Question

我有以下代码在获取数组元素时生成错误

 $task = new task();
    $task->connect();
    $services = $task->viewTask_front_android($_GET['pno']); 
    print_r($services);
     while($info =  mysqli_fetch_assoc($services)) 
     { 
     Print "<b>Name:</b> ".$info['starttime'] . " "; 

     } ?>

来自print_r（服务）的

获得

Array ( [task_id] => 14 [user_id] => 123 [employee_id] => 456 [service_id] => 2 [starttime] => 2:00 AM [endtime] => 4:00 AM [servicename] => se a [servicedescription] => ddsdsd [employeename] => dsd [employeepicture] => pictures/noimage.gif [pic_path] => pictures/noimage.gif )

有什么问题吗？ while（$ info = mysqli_fetch_assoc（$ services））它生成错误Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\wamp\www\proj\android\services.php on line 70

Answer 1

看起来像是

$services = $task->viewTask_front_android($_GET['pno']);

已经返回一个数组（从外观开始的一行）。你没有/无法从中获取数组，就好像它是mysql_query的结果一样，因为它不是。

相反，只需打印$services['starttime']：

print "<b>Name:</b> ".$services['starttime'] . " ";

编辑：如果我不清楚，也完全删除while循环。

Answer 2

不需要循环，它返回一维数组......

Print "<b>Name:</b> $services[starttime] ";

获取数组时出错

2 个答案: