PHP - 在控制台中发送空值并获取未定义索引

Question

I"m sending to a php code a json string named- student

$scope.student = {name: "Joe", grades: "85", info: ""};

Now the php code is simple -

    <?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
    //read the json file contents
    $jsondata = file_get_contents("php://input");

    //convert json object to php associative array
    $data = json_decode($jsondata, true);

    $studentname = $data['studentname'];
    $stuedentgrades = $data['stuedentgrades'];
    $studentinfo = $data['studentinfo'];

$sql = "INSERT INTO students (name, grades, info)
VALUES ('$studentname', '$stuedentgrades', '$studentinfo')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

The thing is that sometimes the "info" can be empty and i would like to pass it empty to the server - but I'm getting -

 Undefined index: info

Now i tried to add to the code the isset -

  `if(isset($_POST('info'))){

     $info= $data['info'];
}else{
    echo "NOOOOOOOOOO";
}   ` 


$studentname = $data['studentname'];
    $stuedentgrades = $data['stuedentgrades'];
    $studentinfo = $data['studentinfo'];

$sql = "INSERT INTO students (name, grades, info)
VALUES ('$studentname', '$stuedentgrades', '$studentinfo')";

but it didn't seems to do the work.

So what am I doing wrong? How can I pass empty value into the table?

As you can see I'm novice when it comes to PHP so any help would be nice

Answer 1

Hope it will help you (please read comments (//...) carefully and check changes):-

$scope.student = [name: "Joe", grades: "85", info: ""]; // `.` from variable name eed to be removed in any manner

Now i assume it's:-

$scope = [name: "Joe", grades: "85", info: ""];

Now the php code is simple -

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 
    //read the json file contents
    $jsondata = file_get_contents("php://input");

    //convert json object to php associative array
    $data = json_decode($jsondata, true);
     // here i assume that $data is exactly equals what you shown above that means $data = [name: "Joe", grades: "85", info: ""];

    //Now  change here:-

    $studentname = (!empty($data['name']))? $data['name'] : ""; //check change here
    $stuedentgrades = (!empty($data['grades']))? $data['grades'] : "";
    $studentinfo = (!empty($data['info']))? $data['info'] : "Nooooo";

$sql = "INSERT INTO students (name, grades, info)
VALUES ('$studentname', '$stuedentgrades', '$studentinfo')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>