php - "Notice: Undefined variable" -
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i'm getting error when run following change password script:
notice: undefined variable: $insert in c:\xampp\htdocs\oh\change.php on line 21 password not changed
<?php include('db_connection.php'); include('crypt.php'); session_start(); if (isset($_post['submit'])) { $oldpassword = $_post['current_password']; $newpassword= $_post['new_password']; $confirm_password =$_post['confirm_password']; $user_name = $_session['username']; $old = decrypt($oldpassword); $select = mysql_query("select * staff username='$user_name'"); $fetch = mysql_fetch_array($select); $data_password = $fetch['password'; if ($newpassword == $confirm_password && $data_password == $old) { $pass = encrypt($confirm_password); $insert = mysql_query("update staff set password='$pass' username='$user_name'"); } if ($insert) { echo "password changed"; } else { echo "password not changed"; } } mysql_close($con); ?>
well, because might not in if
statement $insert
gets set. if case $insert
not exist en notice.
try check if variable exists like:
if (isset($insert)) { echo "password changed"; } else { echo "password not changed"; }
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