为什么不执行此echo命令？

Question

Please could someone help me understand why the echo command, 'Incorrect Membership Number, please try again.' isn't working?

Everything else seems to be functioning okay.

<?php
define('DB_HOST', 'localhost');
define('DB_NAME', 'DBNAME');
define('DB_USER', 'USER');
define('DB_PASSWORD', 'PASS');
$con = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
$db = mysql_select_db(DB_NAME, $con) or die("Failed to connect to MySQL: " . mysql_error());
$mem_no = $_POST['mem_no'];

function SignIn()
{
    session_start();
    if (!empty($_POST['mem_no'])) {
        $query = mysql_query("SELECT * FROM members where mem_no = '$_POST[mem_no]'") or die(mysql_error());
        $row = mysql_fetch_array($query) or die(mysql_error());
        if (!empty($row['mem_no'])) {
            $_SESSION['mem_no'] = $row['mem_no'];
            header("Location: " . $_POST['cat_link']);
        } else {
            echo "Incorrect Membership Number, please try again.";
        } // This line is not executing
    } else {
        echo "Please go back and enter a Membership Number";
    }
}

if (isset($_POST['submit'])) {
    SignIn();
}

HTML Form as follows:

<form method="post" action="/check.php">
<p>Membership No.</p>
<input name="mem_no" type="text" id="mem_no">
<input name="cat_link" type="hidden" value="https://www.redirectlink.com">
<input name="submit" type="submit" id="submit" value="AELP Member Rate">
</form>

Link to test: https://www.eiseverywhere.com/ehome/index.php?eventid=106953&tabid=239372 and a valid 'Membership Number' is 1234 if you wish to test.

Leaving the form blank does give the correct error message and entering a valid number does redirect me correctly, but inputting an invalid number (9999 for eg) doesn't give me the correct output message.

Thank you in advance for any responses.

Regards, Ash

Answer 1

You need to count rows, because even when a sql query has no results it is not empty. So count it.

function SignIn()
{
    session_start();
    if (!empty($_POST['mem_no'])) {
        $query = mysql_query("SELECT * FROM members where mem_no = '". $_POST['mem_no'] ."'") or die(mysql_error());

        #count rows
        $count = mysql_num_rows($query);
        $row = mysql_fetch_array($query) or die(mysql_error());

        #check count
        if ($count != 0) {
            $_SESSION['mem_no'] = $row['mem_no'];
            header("Location: " . $_POST['cat_link']);
        } else {
            echo "Incorrect Membership Number, please try again.";
        } 

    } else {
        echo "Please go back and enter a Membership Number";
    }
}

Just a small reminder. mysql_ class in PHP is deprecated and will be removed in the next versions, I suggest you going to use mysqli_ or work with PDO 's

Answer 2

Your script is hard to debug for several reasons:

1. It's unindented (I've fixed that for you in the question)
2. You're using deprecated mysql functions (see this ). Switch to PDO or mysqli.
3. You have several exit points in your script. Everytime you make a call to the database, you "die" if you fail. that's not good

regardless, I suspect that one of those "die" making your script end prematurely. Instead of using die, handle the errors yourself.

Answer 3

I think this should work:

(Also put error reporting at the top!)

<?php
    error_reporting(E_ALL);
    ini_set("display_errors", 1);
?>

<?php
    session_start();

    define('DB_HOST', 'localhost');
    define('DB_NAME', 'DBNAME');
    define('DB_USER', 'USER');
    define('DB_PASSWORD', 'PASS');

    $con = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error());
    //$db = mysql_select_db(DB_NAME, $con) or die("Failed to connect to MySQL: " . mysql_error()); //Useless

    $mem_no = $_POST['mem_no'];

    function SignIn() {

        if (!empty($_POST['mem_no'])) {
            $query = mysql_query("SELECT * FROM members WHERE mem_no = '" . $_POST['mem_no'] . "'") or die(mysql_error());
            $row = mysql_fetch_array($query) or die(mysql_error());
            $count = mysql_num_rows($query);

            if ($count >= 1) {
                $_SESSION['mem_no'] = $row['mem_no'];
                header("Location: " . $_POST['cat_link']);
            } else {
                echo "Incorrect Membership Number, please try again.";
            } // This line is not executing
        } else {
            echo "Please go back and enter a Membership Number";
        }
    }

    if (isset($_POST['submit'])) {
        SignIn();
    }
?>

Answer 4

It doesnt make sense to put two else statements in a row. Take the last else block out of the inner if block

   if(!empty($_POST['mem_no'])) {
        //code

         }else{

             echo "Please go back and enter a Membership Number";
         } 

That should execute just fine