How to pass a variable/s into while loop MySQLi - php

Question

I have a simple function to SELECT from MySQL:
function dbSelect($query) {
$db = db(); // refers to a database connect function
$result = mysqli_query($db, $query);
while ($row = mysqli_fetch_array($result)) {
echo $row['f_name'] . ' : ' . $row['s_name'] . '</br>';
}
}
as you can see it accepts the $query var which would be:
$query = "SELECT * FROM user"; // or whatever
What I would like to do is pass the line:
echo $row['f_name'] . ' : ' . $row['s_name'] . '</br>';
as an arg in the function call, eg:
function dbSelect($query, $display)
and the $display be something like (which I would echo out under the while statement):
$display = $row['f_name'] . " : " . $row['s_name'] . "</br>";
The above line will give an undefined variable error, I know why.
How would I go about passing the $display to the function or defining it correctly?
ps. I know there is not any error checking, I'm just trying to figure this out, will add that later. :-)
Thanks for any help

Answer 1

Try this, to maintain separation from program logic to program output:
// some configuration file you include
$db = db(); // refers to a database connect function
// some common functions file
function dbSelect($query)
{
global $db;
$result = mysqli_query($db, $query);
return $result;
}
// output script
$records = dbSelect('SELECT * FROM somewhere');
// loop until cannot fetch any more rows from the record set
while ($row = mysqli_fetch_array($records))
{
echo $row['f_name'] . ' : ' . $row['s_name'] . '</br>';
}

Answer 2

Personally I would pass a template name into the function and have the template stored somewhere as a snippet of html which has standard replacement vars
So you may have something like this (note this won't work out the box will need some fiddling but its the concept)
define TEMPLATE_NAME = "<tr><td>%COL1%</td><td>%COL2</td></tr>";
$colstospitout = array('COL1'=>'f_name','COL2'=>'s_name');
function dbSelect($query,$reportcols,$templatename) {
while ($row = mysqli_fetch_array($result))
{
$template = str_replace("%COL1%",$reportcols['COL1'],$templatename);
$template = str_replace("%COL2%",$reportcols['COL2'],$templatename);
echo $template;
}
}
dbSelect($inputquery,$colstospitout,TEMPLATE_NAME);

Answer 3

There are two ways to do it. The proper way to do it would be to have this function only run the query and return the results. You could make another function that outputs the results. This way you have a clear separation of logic in your code.
However, if you know that you want it done in this way, something like this should work:
function dbSelect($query, $format) {
$db = db();
$result = mysqli_query($db, $query);
while ($row = mysqli_fetch_array($result))
{
echo preg_replace_callback("/\{\{\s*(\w+)\s*\}\}/", function($matches) use ($row) { $column = trim($matches[1]); return isset($row[$column]) ? $row[$column] : $matches[0]; }, $format);
}
}
Where the arguments could look something like:
dbSelect("SELECT * FROM user", "{{ f_name }} {{ s_name }}");
You can't pass it the way that you're passing it because $row doesn't exist when you call the function (because it gets defined inside of the function). In addition, your variables in would be evaluated once when the function is called, not each time in the while loop.
And before anyone makes a "suggestion": eval is not an option.