Select From table results that don't equal to $var?

Question

How do I make it pick all results that are not equal to the $var , here's my code.

$opti=mysql_query("SELECT * FROM table1 WHERE imageid=$image_id");
    while ($vari = mysql_fetch_array($opti)) {
        $var = $vari['tagid'];
    $options=mysql_query("SELECT * FROM table WHERE id!=$var");
    while ($taghe1 = mysql_fetch_array($options)) {
     $tagname = $taghe1['name'];
     echo "".$tagname.", ";
    } }

thought so. you've an answer below. time to give 'em the "tick" ;-) — Funk Forty Niner, May 19 '15 at 17:18
^^ Youre right ^^ Yet why is it giving the ones that do equal too ^^? — PHP Nexter, May 19 '15 at 17:20
`$var` is a bare variable, so you're doing `id != foo`. unless you have a field named `foo` in your table, you'll never match anything. you'll just be getting sql errors: no such field. — Marc B, May 19 '15 at 17:24
Um should I edit the question to show where does $var come from? — PHP Nexter, May 19 '15 at 17:27
no use for me to submit an answer, the one below is soOOoooo close ;-) — Funk Forty Niner, May 19 '15 at 17:37
Please, [stop using `mysql_*` functions](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php). They are no longer maintained and are [officially deprecated](https://wiki.php.net/rfc/mysql_deprecation). Learn about [prepared statements](http://en.wikipedia.org/wiki/Prepared_statement) instead, and consider using PDO, [it's not as hard as you think](http://jayblanchard.net/demystifying_php_pdo.html). — Jay Blanchard, May 19 '15 at 17:39
*You sir Sam, are indeed right on zee monnaie* - @JayBlanchard — Funk Forty Niner, May 19 '15 at 17:48
It looks like you are trying to get the names of all the tags that are not yet applied to the image. If that is the case, you could do it with one query and avoid this problem. — Don't Panic, May 19 '15 at 17:53
*after-code service* isn't like it used to be. They just *upped and got their money, and then they were gone*. Sounds like a song, don't it? — Funk Forty Niner, May 19 '15 at 17:53
hum ... @Don'tPanic You are right, How can I do this Then :? — PHP Nexter, May 19 '15 at 18:21

score 1 · Answer 1 · answered May 19 '15 at 17:17

1

Try:

$options=mysql_query("SELECT * FROM table WHERE id<>{$var}");

answered May 19 '15 at 17:17

Michal Przybylowicz

1,558
3
16
22

Thankyou but it picks those that do equal to $var as well – PHP Nexter May 19 '15 at 17:20
Isn't `<>` the same as `!=` ? – Don't Panic May 19 '15 at 17:23
Um , Excuse me ^^ It shows the results that equal to $var first then the ones that do not ^^ – PHP Nexter May 19 '15 at 17:23
1

@Don'tPanic yeah but `<>` looks so much cooler, don't it? ;-) – Funk Forty Niner May 19 '15 at 17:29
2

@Fred-ii- Good point. But also, isn't `"SELECT * FROM table WHERE id<>{$var}"` the same as `"SELECT * FROM table WHERE id<>$var"`? – Don't Panic May 19 '15 at 17:32
@Don'tPanic indeed you are right. OP did state to me earlier that it is a string, so that will fail. Good catch ;-) which is why I asked them "first" before I submitted an answer. Too many unknowns. – Funk Forty Niner May 19 '15 at 17:32
Thanks. I guess what I'm getting at is this is pretty much the same code from the question. No offense intended, Michal. – Don't Panic May 19 '15 at 17:34
@Don'tPanic I was going to pop in my usual "here, use this to catch errors" codes ;-) but I'll just wait this one out. – Funk Forty Niner May 19 '15 at 17:34
Check edit to the question maybe Your code just does not fit into the script because Of anything above? – PHP Nexter May 19 '15 at 17:37
@PHPNexter 22 mins. and still nothing, *eh?* - ah, peccato. – Funk Forty Niner May 19 '15 at 18:00

score 0 · Answer 2 · answered May 19 '15 at 18:03

0

You could use this:

$options=mysql_query("SELECT * FROM table WHERE id not in ('$var')");

You could have multiple values here, e.g.

$options=mysql_query("SELECT * FROM table WHERE id not in ('$var1', '$var2', '$var3')");

answered May 19 '15 at 18:03

Robert Ostrowicki

304
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score 0 · Accepted Answer · answered May 19 '15 at 20:01

You can probably see from the answer you accepted that adding the quotes solved your problem. Another way to do this is to just use one query. I will show an example using mysqli instead of the deprecated mysql, but the same query should work in mysql if you must use it. I added a couple of other suggestions that aren't really addressing your question, but make me feel better about my answer.

// Please be sure to escape $image_id before using it like this
$unused_tags = mysqli_query($db, "SELECT `name` FROM `table` AS t 
    LEFT JOIN (SELECT tagid FROM table1 WHERE imageid=$image_id) AS t1 
    ON t.id = t1.tagid WHERE t1.tagid IS NULL;");

while ($tag = mysqli_fetch_array($unused_tags)) {
    $tags[] = htmlspecialchars($tag['name']); // escape your output
}
echo implode(", ", $tags); // doing it this way eliminates the trailing comma

Select From table results that don't equal to $var?

3 Answers3