PHP: "... variables can be passed by reference" in str_replace()?

Question

I created a function to print a prepared-statement-sql-string with the variables in it, based on what I found in this other StackOverflow question.

Here is my code:

foreach($params as $idx => $param) {
    if ($idx == 0) continue;
    $sql = str_replace('?', "'" . $param . "'", $sql, 1);
}
printError($sql);

When I run this I get: Fatal error: Only variables can be passed by reference for line 3. However when i use

$sql = preg_replace('/\?/', "'" . $param . "'", $sql, 1);

for line 3 it works fine.

Any idea why?

VoteyDisciple · Accepted Answer · 2011-04-30T14:35:41.743

42

The very last parameter, count, is passed by reference. You can see this in the description at http://us.php.net/str_replace where there's a & in front of the variable.

This means you cannot use a literal 1 there. You'd have to do:

$sql = str_replace('?', "'" . $param . "'", $sql, $count);
echo $count;

You'll now have displayed on the screen how many instances were replaced.

edited Apr 30 '11 at 14:35

answered Apr 30 '11 at 14:30

VoteyDisciple

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Indeed, the OP has misunderstood what the `$count` parameter actually does. It's an "out" parameter, not an "in" parameter. – Lightness Races in Orbit Apr 30 '11 at 14:33
1

Quite right; changed my example to illustrate that and added an explanation. – VoteyDisciple Apr 30 '11 at 14:36
Oh, I see. So there is no way to limit the amount of replacings taking place with str_replace()? – Dexter Apr 30 '11 at 14:41
2

Nope. `str_replace()` does not have that option. As you've already noted, `preg_replace()` does. – VoteyDisciple Apr 30 '11 at 14:42

Explosion Pills · Answer 2 · 2011-04-30T14:41:08.737

2

Look at the documentation for preg_replace and str_replace and you will see why. str_replace's fourth argument must be passed by reference, but this is not the case for preg_replace.

edited Apr 30 '11 at 14:41

answered Apr 30 '11 at 14:28

Explosion Pills

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This is _also the case for `$count` with `preg_replace`_, but the `$count` parameter for `preg_replace` is fifth, not fourth. The OP was using `$limit` with `preg_replace`, which does something else. – Lightness Races in Orbit Apr 30 '11 at 14:33

nobjta_9x_tq · Answer 3 · 2018-02-28T18:06:06.007

I rewrite from VoteyDisciple

$sqlLogin = "SELECT * FROM users inner join role on users.roleId = role.id WHERE email=?1 and password=?2";
function makeSql() {
    $args = func_get_args();
    if(isset($args[1])) {
        $len = sizeof($args);
        //var_dump($args);
        $sql = $args[0];
        for ($index = 1; $index < $len; $index++) {
            $sql = str_replace('?'.strval($index), "'" . $args[$index] . "'", $sql);
        }
        return $sql;
    }
    return $args[0];
}
$sql = makeSql($sqlLogin, $myusername1, $mypassword);
$result = mysqli_query($con, $sql);

PHP: "... variables can be passed by reference" in str_replace()?

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