[英]Grab id from a select option field and update it in the selected user
嗨,我在创建程序时遇到一些问题,我有一个选择字段,该字段显示表用户中的所有用户,我希望当我选择一个用户并按链接时,我希望它可以将用户userid更新为puserid。表属性
php代码:
//connect with the database
$con = mysql_connect("localhost","root","");
$db = mysql_select_db("ocm2inf_ocm",$con);
$get=mysql_query("SELECT * FROM users WHERE PortalId = '14'");
$option = '';
//create rows of the users
while($row = mysql_fetch_assoc($get))
{
$option .= '<option value = "'.$row['UserId'].'">'.$row['firstName']. str_repeat(' ', 1). $row['UserId'].'</option>';
}
//update the row in the database after pressing the button
if(isset($_POST['linkownersandproperties']))
{
$UserId = $_POST['UserIdOwner'];
$sql = "UPDATE properties SET puserId='$UserId' WHERE propertyId=$id";
}
html代码:
<form class="create section" action="" method="post">
<select class="form-control" name="UserIdOwner">
<?php echo $option; ?>
</select>
<br />
<button style="width:110px; margin-left:10px;" type="submit" name="linkownersandproperties" value="Submit" class="btn btn-primary btn-lg submit-button1" required="required">Link</button>
</form>
提前致谢 :)
更改此规则
$sql = "UPDATE properties SET puserId='$UserId' WHERE propertyId=$id";
遵守这条规则
$sql = mysql_query("UPDATE properties SET puserId='$UserId' WHERE propertyId='$id'");
将其放在if语句的$sql
下
$query = mysql_query($sql, $db);
if (!$sql) {
echo "Failed";
} else {
echo "Executed";
}
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