php使用mysqli将结果集转换为数组，然后转换为json - php using mysqli convert resultset to an array and then to json

php using mysqli convert resultset to an array and then to json

我以为我想要的很简单，但由于某种原因，我被卡住了。我有以下内容：

$sql = "...";
if ($stmt = $con->prepare($sql)) {          
   $stmt->bind_param("sss", $x,$y,$z);
   if ($stmt->execute()) {
      $result = array(); //not sure if needed
      $stmt->bind_result($x1,$y1,$z1); //not sure if needed
      //loop resultset and put it in an array ($result);
      echo json_encode($result); // convert to json
      $stmt->close();
   }
}

我看到了fetchAll、fetch_assoc等等，但这些调用/函数的错误一直是未定义的。其他例子是未经准备的发言。无论我尝试了什么，都没能用结果集创建数组，我缺少什么？

感谢

使用bind_result后，您仍然需要获取这些：

$sql = "SELECT col1, col2, col3 FROM table_name WHERE col4 = ? AND col5 = ? AND col6 = ?";
if ($stmt = $con->prepare($sql)) {          
   $stmt->bind_param('sss', $x,$y,$z);
   if ($stmt->execute()) {
      $result = array();
      $stmt->bind_result($x1,$y1,$z1);
      // You forgot this part
      while($stmt->fetch()) {
          $result[] = array('col1' => $x1, 'col2' => $y1, 'col3' => $z1);
      }
      echo json_encode($result); // convert to json
      $stmt->close();
   }
}

或者，如果系统上有mysqlnd驱动程序：

$sql = "SELECT col1, col2, col3 FROM table_name WHERE col4 = ? AND col5 = ? AND col6 = ?";
if ($stmt = $con->prepare($sql)) {          
   $stmt->bind_param('sss', $x,$y,$z);
   if ($stmt->execute()) {
      $data = $stmt->get_result();
      $result = array();
      while($row = $data->fetch_assoc()) {
          $result[] = $row;
      }
      echo json_encode($result); // convert to json
      $stmt->close();
   }
}