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[英]Select data from mysql with php, display in html dropdown list and insert selected value as URL parameter via javascript
[英]How to insert selected checkbox data from dropdown list in MYSQL database with PHP
我有一個注冊表,用戶可以在其中通過復選框選擇多個選項。 但是現在當我嘗試將選擇的數據插入數據庫時,它不起作用。 以下是一些代碼:
HTML代碼 :
<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>
JS
<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
});
$('#langOptgroup').multiselect({
columns: 4,
placeholder: 'Select Languages',
search: true,
selectAll: true
});
</script>
和PHP:這是將數據插入數據庫的代碼。 我正在使用舊的php版本插入數據。 只是練習。
<?php
if(isset($_POST['submit'])){
include("test/admin/includes/db.php");
$skills=$_POST['langOpt2'];
foreach($_POST['langOpt2'] as $skills){
mysql_query("insert into `job_seeker_reg` (`j_skills`) values('','$skills')");
}}
?>
貝婁代碼有助於解決您的問題。 我有一個例子是選擇學科。 您可以使用選項菜單執行任務。
HTML代碼:
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代碼:
<?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.........";
}
?>
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