-top:2px; width:200px; height:30px;"></input>
    <input type="submit" class="get_file" id="K_ekle" name="add" onclick="test()" value="Kişiye Ekle" style="float:left;"></input>                    
    <!-- more content here -->
 </form>

edited Feb 04 '18 at 14:07

answered Feb 04 '18 at 13:16

Herco

377
2
9

Thank you for answer Herco but there isn't any problem with $_POST["add"] because i tried echo "message" code and it worked. both of add and denad in same form. – Hüseyin Çiftçi Feb 04 '18 at 13:21
i have
parts. when i use $_POST["denad"] in the
part where i defined, my codes working. but the other
part of same form is giving error.
– Hüseyin Çiftçi Feb 04 '18 at 13:22
Yes I understand there's no problem with "add", otherwise you can't get to the notice I quoted. On your form, do you also have something like ``? – Herco Feb 04 '18 at 13:25
Yes; – Hüseyin Çiftçi Feb 04 '18 at 13:27
Please show the HTML contents of `
` to `
` in your question – Herco Feb 04 '18 at 13:28
i tried to get value of denad with javascript. when i use var denad = document.getElementById("denad").value i am getting value with an alert("denad"). – Hüseyin Çiftçi Feb 04 '18 at 13:30
1

Your form is not posting both `add` and `denad` at the same time. I can not help you without seeing how you post your form. – Herco Feb 04 '18 at 13:34
my form is very long. can i send you with e-mail or something? – Hüseyin Çiftçi Feb 04 '18 at 13:35
Can't you filter out the unrelated `` tags? Or use justpaste.it or something? – Herco Feb 04 '18 at 13:46
i deleted some parts and posted. Thank you for your help. – Hüseyin Çiftçi Feb 04 '18 at 13:54

about $_POST warning

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