mysql query works with plain text, but not with variable

Question

I am trying to print out some topic information, but it is not going so well. This is my query:

SELECT * FROM topics WHERE id='$read'

This doesn't work. I've echo'ed the $read variable, it says 1. So then if I do like this:

SELECT * FROM topics WHERE id='1'

It works perfectly. I don't get what is the problem. There's no hidden characters in $read or anything else like that.

When you say it doesn't work, what exactly isn't working? Are you getting an error? Is it not returning results? — Jared, Dec 27 '12 at 16:46
Do not `echo` variable for debugging, use `var_dump`. `echo "1 ";` will give you the same output as `echo "1";` and you will be unaware of any non printable characters, while `var_dump` will tell you type and size in bytes. — dev-null-dweller, Dec 27 '12 at 17:12
@Jrod It just doesnt show anything, when i do mysql_num_rows it says 0 — finst33, Dec 27 '12 at 17:14

score 1 · Answer 1 · answered Dec 27 '12 at 16:46

1

Try like this:

$query = "SELECT * FROM topics WHERE id='" . $read . "'"

answered Dec 27 '12 at 16:46

Danilo Radenovic

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Akash · Answer 2 · 2012-12-27T16:46:20.450

0

ID is normally a numeric field, it should be

$id = 1;
$query = "SELECT * FROM topics1 WHERE id = {id}"

If you are using strings for some reason, fire a query like

$id = '1';
$query = "SELECT * FROM topics1 WHERE id = '{$id}'"

edited Dec 27 '12 at 16:46

answered Dec 27 '12 at 16:44

Akash

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score 0 · Answer 3 · answered Dec 27 '12 at 16:46

0

SELECT * FROM topics WHERE id=$read

it consider it as string if you put i single quotes

answered Dec 27 '12 at 16:46

Sivagopal Manpragada

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score 0 · Answer 4 · answered Dec 27 '12 at 16:55

I wonder why all the participants didn't read the question that clearly says that query with quotes

SELECT * FROM topics WHERE id='1'

works all right.

As for the question itself, it's likely some typo. Probably in some other code, not directly connected to $read variable

score 0 · Answer 5 · answered Dec 27 '12 at 17:12

0

try

$query = sprintf("SELECT * FROM topics WHERE id='%s';",$read);

Also remember to escape the variable if needed.

answered Dec 27 '12 at 17:12

Lars

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score 0 · Answer 6 · answered Dec 27 '12 at 17:29

Looks like you might have an issue with the query generation as everyone else is pointing to as well. As Akash pointed out it's always good to build your query in to a string first and then feed that string to the MySQL API. This gives you easy access to handy debugging techniques. If you are still having problems try this.

$id = 1;
$query = "SELECT * FROM `topics1` WHERE `id`={$id}";
echo ": Attempting Query -> {$query}<br />";

$res = mysql_query($query, $dblink);
if($res <= 0)
    die("The query failed!<br />" . mysql_error($dblink) . "<br />");

$cnt = mysql_num_rows($res);
if($cnt <= 0)
{
    $query = "SELECT `id` FROM `topics1`";
    echo "No records where found?  Make sure this id exists...<br />{$query}<br /><br />";

    $res = mysql_query($query, $dblink);
    if($res <= 0)
        die("The id listing query failed!<br />" . mysql_error($dblink) . "<br />");

    while($row = mysql_fetch_assoc($res))
        echo "ID: " . $row['id'] . "<br />";
}

This will at least let you monitor between calls, see what your query actually looks like, what mysql says about it and if all else fails make sure that the ID you are looking for actually exists.

score -1 · Answer 7 · answered Dec 27 '12 at 16:47

-1

try with this : SELECT * FROM topics WHERE id=$read

answered Dec 27 '12 at 16:47

haffane hatim

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mysql query works with plain text, but not with variable

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