[SOLVED] if else and mysql

[lyax]

i want to check my mysql database for the row if it is 0 or 1 or whatever it is and then print a word/words into a sentence according to the result..

for ex: my TABLE has a row named HEALTH and its value is 3. i want to use a variable in a sentences that will print GOOD. but if it were 2, script would print NOT BAD and for 1, print BAD.. my sentences is like this: Hi bla bla, your health is $var now.. how can i do this.. with if else tag i can make it print a sentences but cant do as a $var in a sentence???

[suttercain]

You should be able to use the variable in the sentence, as you wrote above if you use double quotes.

 

"Wow that is so $var up!"

 

or in single using .

 

'Wow that is so ' . $var . ' up!'

[marf]

I was about to say the same thing.

 

Also it sounds like from what you wrote you have code. Post the code and we can help you fix it that much faster.

 

[lyax]

after connected to the database, the code must be something like this : (gives parse error, couldnt do that)

 

if (row['health']=="3")

  $var = 'good';

elseif (row['health']=="2")

  $var = 'normal';

elseif (row['health']=="1")

  $var = 'bad';

else

  echo "";

 

 

your health is $var .

 

 

[suttercain]

Can you post the error?

 

I think the error is going to tell you that you need brackets:

 

if (row['health']=="3") {
  $var = 'good'; }
elseif (row['health']=="2") {
  $var = 'normal'; }
elseif (row['health']=="1") {
  $var = 'bad'; }
else {
  echo ""; }

echo "Your health is $var";

[lyax]

$result = mysql_query("SELECT * FROM utopia
WHERE kullaniciadi='$session->username'");

if (row['disgorunus']=="3") {
  $var = 'good'; }
elseif (row['disgorunus']=="2") {
  $var = 'normal'; }
elseif (row['disgorunus']=="1") {
  $var = 'bad'; }
else {
  echo ""; }

echo "Your health is $var";  //this is not here originally.

while($row = mysql_fetch_array($result))
  {
  echo $row['karakteradi'] . " şu an " . $row['disgorunus'] . " görünüyor ve kendini " . $row['ruhhali']. " hissediyor. " . $row['karizma']. " karizmaya sahip ve sağlığı " . $row['saglik']. ".";
  }

 

this is the code i am using.

gives parse error for the line if (row['disgorunus']=="3") { says: parse error on this line ..

 

[chigley]

Put a $ before row

[kenrbnsn]

Also, you do the query, but you never fetch the row information. Try something like this:

<?php
$result = mysql_query("SELECT * FROM utopia WHERE kullaniciadi='$session->username'");
$row = mysql_fetch_assoc($result);
switch ($row['disgorunus']) {
     case '1':
          $var = 'bad';
          break;
     case '2':
          $var = 'normal';
          break;
     case '3':
          $var = 'good';
          break;
      default:
          $var = '';
}
echo "Your health is $var";  //this is not here originally.
?>

 

Ken

[lyax]

yes that gives no error, thats the answer i think.  thanks so much a lot.