Running an If/Else PHP function with SQL

Question

Thank you for reading my question.

I am trying to make a site where information from a database is displayed onto a webpage. The end result will look like this, but for a different game.

Here is a plain HTML page of what I want it to look like.

So far I know that my connection to the database works. When I run:

mysql_select_db("DATABASE", $con);
$result = mysql_query("SELECT * FROM DATABASE");
while($row = mysql_fetch_array($result)) {
    echo $row['Title'] . " " . $row['Type'];
    echo "<br />";
}

It returns the Title and Type.

What I want to do is run an If/Else statement that runs a different that block of code depending on the card type.

while($row = mysql_fetch_array($result)) {
    if ($row['Title'] == 'Hero') { 
        echo "<div>";
    }
}

I tried this based on the tutorials at w3schools.com but it doesn't work.

Do any of you have any ideas for what I should do?

EDIT:

Here is what I tried running:

while($row = mysql_fetch_assoc($result)) {
    if ($row['Title'] == 'Hero') {
        echo $row['Title'] . " Hero.<br>";
    } else {
        echo $row['Title'] . " Who cares.<br>";
    }
}

Here is the output (Gimli should show up as a Hero):

For Gondor! Who cares.<br>
Bilbo Baggins Who cares.<br>
Ungoliant's Spwan Who cares.<br>
Gimli Who cares.

EDIT 2: Thank you Phil for spotting the error, I now get the result I wanted using Mikushi's method. Thank you all so much.

first line in the loop do a var_dump($row), paste the output here. — Mikushi
– Mikushi, Commented Feb 21, 2011 at 2:25
array(5) { ["Title"]=> string(11) "For Gondor!" ["Set"]=> string(4) "Core" ["Chapter Pack"]=> string(6) " " ["Number"]=> string(2) "22" ["Type"]=> string(5) "Event" } array(5) { ["Title"]=> string(13) "Bilbo Baggins" ["Set"]=> string(19) "Shadows of Mirkwood" ["Chapter Pack"]=> string(0) "" ["Number"]=> string(1) "1" ["Type"]=> string(4) "Hero" } array(5) { ["Title"]=> string(17) "Ungoliant's Spwan" ["Set"]=> string(4) "Core" ["Chapter Pack"]=> string(6) " " ["Number"]=> string(2) "76" ["Type"]=> string(5) "Enemy" } — Mason240
– Mason240, Commented Feb 21, 2011 at 2:34
array(5) { ["Title"]=> string(12) "• Gimli" ["Set"]=> string(4) "Core" ["Chapter Pack"]=> string(6) " " ["Number"]=> string(2) "11" ["Type"]=> string(4) "Hero" } Who cares. — Mason240
– Mason240, Commented Feb 21, 2011 at 2:35
@Mason240 Code doesn't work very well in comments. You should edit your question to add the new information — Phil
– Phil, Commented Feb 21, 2011 at 3:13
@Mason240 You'd get the same results using mysql_fetch_assoc() as you would using mysql_fetch_array() — Phil
– Phil, Commented Feb 21, 2011 at 3:46

Community · Accepted Answer · 2023-11-17 19:22:09Z

3

The fetching of your mysql result seems wrong, should be like this:

while($row = mysql_fetch_assoc($result)) {
  if ($row['Title'] == 'Hero') { 
    echo ""; }
  }

mysql_fetch_array fetch the result as an indexed array (1=> data, 2=> thing) , which explains why $row['Title'] doesn't work. The difference:

https://www.php.net/mysql_fetch_array

https://www.php.net/mysql_fetch_assoc

Please, always refer to the documentation, it's very well done and a better source than w3cschools.

edited Nov 17, 2023 at 19:22

CommunityBot

11 silver badge

answered Feb 20, 2011 at 23:12

Mikushi

3,3614 gold badges22 silver badges22 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Phil Over a year ago

mysql_fetch_array defaults to MYSQL_BOTH which includes both numeric and associative indexes

Mikushi Over a year ago

@Phil Brown : Absolutely right, didn't think of that, always make use of mysql_fetch_assoc. So yeah, i'll approve your question: "Define 'doesn't work'".

Mason240 Over a year ago

I tried mysql_fetch_assoc,but got the same result. I posted the test function, along with the result in the question text above.

Phil Over a year ago

@Mason240 Not anything against Mikushi but if this doesn't answer your question, you should not have accepted the answer

Phil · Accepted Answer · 2011-02-21 03:11:25Z

2

Maybe it's one of those all too obvious things but...

Shouldn't it be

if ($row['Type'] == 'Hero') // "Type", not "Title"

answered Feb 21, 2011 at 3:11

Phil

166k25 gold badges265 silver badges269 bronze badges

3 Comments

Mason240 Over a year ago

No, because need execute a different block of code for based on the Type of card - this is just a test. Here is a plain HTML mock up of what I am trying to create.

Phil Over a year ago

@Mason240 Sorry, not following you there. The only time "Hero" shows up in your data dump is for record "Type" fields. None of your "Title" fields contain "Hero"

Mason240 Over a year ago

Yes, I see what you mean, I fixed it and it does work now. I triple checking everything to make sure I wasn't wasting your time with simple errors, but I have been staring at this all day. I think next time I will take a looong break first. Thank you all.

Collectives™ on Stack Overflow

Running an If/Else PHP function with SQL

2 Answers 2

4 Comments

3 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

4 Comments

3 Comments

Your Answer

Sign up or log in

Post as a guest

Related