php function array argument manipulation

Question

i am making insert function that takes $table argument and $cols(as array)argument. it inserts into given table given values:

$db->query("insert into $table({$cols[0]},{$cols[1]}) values('{$_POST[{$cols[0]}]}','{$_POST[{$cols[1]}]})");

this is all nice except i don't how long array is. how to do this??

alex · Accepted Answer · 2011-03-27 23:51:37Z

5

One thing you haven't done is escaped the SQL using the correct escaping mechanism.

$postCols = $_POST['cols']; 

foreach($postCols as &$col) {
    $col = '"' . mysql_real_escape_string($col) . '"';
}

$db->query("insert into $table(" . implode(',', $cols) . ") values(" . implode(',', $postCols . ");

edited Mar 27, 2011 at 23:51

answered Mar 27, 2011 at 23:38

alex

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4 Comments

Jess Over a year ago

Haha, I've been using C# to much. Completely forgot about the implode function. Your code is a lot cleaner.

jon_darkstar Over a year ago

looks great, just remember to escape the values and wrap them in quotes.

$escAndQuote = function($x) {return "'" . mysql_real_escape($x) . "'";}; $colVals = array_map($escAndQuote, $_POST['cols']);

daniel.tosaba Over a year ago

what does $_POST['cols'] represent?

alex Over a year ago

@Daniel It looks like you get the POST variables directly, in which case you could change the loop to suit.

Jess · Accepted Answer · 2011-03-27 23:37:14Z

2

I would just use some foreach loops

<?php
  $sql = "INSERT INTO $table (";
  foreach ($cols as $col)
      $sql .= "`$col`,";
  $sql = substr($sql,0,-1);
  $sql .= ") VALUES(";
  foreach ($cols as $col)
      $sql .= "'".$_POST[$col]."',";
  $sql = substr($sql,0,-1);
  $sql .= ");";

  echo $sql;
?>

answered Mar 27, 2011 at 23:37

Jess

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