php sanitize output form array

Question

I have an array that takes data from a mysql query.

$mysql = "select distinct name from software";
$result = mysqli_query($conn, $mysql);
    while($myrow = mysqli_fetch_assoc($result)) {
      $array[] = $myrow;
}

What I would like to do is to have the data output like:

item1, item2, item3

In the end what I want to do is to be able to store that info in a variable to use for a mysql query

$array = item1, item2, item3

select * from table where item != ($array)

However I cannot get the data to output correctly,

If I use print_r the data is shown and the array is good.

If i use array_key() I only get the numbers, how can I get the names?

print_r output:

( [0] => Array ( [name] => Skype for Business Basic 2016 - en-us ) [1] => Array ( [name] => Microsoft Visual C++ 2008 Redistributable - x64 9. ) [2] => Array ( [name] => Microsoft Office 365 Business - en-us ) [3] => Array ( [name] => Microsoft Silverlight ) [4] => Array ( [name] => NVIDIA 3D Vision Driver 341.92 ) [5] => Array ( [name] => NVIDIA Graphics Driver 341.92 ) [6] => Array ( [name] => NVIDIA Update 10.4.0 ) [7] => Array ( [name] => NVIDIA HD Audio Driver 1.3.30.1 ) [8] => Array ( [name] => FortiClient ) )

Array, Array, Array, Array, Array, Array, Array, Array, Array is all i get from implode(', ', $array) — Rick James
– Rick James, Commented Apr 6, 2016 at 22:10

Don't Panic · Accepted Answer · 2016-04-06 22:16:01Z

1

When you do this:

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

You are appending the entire row array to $array for each row you fetch. You can change it to

while($myrow = mysqli_fetch_assoc($result)) {
    $array[] = $myrow['name'];
}

to only append the name string. Then you will have an array of strings rather than an array of arrays, and implode will give you the results you expect.

answered Apr 6, 2016 at 22:16

Don't Panic

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Blablaenzo · Accepted Answer · 2016-04-06 22:31:18Z

0

You can use array_column (PHP 5.5+) to get the values you want:

print_r(array_column($array, 'name'));

edited Apr 6, 2016 at 22:31

answered Apr 6, 2016 at 22:24

Blablaenzo

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Ben Holness · Accepted Answer · 2016-04-07 01:55:41Z

0

Are you just using the output for the second query and nothing else? If so then why store it in an array? Instead you could do something like

$namelist = "";
$mysql = "select distinct name from software";
$result = mysqli_query($conn, $mysql);
while($myrow = mysqli_fetch_assoc($result)) {
  $namelist .= mysqli_real_escape_string($conn, $myrow['name']).",";
}
$namelist = substr($namelist, 0, -1);

Then it's directly ready for the second query, which will need a NOT IN:

select * from table where item NOT IN ($array)

Note that it's better practice to use prepared statements over mysqli_real_escape_string, but in this case it should be fine as $namelist comes directly from your database already.

answered Apr 7, 2016 at 1:55

Ben Holness

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