php function to select data from mysql table

Question

I use the following function to select data from a MYSQL database:

if (!function_exists('select_array_where')) {
function select_array_where($what, $from, $where){

$mysqli = new mysqli("host", "user", "password", "db");
if ($mysqli->connect_errno)
{
    echo "<br>Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error . "<br>";
}

$result = $mysqli ->query("SELECT $what FROM $from WHERE $where");

$array = Array();

while ($row = mysqli_fetch_assoc($result)) {
  $array[] =  $row[$what];
}

print_r ($array);
return $array;
}
}

Which works perfectly well if $what has only one value:

$name_array = Array();
    $name_array =& select_array_where("name", "inter_company", "transaction_type='interest'");

But it does not work in the following case:

  $company_int_array = array();
  $company_int_array =& select_array_where("amount, perc", "inter_company", "transaction_type='interest'");

I've a feeling it relates to punctuation and have tried movies comma's, " and ' around, in and out but nothing helps...

Anyone an idea what I'm doing wrong?

Anyway, could you please copy the error message you get? Adding a try..catch would be a good idea also — Yazid Erman
– Yazid Erman, Commented Jan 2, 2016 at 17:24
"amount, perc", these would need to be separated "amount", "perc", — Funk Forty Niner
– Funk Forty Niner, Commented Jan 2, 2016 at 17:27
Doesn't work, believe that I can only have three things between " and " because of the way the function is defined ($what, $from and $where) — dathiezer
– dathiezer, Commented Jan 2, 2016 at 17:31
Try to use ticks ` around each word the amount, perc with your existing quotes, or wrapping the whole thing. — Funk Forty Niner
– Funk Forty Niner, Commented Jan 2, 2016 at 17:36

Sean · Accepted Answer · 2016-01-02 17:46:48Z

2

Your issue is that

 $array[] = $row[$what];

is evaluating to

 $array[] = $row['amount perc'];

which is invalid as you don't have a single column named 'amount, perc'.

The simplest solution is to change to

$array[] = $row;

which is equivalent to doing

$array[] = array('amount'=>$row['amount'], 'perc'=>$row['perc']);

but does not require you to know/specify the actual column names/keys

answered Jan 2, 2016 at 17:46

Sean

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1 Comment

Funk Forty Niner Over a year ago

Luvin' that green ;-) Good job Sean.

Yazid Erman · Accepted Answer · 2016-01-02 17:42:20Z

-1

Well, you have to replace the line: $array[] = $row[$what]; With the lines:

 $Whatsarr = explode ($what);
 Foreach ($what as $col)
 $aux[$col] = $row [$col];
 $array []= $aux:

edited Jan 2, 2016 at 17:42

answered Jan 2, 2016 at 17:22

Yazid Erman

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5 Comments

Sean Over a year ago

How does this address the OP's issue on this line - array[] = $row[$what];? That would evaluate to array[] = $row['amount', 'perc'];, which is invalid.

dathiezer Over a year ago

Correct, this doesn't help

Yazid Erman Over a year ago

I have fixed my answer

Sean Over a year ago

Your edit is closer, but this would create a new array element for each column, each time. So if you had 2 columns and 3 rows in the database, then instead of 3 array elements with 2 key/value pairs in each, you will have 6 array elements with 1 key/value pair in each.

Yazid Erman Over a year ago

Did you execute it to know this?! Notice that the Foreach loop includes only the first line after it

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