mysqli_fetch_assoc仅返回数组的第一个元素

sklrboy 发表于 Dev

Sklrboy

我正在做一个项目，已登录的用户应该能够在首页上发布笔记，并且正在登录的特定用户的笔记应该打印在新笔记表单上方。我已经为此编写了一个函数，该函数可以mysqli_query识别我拥有的所有6个条目，但是mysqli_fetch_assoc仅打印出6个中的第一个音符。我该怎么办？这是我的代码：

<?php    
    function find_notes_by_id($user_id) {
	global $connection;
	
	$safe_user_id = mysqli_real_escape_string($connection, $user_id);
	
	$query = 'SELECT content ';
	$query .= 'FROM notes ';
	$query .= 'WHERE user_id = '.$safe_user_id;
	$result = mysqli_query($connection, $query);
 
	//mysqli_result Object ( [current_field] => 0 [field_count] => 1 [lengths] => [num_rows] => 6 [type] => 0 )

	confirm_query($result);

	$row = mysqli_fetch_assoc($result);
	//Array ( [content] => First! ) = it only shows the very first element
	return $row;
}
?>

<?php
$notes_set = find_notes_by_id($userRow['id']);
  foreach($notes_set as $note){
	echo $note;
	echo "<br />";
  }
?>

生死

1.您需要遍历整个result-set object过程while()

2.将所有数据保存到array，然后返回array以获取所有记录

如下所示：

<?php    
  function find_notes_by_id($user_id) {
   global $connection;

    $safe_user_id = mysqli_real_escape_string($connection, $user_id);

    $query = "SELECT `content` FROM `notes` WHERE `user_id` = $safe_user_id";
    $result = mysqli_query($connection, $query);

    confirm_query($result);

    $final_data = array(); // create an array
    while($row = mysqli_fetch_assoc($result)){ // iterate over the result-set object to get all data
        $final_data[] = $row; //assign value to the array
    }
    return $final_data; // return array
  }
?>

现在：-

<?php
$notes_set = find_notes_by_id($userRow['id']);

print_r($notes_set) ; // print result to check array structure so that you can use it correctly in foreach
  foreach($notes_set as $note){
    echo $note['content'];
    echo "<br />";
  }
?>

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