Question : if sentence returns weired value

The code snippet attached doesn't give me the answer I would have expected.

This piece of code seems not to work:

if($max_upptagnings_turnummer >= $max_upptagnings_turnummer_bat_2) { $nytt_turnummer = $max_upptagnings_turnummer + 1; }
if($max_upptagnings_turnummer < $max_upptagnings_turnummer_bat_2) { $nytt_turnummer = $max_upptagnings_turnummer_bat_2 + 1; }

In my practical example, "$max_upptagnings_turnummer " returns 1 from the database. and "$max_upptagnings_turnummer_bat_2" returns 2.

Thus $max_upptagnings_turnummer < $max_upptagnings_turnummer_bat_2 which leads to
{ $nytt_turnummer = $max_upptagnings_turnummer_bat_2 + 1; }.
Since $max_upptagnings_turnummer_bat_2 returns 2, and it should be increased by 1, I expect $nytt_turnummer to return 3. But it doesn't. It returns 2.

Please gurus of php and MySQL, could you help me identify where I go wrong.

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<?php
if(!empty($_POST["upptagningsdatum_varv1"])) {
$SQL17 = " SELECT MAX(upptagnings_turnummer) as max_upptagnings_turnummer FROM medlem WHERE upptagningsdatum = '$upptagningsdatumet_varv1' AND upptagningsupplagsplats = '$varv_3_namn' ";  }

if(!empty($_POST["upptagningsdatum_varv2"])) {
$SQL17 = " SELECT MAX(upptagnings_turnummer) as max_upptagnings_turnummer FROM medlem WHERE upptagningsdatum = '$upptagningsdatumet_varv2' AND upptagningsupplagsplats = '$varv_4_namn' ";  }

$ret17 = mysql_query($SQL17);
if (!$ret17) { echo("Nr 2 (här): " .mysql_error()); }
else {
	while ($row = mysql_fetch_array($ret17)) {
        $max_upptagnings_turnummer  = $row["max_upptagnings_turnummer"];
}}

// Båt nr 2:
//if($_POST['bat_nr'] == "2") {
if(!empty($_POST["upptagningsdatum_varv1"])) {
$SQL18 = " SELECT MAX(upptagnings_turnummer_bat_2) as max_upptagnings_turnummer_bat_2 FROM medlem WHERE upptagningsdatum = '$upptagningsdatumet_varv1]' AND upptagningsupplagsplats = '$varv_3_namn' ";  }

if(!empty($_POST["upptagningsdatum_varv2"])) {
$SQL18 = " SELECT MAX(upptagnings_turnummer_bat_2) as max_upptagnings_turnummer_bat_2 FROM medlem WHERE upptagningsdatum = '$upptagningsdatumet_varv2]' AND upptagningsupplagsplats = '$varv_4_namn' ";  }

$ret18 = mysql_query($SQL18);
if (!$ret18) { echo("Nr 4: " .mysql_error()); }
else {
	while ($row = mysql_fetch_array($ret18)) {
        $max_upptagnings_turnummer_bat_2  = $row["max_upptagnings_turnummer_bat_2"];
}}

//------------------------------
if($max_upptagnings_turnummer >= $max_upptagnings_turnummer_bat_2)  {  $nytt_turnummer = $max_upptagnings_turnummer + 1;  }
if($max_upptagnings_turnummer < $max_upptagnings_turnummer_bat_2)  {  $nytt_turnummer = $max_upptagnings_turnummer_bat_2 + 1;  }
//------------------------------

Answer : if sentence returns weired value

I would first do an 'echo' to make sure you are getting the info you're expecting.

echo "$max_upptagnings_turnummer - $max_upptagnings_turnummer_bat_2<br>";

Question : if sentence returns weired value

Answer : if sentence returns weired value

I would first do an 'echo' to make sure you are getting the info you're expecting. 1: echo "$max_upptagnings_turnummer - $max_upptagnings_turnummer_bat_2<br>";

I would first do an 'echo' to make sure you are getting the info you're expecting.

1:

echo "$max_upptagnings_turnummer - $max_upptagnings_turnummer_bat_2<br>";