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Found 1 result

  1. Hi All, I'm looking for a solution in using a function. The function fetches data from a database. In a column multiple results are found. I want to have all the records outisde the function. Inside the function all records are shown, but outside only the last record. I thought I could use return array, but it only returns multiple variables and not multiple rows. Not sure if fetch() is the solution for arrays. example: <?php function test_func($val4) { global $val1; global $val2; global $val4; $servername = "host"; $username = "johndoe"; $password = "admin"; $dbname = "dbase_db"; // Create connection $conn = new mysqli($servername, $username, $password, $dbname); // Check connection if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } // if ($stmt = $conn->prepare("SELECT col1, col2 FROM table WHERE col4 = ? ; ")) { $stmt->bind_param("s", $val4); $stmt->execute(); $stmt->store_result(); $num_of_rows = $stmt->num_rows; var_dump($num_of_rows); $stmt->bind_result($val1, $val2); if ($num_of_rows > 0){ while ($stmt->fetch()) { // managed to output multiple rows here, but that's not the goal } return array ($val1, $val2); } } } $val4='form_value'; test_func($val4); // return multiple rows here, outside the function How should I solve this?