我试图验证我的选择标记,以便如果用户没有选择会话,则在按下提交按钮时出现警报。我有我的html和php代码下面,谢谢
<select name="ses">
<option> Select a Session </option>
<?php
$get_sess = "select * from sessions where course_id='1'";
$run_sess = mysqli_query($con, $get_sess);
while ($sess_row=mysqli_fetch_array($run_sess)){
$ses_id=$sess_row['session_id'];
$ses_title=$sess_row['session_title'];
echo "<option value='$ses_id'>$ses_title</option>";
} ?>
</select>
我的PHP代码是:
<?php
//if submit button is set i.e 'publish exercise now' pressed then:
if(isset($_POST['submit1'])){
$exercises_ses = $_POST['ses'];
$exercise_text = $_POST['exercise_text'];
if($_POST['ses'] == 'NULL'){
echo "<script>alert('Please select a session for the exercise')</script>";
}
}
您应该只使用空函数。这将检查值是否为"空",检查手册中所有符合"空"的内容,每个评论我认为这符合您的标准。
if(isset($_POST['submit1'])){
$exercises_ses = $_POST['ses'];
$exercise_text = $_POST['exercise_text'];
if(empty($_POST['ses'])){
echo "<script>alert('Please select a session for the exercise')</script>";
} else {
// execute functions form value is correct
}
}
'NULL'
是一个字符串,如果您想与空值进行比较-使用NULL
。我不认为你的db中的$ses_id
有'NULL'
的价值。检查是否设置了值可以用isset
来完成,用empty
来检查是否为空:
if (isset($_POST['ses']) && !empty($_POST['ses'])) {