Echo doesnt appears in my pop up

Question

I have a popup with a drop down but the echo don't work, why?

if (($aData2['rennen'] == 6)) {
    print"Es nehmen bereits 6 Fahrer an diesem Rennen teil, bitte Name des Rennnens ändern";
    exit;
}

if () {.......... try if ($aData2['rennen'] == 6) { print "whatever";} — Mark Ng, Dec 09 '15 at 14:40
it is returning true they code is running inside the if but its not given out — j.m.l, Dec 09 '15 at 15:13
how about you add another line before your if statement, just try $aData2['rennen'] = 6; then followed by your original if statement and see what comes out — Mark Ng, Dec 09 '15 at 15:15

score 0 · Answer 1 · answered Dec 09 '15 at 14:22

0

Before your condition, add a echo $aData2['rennen']; to check if the value is 6. Then in this case use echo instead of print. Print can be used for tests.

answered Dec 09 '15 at 14:22

J. Grunder

143
2
13

1

print can (and frequently is) be used as an alternative to echo..... it is not simply for tests (in fact it isn't even particularly useful for tests `$x = false; print $x;` is no different to `$x = false; echo $x;` and both will display nothing.... but at least you're suggesting displaying the value of `$aData2['rennen']` as a basic debugging – Mark Baker Dec 09 '15 at 14:32
OK, I understand what you mean, I never use print and use echo. I wrote test but I meant function. I based by answer with this topic : http://stackoverflow.com/questions/234241/how-are-echo-and-print-different-in-php – J. Grunder Dec 09 '15 at 15:06

score 0 · Answer 2 · answered Dec 09 '15 at 15:22

$sSql = "SELECT Auto FROM Fahrer WHERE Auto = '$Auto'AND Rennen='$Rennen'";
$stmt = $pdo->query($sSql);
$aData2 = $stmt->fetch(2);

if(!empty($aData2)) {
    exit;
}
$sSql = "SELECT COUNT(Rennen) AS rennen FROM Fahrer WHERE Rennen='$Rennen'";
$stmt = $pdo->query($sSql);
$aData2 = $stmt->fetch(2);

if($aData2['rennen']==6){
    echo "Es nehmen bereits 6 Fahrer an diesem Rennen teil, bitte Name des Rennnens ändern";
    exit;
}


$sql = "INSERT into Fahrer (Auto,Vorname,Rennen) VALUES('$Auto','$Name','$Rennen');";
$pdo->query($sql);

this more code of my site

score -1 · Answer 3 · answered Dec 09 '15 at 14:22

-1

Try this one:

if(($aData2['rennen']==6)) 
{ 
echo 'Es nehmen bereits'.$aData2['rennen'].'Fahrer an diesem Rennen teil, bitte Name des Rennnens ändern';
exit; 
}

answered Dec 09 '15 at 14:22

Herolind

43
5

2

And what's the difference? Besides using `echo` instead of `print`? Why should this display output when OP's doesn't? – Mark Baker Dec 09 '15 at 14:25
echo and print are more or less the same. They are both used to output data to the screen. The differences are small: echo has no return value while print has a return value of 1 so it can be used in expressions. echo can take multiple parameters (although such usage is rare) while print can take one argument. – Herolind Dec 09 '15 at 14:26
1

I know what the differences are between `echo` and `print`..... what I don't see is why this code should do anything different than the OP's...... if you suggest something to try, explain ___why___ the OP should try it.... why might this work when their own code doesn't? – Mark Baker Dec 09 '15 at 14:29
print only takes one parameter, while echo can have multiple parameters. print returns a value (1), so can be used as a function. echo is slightly faster. Speed. There is a difference between the two, but speed-wise it should be irrelevant which one you use. echo is marginally faster since it doesn't set a return value if you really want to get down to the nitty gritty. Expression. print() behaves like a function in that you can do: $ret = print "Hello World"; And $ret will be 1. That means that print can be used as part of a more complex expression where echo cannot – Herolind Dec 09 '15 at 14:33
1

you're beginning to repeat yourself, and still not answering my question - there's no reason why using `echo` will make any difference to using `print` here – Mark Baker Dec 09 '15 at 14:35

Echo doesnt appears in my pop up

3 Answers3