如何编辑下拉选择的值？
[英]How to edit drop-down selected value?
问题描述
I have created a drop-down menu and many other text fields on my html page which fetches the option values from database. 
我在html页面上创建了一个下拉菜单和许多其他文本字段，可从数据库中获取选项值。 It is inserting the selected values in my database table. 它将选定的值插入我的数据库表中。 I want to edit the selected values from the drop-down menu and update them in database. 我想从下拉菜单中编辑选定的值，然后在数据库中更新它们。 When I open the edit page, the form shows the previous saved values to edit them. 当我打开编辑页面时，该表单显示先前保存的值以对其进行编辑。 But the problem is; 但是问题是； When I don't edit the drop-down values and submit the form, It shows me error of undefined index and when I edit or select different value of drop-down then it works fine and don't show any error. 当我不编辑下拉列表的值并提交表单时，它向我显示未定义索引的错误；当我编辑或选择不同的下拉列表的值时，它可以正常工作，并且不显示任何错误。 Here is my code: HTML 这是我的代码： HTML
<label>Courses</label><select name="courses" id="courses" class="dropdownclass"><option selected="selected" value="" disabled selected hidden>-- Select an option --</option><?php mysql_connect('localhost', 'root', ''); mysql_select_db('db'); $sql = "SELECT courses FROM table"; $result = mysql_query($sql); while ($row = mysql_fetch_array($result)) { echo "<option value=' " . $row['courses'] ."'>" . $row['courses'] ."</option>"; } ?> </select> <!-- begin snippet: js hide: false console: true babel: false -->
EDITRECORD 编辑记录
<?php include("connection.php"); $Sno = (int)$_GET['Sno']; $query = mysql_query("SELECT * FROM table WHERE Sno = '$Sno'") or die(mysql_error()); while($row = mysql_fetch_array($query)) { if (isset($_POST)) { echo ""; $id=$row['id']; $courses=$row['courses']; } } ?> <!DOCTYPE html> <html> <head> <title>Edit Record</title> </head> <body> <form id="form" action="update.php" method="post" enctype="multipart/form-data"> <fieldset> <input type="hidden" name="new" id="Sno" value="<?=$Sno;?>" /> <label>Courses</label><select name="courses" id="courses" class="dropdownclass" ><option selected="selected" value="" disabled selected hidden><?php echo $courses; ?></option><?php mysql_connect('localhost', 'root', ''); mysql_select_db('db'); $sql = "SELECT courses FROM table"; $result = mysql_query($sql); while ($row = mysql_fetch_array($result)) { echo "<option value=' " . $row['courses'] ."'>" . $row['courses'] ."</option>"; } ?> </select> </fieldset> </form> </body> </html>
UPDATE 更新
<?php include("connection.php"); $Sno =''; if( isset( $_POST['new'])) { $Sno = (int)$_POST['new']; } $id = mysql_real_escape_string($_POST['id']); if(isset($_POST['courses'])){ $courses = mysql_real_escape_string($_POST['courses']); }else{ $courses=$_POST['courses']; } query="UPDATE technicalsol SET id= '$id', courses = '$courses' WHERE Sno=$Sno"; $res= mysql_query($query); if($res){ echo "<div style ='font-size:20px; margin-left:140px;'>Records updated Successfully</div>"; include("search.php"); }else{ echo "Problem updating record. MY SQL Error: " . mysql_error(); } ?>
I want that; 我要那个; If I don't edit the drop-down selected value, It just takes the previous value and saves it. 如果我不编辑下拉选择的值，它将仅使用先前的值并保存。
1 个解决方案
解决方案1
-1 2018-04-05 05:10:36
What you are doing right now is listing the options in dropdown list. 您现在正在做的是在下拉列表中列出选项。 You are not setting up any answer 您没有设置任何答案
In your HTML: 在您的HTML中：
while ($row = mysql_fetch_array($result)) {
echo "<option value=' " . $row['courses'] ."'>" . $row['courses'] ."</option>";
}
make sure you make the old value selected. 确保您选择了旧值。 may be by using an IF statement 可能是通过使用IF语句
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