Question

我认为我想要的很简单，但出于某种原因，我被困住了。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以及更多内容，但我不断将这些调用/函数的错误视为未定义。其他例子是未准备好的陈述。无论我尝试过什么，我都无法使用结果集创建一个数组，我缺少什么？

由于

Answer 1

使用bind_result后，您仍然需要获取它们：

$sql = "SELECT col1, col2, col3 FROM table_name WHERE col4 = ? AND col5 = ? AND col6 = ?";
if ($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();
   }
}

php使用mysqli将结果集转换为数组，然后转换为json

1 个答案: