A small PHP question about logon

{Code ...} the above code is used to verify whether the user name and password entered by the user are correct. for convenience, I simplified the code and removed the content for obtaining database data, why does the output & quot; fail to find the user & quot; after opening the page?

If ($ row) {// Determine whether the password is correct if ($ row ['User _ password'] ===$ password) {echo 1 ;} else {echo "incorrect password" ;}} else {echo "no user name found ";}

The above code is used to verify whether the user name and password entered by the user are correct. for convenience, I simplified the code and removed the content for obtaining database data, why does "no user found" output when nothing is lost after the page is opened?

Reply content:

If ($ row) {// Determine whether the password is correct if ($ row ['User _ password'] ===$ password) {echo 1 ;} else {echo "incorrect password" ;}} else {echo "no user name found ";}

The above code is used to verify whether the user name and password entered by the user are correct. for convenience, I simplified the code and removed the content for obtaining database data, why does "no user found" output when nothing is lost after the page is opened?

If ($ row) code must be written in the POST request code block. In this way, the code will not be executed.

For example, after clicking Login, the submitted user and password fields are: username password

$ Username = isset ($ _ POST ['username'])? $ _ POST ['username']: false; $ password = isset ($ _ POST ['password'])? $ _ POST ['password']: false; if ($ username & $ password) {if ($ row) {// Determine whether the password is correct if ($ row ['User _ password'] ===$ password) {echo 1 ;}else {echo "incorrect password ";}} else {echo "user name not found ";}}

You print $ row to see what it is.

It indicates that the code is executed when you open the webpage, and if it is false

Note:$rowIf it is false, you should focus on obtaining$rowInstead of focusing on the posted code.