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$sqlName = "SELECT idStudent,
STUFF(
(
SELECT ', ' + b.CONCAT(prefix, firstname, ' ' , lastname)
FROM student AS b
WHERE student.idProject = '$idProject'
FOR XML PATH(''), TYPE
).value('.[1]', 'nvarchar(max)'), 1, 1, ''
) AS [FullName]
FROM student ";
$objQueryName = mysqli_query($link,$sqlName);
while($objResultName = mysqli_fetch_array($objQueryName))
$name = $objResultName["FullName"];
{echo $name;" ,";
?> </td>
มันขึ้น error แบบนี้ค่ะ
Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, bool given in C:\xampp\htdocs\projectpresent\show-Project-detail-PosterName.php on line 111
Tag : PHP, MySQL
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| Date :
2021-04-01 17:31:15 |
By :
siriwimon2559 |
View :
582 |
Reply :
1 |
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