CODEWITHSUNDEEP
phpmysql
funnyfish
funnyfish

Reputation: 143

PHP Replace character with string from an array

For mysql I use format:

$sql = "select * from table where area_id = ? and item_id = ?";

Then prepare and bind the parameters etc. If the query fails and I log the $sql variable then I get exactly the string above which isn't that useful. What I want is the sql string with the bound values in. As I understand it there is no easy way to do this so I am thinking I can do something like:

sql_log(str_replace('?', array($area_id, $item_id), $sql));

To get something like this in my log: 

"select * from table where area_id = West and item_id = West" (spot the error!)

So I know what my error is. But it doesn't work. I get this:

"select * from table where area_id = Array and item_id = Array"

Upvotes: 2

Views: 1659

Answers (4)

bmatovu
bmatovu

Reputation: 4074

Laravel has a beautiful helper for just that.

/**
  * Replace a given value in the string sequentially with an array.
  *
  * @param  string  $search
  * @param  array   $replace
  * @param  string  $subject
  * @return string
  */
function replaceArray($search, array $replace, $subject)
{
    $segments = explode($search, $subject);

    $result = array_shift($segments);

    foreach ($segments as $segment) {
        $result .= (array_shift($replace) ?? $search).$segment;
    }

    return $result;
}

$sql = 'SELECT * FROM tbl_name WHERE col_b = ? AND col_b = ?';

$bindings = [
  'col_a' => 'value_a',
  'col_b' => 'value_b',
];

echo replaceArray('?', $bindings, $sql);

// SELECT * FROM tbl_name WHERE col_b = value_a AND col_b = value_b

Source: Str::replaceArray()

Upvotes: 0

Carlos Bello
Carlos Bello

Reputation: 21

try with this:

sprintf('select * from table where area_id = %s and item_id = %s', $area_id, $item_id);

or

sprintf('select * from table where area_id = "%s" and item_id = "%s"', $area_id, $item_id);

if your fields in the database are integers the %s you have to replace it with %d and do not use quotes

Upvotes: 1

Justin Pearce
Justin Pearce

Reputation: 5097

Unfortunately, mysqli does not have a good way to get just the query. You could use a method to do the replacement of your parameters:

function populateSql ( string $sql, array $params ) : string {
    foreach($params as $value)
        $sql = preg_replace ( '[\?]' , "'" . $value . "'" , $sql, 1 );
    return $sql;
}

Upvotes: 2

splash58
splash58

Reputation: 26153

Use preg_replace_callback function

$sql = "select * from table where area_id = ? and item_id = ?";
$replace = array('area_id', 'item_id');
echo preg_replace_callback('/\?/', function($x) use(&$replace) { return array_shift($replace);}, $sql);
// select * from table where area_id = area_id and item_id = item_id

Upvotes: 3

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