skipe $_POST in php

Question

i have a php page which i should post some data to it like this:

$player=$_POST['player']; $age=$_POST['age']; $data=$_POST['data'];

but some times my page posts data and some times it shouldn't, when i dont post data i got Undefined index: error, is there any way to skip $_POST['data'] when no data is posted?

score 3 · Answer 1 · answered Feb 09 '15 at 14:10

3

$player = isset($_POST['player'])?$_POST['player']:"";
$age = isset($_POST['age'])?$_POST['age']:"";
$data = isset($_POST['data'])?$_POST['data']:"";

The isset first checks if it's defined, if yes, it assigns the variable else it assigns empty string.

answered Feb 09 '15 at 14:10

Rikudou_Sennin

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thanks man i didnt know about isset, it is working now haha :D – Sina Nouri Feb 09 '15 at 14:11
if it helped, will you accept my answer? :) – Rikudou_Sennin Feb 09 '15 at 14:12
i just click check sing to accept it but it said not until 12 minutes – Sina Nouri Feb 09 '15 at 14:12
1

@Rikudou_Senninthere is a 15minute wait before OP can accept any answer. – ʰᵈˑ Feb 09 '15 at 14:12

score 2 · Accepted Answer · answered Feb 09 '15 at 14:12

2

Use simple if (isset($_POST['key'])):

$player = isset($_POST['player']) ? $_POST['player'] : '';
$age = isset($_POST['age']) ? $_POST['age'] : '';
$data = isset($_POST['data']) ? $_POST['data'] : '';

Or better, but every value to some 'refix':

<input name="Data['player']"/>

and in php just do:

if (isset($_POST['Data'])) {
     $player = $_POST['Data']['player'];
     /* ... */
}

answered Feb 09 '15 at 14:12

Justinas

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1

I didn't downvote, by why "better, put every value to some 'refix'". Why is this better? Surely you should have used "alternatively". Maybe I'm being pedantic – ʰᵈˑ Feb 09 '15 at 14:14
@ʰᵈˑ Because if prefix is set, than your form is sent and every other values will be set too (unless you are missing html form element, than it will not work). – Justinas Feb 09 '15 at 14:15
$player = $_POST['Data']['player']; i just dont get this data is not related to player so why $_POST['Data']['player']? – Sina Nouri Feb 09 '15 at 14:19
@SinaNouri It's not `data` from your example, it's wrapping key, that holds all information from your form. It's example from Yii framework, where all form data is inside key same as model name. Like you you have contacts form, it will be `$_POST['Contacts']['email']` / `$_POST['Contacts']['content']`... – Justinas Feb 09 '15 at 14:22

score 0 · Answer 3 · answered Feb 09 '15 at 14:12

0

if(isset($_POST['DATA'])){
 $data = $_POST['DATA']
}

answered Feb 09 '15 at 14:12

Epodax

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score -2 · Answer 4 · answered Feb 09 '15 at 14:10

-2

if(isset($_POST['player'])):
    $player = $_POST['player'];
endif;

answered Feb 09 '15 at 14:10

Mr Pablo

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This will still show a warning – CᴴᵁᴮᴮʸNᴵᴺᴶᴬ Feb 09 '15 at 14:11
That syntax is encouraged for templating: http://php.net/manual/en/control-structures.alternative-syntax.php – ʰᵈˑ Feb 09 '15 at 14:11
because you're checking if against an undefined variable and/or key, you should check that it is set, with `isset` first – CᴴᵁᴮᴮʸNᴵᴺᴶᴬ Feb 09 '15 at 14:13
it will still throw the notice... you must use `isset()` – Rikudou_Sennin Feb 09 '15 at 14:13

score -2 · Answer 5 · answered Feb 09 '15 at 14:11

-2

Check if data is posted

if(isset($_POST['player']))
$player=$_POST['player'];
}
if(isset($_POST['age']))
$age=$_POST['age'];
}
if(isset($_POST['data']))
$data=$_POST['data'];
}

answered Feb 09 '15 at 14:11

rack_nilesh

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5
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that's actually only false in CLI :D – Aitch Feb 09 '15 at 14:12
$_POST is always set (in webserver) – Rikudou_Sennin Feb 09 '15 at 14:12
it should work now. please remove downvote – rack_nilesh Feb 09 '15 at 14:16

skipe $_POST in php

5 Answers5