我有一个注册表单,用户可以通过复选框选择多个选项。但是现在当我尝试将所选数据插入数据库时,它无法正常工作。以下是一些代码:
HTML CODE :
app.post("/api/posts", function(req, res) {
console.log(req.body);
db.Post.create({
image: req.body.image,
caption: req.body.caption,
tags: req.body.tags
}).then(function(dbPost) {
res.redirect('/');
console.log("posted!")
}).catch(function(error) {
// Handle the error here
console.error(error);
});
JS
<div class="col-md-6">
<div class="form-group">
<label for="decisions3">Skills</label>
<select name="langOpt2[]" id="langOpt2" multiple="multiple" class="form-control" required data-validation-required-messge="This field is required">
<?php $selectskill = 'select * from skills where status=1';
$dataskill = mysql_query($selectskill);
while($resultskill = mysql_fetch_object($dataskill))
{?>
<option value="<?=$resultskill->skill_name?>"<?php if($result001->skill_name==$resultskill->skill_name){?> selected="selected"<?php } ?> >
<?=$resultskill->skill_name?>
</option>
<?php }
?>
</select>
</div>
和PHP: 这是将数据插入数据库的代码。我使用旧的php版本插入数据。只是练习。
<script>
$('#langOpt').multiselect({
columns: 1,
placeholder: 'Select Languages'
});
$('#langOpt2').multiselect({
columns: 1,
placeholder: 'Select Languages',
search: true
});
$('#langOpt3').multiselect({
columns: 1,
placeholder: 'Select Languages',
search: true,
selectAll: true
});
$('#langOptgroup').multiselect({
columns: 4,
placeholder: 'Select Languages',
search: true,
selectAll: true
});
</script>
答案 0 :(得分:0)
Bellow代码有助于解决您的问题。我举了一个例子就是选择主题。您可以使用选项菜单来完成任务。
HTML code:
select subject: <select name="Subject" id="dob-day">
<option ">-----</option>
<option value="Maths">Maths</option>
<option value="Science">Science</option>
<option value="Computer Science">Computer Science</option>
<option value="Languages">English</option>
<option value="Others">Others</option>
</select>
php代码:
<?php
$dbhost = "localhost";// enter host
$dbuser = "root";// enter root
$dbpass = ""; // enter password
$dberror1 = "Could not connect to the database connection";
$dberror2 = "Could not connect to the database";
$conn = mysqli_connect($dbhost,$dbuser,$dbpass) or die ($dberror1);
$select_db = mysqli_select_db($conn,'your_database_name') or die ($dberror2);
$subject = $_POST['Subject'];
$sql = "INSERT INTO your_table_name (Subject) VALUES ('$subject')";
$result= mysqli_query($conn,$sql);
if($result){
echo "Enter is success.........";
}
?>