我有一些PHP代码可以从数据库中设置变量。我会解释一下,如果没有意义,请告诉我。
因此,如果*
,我有一个查询从表格中选择class = '$class'
。这一切都有效我开始工作了。
然后我设置了像$id = $row["id"];
这样的变量,并且一切正常,在我的HTML代码中我有<p><?php echo $id?></p>
,如果它们class = $class
它会显示它,但是,如果它没有满足那些没有设置变量的要求,所以我得到错误Notice: Undefined variable: id in C:\wamp64\www\studentplanner\account\homework.php on line 73
。
我想要做的只是在满足要求的情况下以HTML格式输出结果。
不知道这是否有意义!
$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
// output data of each row
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
$teacher_set = $row["teacher_set"];
$class = $row["class"];
$name = $row["name"];
$description = $row["description"];
}
}
<p><?php echo $id?></p>
<p><?php echo $teacher_set?></p>
<p><?php echo $class?></p>
<p><?php echo $name?></p>
<p><?php echo $description?></p>
答案 0 :(得分:1)
在while循环之前设置标志变量然后使用该变量来决定是否以html打印数据
$data_exist = false;
if (mysqli_num_rows($result) > 0) {
// output data of each row
$data_exist = true;
while($row = mysqli_fetch_assoc($result)) {
$id = $row["id"];
$teacher_set = $row["teacher_set"];
$class = $row["class"];
$name = $row["name"];
$description = $row["description"];
}
}
if($data_exist)
{
?>
<p><?php echo $id?></p>
<p><?php echo $teacher_set?></p>
<p><?php echo $class?></p>
<p><?php echo $name?></p>
<p><?php echo $description?></p>
<?php
}
?>
答案 1 :(得分:0)
$sql = "SELECT * FROM homework WHERE class = '$class'";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
while($row = mysqli_fetch_assoc($result)) {
?>
<p><?php echo $row["id"];?></p>
<p><?php echo $row["teacher_set"];?></p>
<p><?php echo $row["class"];?></p>
<p><?php echo $row["name"];?></p>
<p><?php echo $row["description"];?></p>
<?php } } ?>