SQL returns non-array value in PHP

Question

    $request = 'SELECT * FROM flight WHERE Id = \''.$_SESSION['LFLightRadio'].'\'';
    $data = mysql_fetch_array(mysql_query($request, $SQL));
    echo '<table class="table">';
    foreach($data as $key => $value) {
        echo '<th class="head" align="center" height="19">'.$key.'</th>';
    }
    echo '<tr>';
    foreach($data as $key => $value) {
        echo '<td class="cell" align="center" height="19">'.$value.'</td>';
    }
    echo '</tr></table>';

I know that the LFlightRadio value is set, and is a value returned by the Id value of a previously returned row from the flight database. So within "flight", a record definitely exists with this Id. But, this still gives me a non-array result, so that when I try to use foreach on it, it errors out. Suggestions?

Perhaps do var_dump($data) and see what it says?

Mitch Dempsey
– Mitch Dempsey

2010-05-07 17:31:36 +00:00
Commented May 7, 2010 at 17:31 — Mitch Dempsey
– Mitch Dempsey, Commented May 7, 2010 at 17:31

Community · Accepted Answer · 2017-05-23 10:27:38Z

2

before the echo '<table class="table">';, add:

echo '<br>$_SESSION[\'LFLightRadio\']="'.$_SESSION['LFLightRadio'].'"<br>';

to make sure you actually have a value to compare to flight.Id in the query. The way you are doing this is a huge SQL injection attack waiting to happen! See this question: mysql_real_escape_string() for $_SESSION variables necessary?

EDIT

add this before the echo '<table class="table">';:

echo '<br>$request="'.$request.'"<br>';

run that query on the database, are any rows returned?

edited May 23, 2017 at 10:27

CommunityBot

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answered May 7, 2010 at 17:31

KM.

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7 Comments

Puppy Over a year ago

Doesn't matter - it's not a public site - and I already made sure that LFlightRadio exists, because I checked it repeatedly in statements that I cut.

KM. Over a year ago

when you do code a public site you will fall back into this habit.

Puppy Over a year ago

Well, that was revealing. $_SESSION['LFlightRadio'] is 8. But the constructed $request doesn't actually contain this value.

Marcus Adams Over a year ago

Better to use parametrized queries for sure, but session variables should be safe as long as you're not putting user input in them.

Puppy Over a year ago

The database is full of junk for demonstration purposes, and the exactitudes of my SQL communication are not the subject of the demo.

|

brandon k · Accepted Answer · 2010-05-07 19:22:48Z

1

You have a mis-capitalization in your code.

$_SESSION['LFLightRadio'] is most likely intended to be $_SESSION['LFlightRadio']

Array keys are case sensitive.

answered May 7, 2010 at 19:22

brandon k

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Comments

SorcyCat · Accepted Answer · 2010-05-07 17:48:40Z

1

Did you mean to use

mysql_fetch_assoc

instead of mysql_fetch_array? Array would have keys like 0,1,2...

answered May 7, 2010 at 17:48

SorcyCat

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1 Comment

Puppy Over a year ago

It has both, actually. I just changed to assoc, cause I was dumping duplicates on the screen.

Puppy · Accepted Answer · 2010-05-07 17:57:31Z

0

    $var = $_SESSION['LFlightRadio'];
    $request = "SELECT * FROM flight WHERE Id = '$var'";
    $data = mysql_fetch_array(mysql_query($request, $SQL));

This works! For some reason, when I concatenated it straight, it didn't actually add the variable into the string. But when I've done that, it got put in and I got the expected.

edited May 7, 2010 at 17:57

answered May 7, 2010 at 17:47

Puppy

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2 Comments

Marcus Adams Over a year ago

Concatenation should have worked. It's most likely a typo. Notice the case difference between 'LFlightRadio' and 'LFLightRadio'.

Puppy Over a year ago

You see what you want to see, rather than what's there, I guess.

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