Question

这是我的代码：

$user = $_SESSION['uname'];
echo $user;
$output=mysqli_query($con,"SELECT pedia_id FROM users WHERE uname = $user");
echo $output;
$result = mysqli_query($con,"SELECT * FROM infant_info where pedia_number = $output ORDER BY last_name ASC");

它回显$user但不回应$output，我收到了错误：

警告：mysqli_fetch_array（）要求参数1为mysqli_result，第69行的C：\ xampp \ htdocs \ Infant \ view \ InfantInfo_table.php中给出布尔值

Answer 1

你需要将$ user放在引号中。

"SELECT pedia_id FROM users WHERE `uname` = '".$user."'"

mysql_query也会返回mysqli_result。你需要从$ output获取pedia_id，然后在下一个查询中使用它。

$row = mysqli_fetch_array($output);

现在在下一个查询中使用$ row [0]而不是$ output。把它放在引号中。

$user = $_SESSION['uname'];
echo $user;
$output=mysqli_query($con,"SELECT pedia_id FROM users WHERE uname = '" . $user . "'");
echo $output;  //its mysqli_result object
//Fetch the row from the result
$row = mysqli_fetch_array($output);
echo $row[0];  //this is pedia_id
$pedia_id = $row[0];  //wrap it in quotes if its a string
$result = mysqli_query($con,"SELECT * FROM infant_info where pedia_number = $pedia_id ORDER BY last_name ASC");

while($infantRow = mysqli_fetch_array($result))
{
   //do something here with individual infant_info row
}

编辑：假设$ user的字符串值可以有引号。在这种情况下你也需要逃避它。一般来说，消毒用于形成查询的任何值都被认为是最佳实践。或者你可以使用Prepared Statements。

如何从查询中获取数据

1 个答案: