如何只更新选定的mysql数据库条目而不会弄乱早期的数据库条目

时间:2018-02-02 11:00:17

标签: php mysql

好的,我真的需要一些帮助,我需要能够只更新数据库的选定部分而不会弄乱任何先前的条目。此代码确实填充了数据库,但最终会弄乱先前的条目

$errors = array();
if(isset($_POST)){
 $v_fname = ((isset($_POST['v_fname']))?sanitize($_POST['v_fname']):'');
 $v_lname = ((isset($_POST['v_lname']))?sanitize($_POST['v_lname']):'');
 $v_dob = ((isset($_POST['v_dob']))?sanitize($_POST['v_dob']):'');
 $v_country = ((isset($_POST['v_country']))?
 sanitize($_POST['v_country']):'');

 $v_city = ((isset($_POST['v_city']))?sanitize($_POST['v_city']):'');

 $v_phone = ((isset($_POST['v_phone']))?sanitize($_POST['v_phone']):'');

 $v_bio = ((isset($_POST['v_bio']))?sanitize($_POST['v_bio']):'');


if($v_bio == "" && $v_phone == "" && $v_city == "" && $v_fname == "" && 
 $v_lname == ""&& $v_country == "" ){
 $errors = "Please Fill all the details";

}
if(!empty($errors)){
//echo display_errors($errors);

}else{


$db-> query("UPDATE `vendors` SET `v_phone` = '$v_phone', `v_fname` 
='$v_fname', `v_bio` = '$v_bio', `v_location` = '$v_city', `v_country` = 
'$v_country', `v_dob` = '$v_dob', `v_lname` = '$v_lname' WHERE v_id = 
'$v_id'");
echo $v_id;

}
}

这是我想要更新的个人资料。我希望用户能够选择性地创建条目

enter image description here

1 个答案:

答案 0 :(得分:0)

您需要设置 $v_id

我在你的代码中没有看到它