MySql column value by searching of another value in particular record

Ionic Framework ionic-v3
6

@lado No. Your backend should be like:

<?php
require 'connection.php';

$name = $_POST['name'];
echo 'Name: ' . $name; // add this line so you can see what you receive as name on your back-end

$sql = "SELECT * from nameTab where name='$name'";

$result = mysqli_query($con, $sql);

while($row = $result->fetch_assoc()) {
    echo "username: " . $row["username"] . "with id: " . $row["ID"];
}
?>

Instead of $_POST['name'] you could try:

// (remove it) $name = $_POST['name'];
$request = json_decode(file_get_contents("php://input"));
echo 'JSON: ' . $request;

$name = $request->name;
echo 'Name: ' . $name;

(Because you are sending a JSON)

Your front-end should be like:

  check() {
    var headers = new Headers();
    headers.append('Content-Type', 'application/x-www-form-urlencoded');
    let options = new RequestOptions({ headers: headers });

   let postParams = { name: this.name };

    this.http.post("http://site.info/php/check.php", JSON.stringify(postParams), options)
      .subscribe(data => {
        console.log('data', data); // first see what you receive as data
      }, error => {
        console.log(error);
      });
  }

If you don’t receive the name (echo 'Name: ' . $name; outputs nothing as the name), try without the stringify:

this.http.post("http://site.info/php/check.php", postParams, options)

Or try with FormData:

let body = new FormData();
body.append('name', this.name);
this.http.post("http://site.info/php/check.php", body , options)

I’m not a http expert though, but you can search examples in the web, like:

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