🔒 Closed Need help php

[D0MIN4T0R]

mga masters, pwede patulong panu mag validate ng form before saving sa database? gamit ang php o javascript. nag research na po ako pero hindi sya nag vavalidate pag sinave sa database.
Thanks in advance po

Hirap pag self study :banghead:

 

[That Poppy]

you mean mag validate ng required fields sa form?

 

[D0MIN4T0R]

you mean mag validate ng required fields sa form?

yes po. ang problema ko is hindi sya masi save sa database boss.

 

[That Poppy]

ing mong sabhin dapat hindi masave sa data base pag ka hindi kinumpleto ang required fields? meron nmn code sa html na required. ex

<form action="demo_form.asp">
Username: <input type="text" name="username" required>
<input type="submit">
</form>


kung gusto mo nmn gmitan ng php

<?php
// define variables and set to empty values
$nameErr = $emailErr = $genderErr = $websiteErr = "";
$name = $email = $gender = $comment = $website = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$nameErr = "Name is required";
} else {
$name = test_input($_POST["name"]);
}

if (empty($_POST["email"])) {
$emailErr = "Email is required";
} else {
$email = test_input($_POST["email"]);
}

if (empty($_POST["website"])) {
$website = "";
} else {
$website = test_input($_POST["website"]);
}

if (empty($_POST["comment"])) {
$comment = "";
} else {
$comment = test_input($_POST["comment"]);
}

if (empty($_POST["gender"])) {
$genderErr = "Gender is required";
} else {
$gender = test_input($_POST["gender"]);
}
}
?>

 

[jokerger25]

Mas mainam na server-based ang gamitin mo for validation wag yung client-based... Gaya ng pinapakita ni kira111101 .for security reasons...gamitin mo lang yung client-based to check if may value yung fields...

Server-based=Php
Client-based=javasript/jquery

 

[jrmendz]

arresultTry mo aralin ang AJAX Yan kasi pinaka used ngayon. gagawa ka ng API Request from AJAX (client) to PHP (server) so mangyayari dalawang validation. Kung ung validation sa backend nagreturn ng false hindi na siya tutuloy. to make a cleaner code.

function getValidated($arrData)
{

if($arrData)
{

if($arrData['name'])
{

return true;​

}​

}

return false;​

}

tapus dun sa action button mo

function actionSubmit()
{
$arrRequest = $_POST;
$arrResult = [];

//If true it will proceed
if($this->getValidated($arrRequest))
{

return $arrResult = [

'data' => $arrRequest,
'success' => true​

];​

}

return false;​

}

then sa ajax

$("#actionSubmit").submit(function (e) {
$.ajax({
type: "post",
dataType: 'json',
url: "sample_function.php",
data: {
'name': name
},
success: function (data) {
if (data.success == true) {
//TODO
}
}
});
e.preventDefault();
});

 

[D0MIN4T0R]

ing mong sabhin dapat hindi masave sa data base pag ka hindi kinumpleto ang required fields? meron nmn code sa html na required. ex

<form action="demo_form.asp">
Username: <input type="text" name="username" required>
<input type="submit">
</form>


kung gusto mo nmn gmitan ng php

<?php
// define variables and set to empty values
$nameErr = $emailErr = $genderErr = $websiteErr = "";
$name = $email = $gender = $comment = $website = "";

if ($_SERVER["REQUEST_METHOD"] == "POST") {
if (empty($_POST["name"])) {
$nameErr = "Name is required";
} else {
$name = test_input($_POST["name"]);
}

if (empty($_POST["email"])) {
$emailErr = "Email is required";
} else {
$email = test_input($_POST["email"]);
}

if (empty($_POST["website"])) {
$website = "";
} else {
$website = test_input($_POST["website"]);
}

if (empty($_POST["comment"])) {
$comment = "";
} else {
$comment = test_input($_POST["comment"]);
}

if (empty($_POST["gender"])) {
$genderErr = "Gender is required";
} else {
$gender = test_input($_POST["gender"]);
}
}
?>

thanks master, ta try ko to mamaya.

Mas mainam na server-based ang gamitin mo for validation wag yung client-based... Gaya ng pinapakita ni kira111101 .for security reasons...gamitin mo lang yung client-based to check if may value yung fields...

Server-based=Php
Client-based=javasript/jquery

ahH. sige po. try ko rin po yung ka kira. beginners pa rin nmn kami boss.